calculus i with precalculus 3rd edition larson solutions manual

Calculus I with Precalculus 3rd Edition Larson Solutions Manual Full Download: http://alibabadownload.com/product/calculus-i-with-precalculus-3rd-edition-larson-solutions-manual/

NOT FOR SALE CONTENTS

Chapter P

Prerequisites............................................................................................1

Chapter 1

Functions and Their Graphs.................................................................89

Chapter 2

Polynomial and Rational Functions...................................................173

Chapter 3

Limits and Their Properties................................................................277

Chapter 4

Differentiation.....................................................................................320

Chapter 5

Applications of Differentiation ..........................................................407

Chapter 6

Integration ...........................................................................................524

Chapter 7

Exponential and Logarithmic Functions ...........................................606

Chapter 8

Exponential and Logarithmic Functions and Calculus.....................675

Chapter 9

Trigonometric Functions....................................................................725

Chapter 10

Analytic Trigonometry.......................................................................813

Chapter 11

Trigonometric Functions and Calculus .............................................885

Chapter 12

Topics in Analytic Geometry.............................................................944

Chapter 13

Additional Topics in Trigonometry .................................................1054

INSTRUCTOR USE ONLY iii

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This sample only, Download all chapters at: alibabadownload.com

NOT FOR SALE Preface The Complete Solutions Manual for Calculus I with Precalculus: A One-Year Course, Third Edition, is a supplement to the text by Ron Larson and Bruce H. Edwards. Solutions to every exercise in the text are given with all essential algebraic steps included.

INSTRUCTOR USE ONLY iv

© Cengage Learning. All Rights Reserved.

CHAPTER P Prerequisites Section P.1

Solving Equations...................................................................................2

Section P.2

Solving Inequalities ..............................................................................17

Section P.3

Graphical Representation of Data........................................................36

Section P.4

Graphs of Equations .............................................................................44

Section P.5

Linear Equations in Two Variables .....................................................55

Review Exercises ..........................................................................................................70 Chapter Test ...............................................................................................................83 Problem Solving ...........................................................................................................85

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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NOT FOR SALE C H A P T E R Prerequisites

P

Section P.1 Solving Equations 1. equation

14.

2. ax b

0

3. extraneous 4. factoring; extracting square roots; completing the square; Quadratic Formula 5. 4 x 1

7x 2

23

7x 2 2

4 x 4 is an identity by the Distributive

15.

2x 4

2 x 2

This is an identity by simplification. It is true for all real values of x. 8. x 2 23x 2

21 7 3

8x 5

x 2 6 x 4 is an identity by

1 4x is conditional. There are real x 1 x 1 values of x for which the equation is not true.

16.

5x 5 x

25 5 5

7x 3

12.

7 x

4x

20

x

5

2 y

7 6y

2 y 6y

7 6y 6y

2 5y

7

2 2 5y

7 2

5y 5 y

4 18.

5 5 1

3 x 3

51 x 1

7 x x

19 x

3x 9

5 5x 1

7

19 x

3x 9

4 5x

7 19 12 13.

3x 17 3 3 x

7 6y

15 11

19

3x 17

17. 4 y 2 5 y

15

x

20 5

7 x 3 3 3x

5 3 10. 24 is conditional. There are real values of x x x for which the equation is not true (for example, x 1 ).

x 11 11

20 25

9. 3

x 11

3 x 3 x 20

5x

simplification. It is true for all real values of x.

11.

3 x 20

5x 5 5

real values of x for which the equation is not true. 4x 4 2x

21

5x 5

2 x 10 is conditional. There are

7. 4 x 1 2 x

7x 7x 7 x

8 x 3x 5

Property. It is true for all real values at x. 6. 6 x 3 5

23 2

7 2x 7 7 2x

19 x 19

3x 9 5 x 9

x 25 25 7

2 x

18

2 x 2 x

18 2 9

4 5x 5x 9

8x

5

x

85

19. x 3 2 x 3

8 5x

x 6x 9

8 5x

5 x 9

8 5x

5 x 5 x 9

8 5x 5x

9 z 8

INSTRUCTOR USE ONLY No solution.

2

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© Cengage Learning. All Rights Reserved.

Chapter P.1 5 x 2 2 x 5

20. 9 x 10

9 x 10

5 x 4 x 10

9 x 10

9 x 10

27.

The solution is the set of all real numbers. 3x 4x 21. 8 3 4x · § 3x 24¨ ¸ 3¹ ©8

3 2

23.

x

5

5

1 4

4 ª¬ 32 z 5

1 4

3 2

1 4

z

z

24 º¼

z 5 4 z 24

4

40 40

6 z 5 z 24

0

6 z 30 z 24

0

30.

5z

6

z

65

24. 0.60 x 0.40100 x

50

0.60 x 40 0.40 x

50

0.20 x

10

x

50

25. x 8

100 y 100 y

1 2

0

2 x 2 x

x 8

2x 4 x

x 8

x 4

5x 4 5x 4 35 x 4

2 3 25 x 4

15 x 12

10 x 8

5x

20

x

4

15 4 x 15 6 x x 9 x 9 9 7

31.

3

3z 6 z

Contradiction; no solution 26. 8 x 2 3 2 x 1

2 x 5

8 x 16 6 x 3

2 x 10

2 x 13

2 x 10

13

10

Contradiction; no solution

32.

98 y y

6 3 x 7 7 7x x

3 z 2

8 z 4

100 y

49 2 y 49

24

100

17 y 32 y

3x 10

29.

30

z

17 y 32 y y y y

y

3x · § 10¨ 3 ¸ 10 ¹ © 30 3 x

6 x

10

17 y 32 y y y

28.

96 23 96 23 3

310

x

96

x

2x 5x

15 x 18 72

31x

23 x 23

x x 5 2 x· §x 10¨ ¸ 2¹ ©5

35 x 6 72

400 16 x

24 4

3

5x 6 6 4 § 5x 6 · 12¨ ¸ 126 © 4 ¹

4100 4 x

4

9 x 32 x

22.

100 4 x 3 § 100 4 x · 12¨ ¸ 3 © ¹

Solving Solvin Equations Solvi

2 z 2 2 · ¸ z 2 z 2¹

2

§ ¨2 © 2z 4 2 0

1 2 x x 5 1 x 5 2 x

0

Multiply both sides by x x 5 .

0

3x 5

0

3x

5

x

5 3

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NOT FOR SALE

Chapter P

4

33.

Prerequisites requisites

4 x 2 x 4 x 4 x 4 2 x 4

Multiply both sides by x 4 .

0 0

x 4 2x 8

0

3 x 12

0

3x

12

x

4

A check reveals that x 4 is an extraneous solution because it makes the denominator zero. There is no real solution. 34.

7 8x 2x 1 2x 1 7 2 x 1 8 x 2 x 1

4 2 x 1 2 x 1

14 x 7 16 x 8 x

16 x 2 4

2

35.

6x

11

x

11 6

3 4 x 3x x 3 4 x x 3 x

1 x 3 1 x 3

3 4 x 3

x

2

3 4 x 12

x

3x

9

x

3

A check reveals that x There is no solution. 36.

6 x 3 2 x

3 x 5

6 x 18 2 x

3 x 15

4 x 18

3 x 15

x

Multiply both sides by x x 3 .

3

A check reveals that x There is no solution. 2 5

x

2

x2 4 x 4 5

3 is an extraneous solution because it makes the denominator zero. 3

2

2 x

0

x

0 4 x 2 x 1

4 x2 4x 1

4x2 4x 4

1

1

3 8x

General form: 2 x 2 8 x 3

6x 9

2

2 x

39. 2 x 2

x2 6x 9

4x 9

38.

3 is an extraneous solution because it makes the denominator zero.

x x 3

x

37.

Multiply both sides by x x 3 .

3 x 5

6 2 x x 3

Multiply both sides by 2 x 1 2 x 1 .

4

40.

13 3 x 7

2

0

13 3 x 2 14 x 49

0

13 3 x 42 x 147

0

2

0

General form: 3 x 2 42 x 134

0

4

INSTRUCTOR USE ONLY This is a contradiction. So, the equation has no solution.

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© Cengage Learning. All Rights Reserved.

NOT FOR SALE Chapter P.1

41.

1 5

3x 2

10

18 x

3x 10 2

General form: 3x 90 x 10 2

42.

5x2 1

x2 2x

5x2 1

2 x 1

6 x 3x

0

3x 2 x 1

0

2

3x x

0 or

12

x

2 3 x 2

x

45.

x

0

0

0 2

0 0

0 x

9

x 1

0 x

1 0 0

x 7

0 or

x 5

0

x

7 or

x

5

4 x 2 12 x 9

0

3 2 x 3

3 x x

32

x

2

0

x 2 8 x 128

0

x 16 x 8

0

53. x 2 2ax a 2

x

54.

a

0

2

0 0 a

x a 2 b 2 ª¬ x a bºª ¼¬ x a bº¼ ¬ª x a b ºª ¼¬ x a b º¼

x

0 0

0 x

a b

x a b

0 x

a b

55. x 2

49

x

r7

56. x 2

32 r 32

x 57. 3x 2

81

2

27

x

r3 3

x

0

0

x a b

58. 9 x 2

0

0 or 1 2 x 3 or

6 or x 16

1 x2 8

x

0

0 x

2 x

0

x

x 9

7 x 5

x 2

x a

x

47. x 12 x 35

or

0

8

4 or

3 x 1

x 6

0

0

x

3 5x 2 x2

6 x 2

16

x 2

9 x 1

0

0 x

x

4 x 2

12

0 x

2 3

2

49.

x2 4x

x

0

0

0

2x 3

11

2

or 3 x 2

x2 2x 8

2 x

0 x

x 8

23

46. x 2 10 x 9

48.

x 11

x 16

0 or

x

32

0

x 4

x

0 x

52.

0

3x 2

2x 3

x

9x2 4

3 x

0

x 4 x 12

10

0 or 2 x 1

44.

0

2 x 3 x 11

51.

0

General form: 4 x 2 2 x 1 43.

0

x x 2 4 x 2 2 x 1

1 4 x 2

2 x 19 x 33 2

90 x

5

19 x 33

2x2

50.

Solving Solvin Equations Solvi

r4 2

36

2

4

x

r 4

r2

0 12

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NOT FOR SALE

Chapter P

6

59.

x

12

2

x

x

16

2

x

r 25

x 13

r5

5

x 4x 2

x 4x 2

2

x

2

2

2 r

14

2

x 4

5r

x

0 27

2 2 32

3

2

7

x 3

r 7

x

3 r

r3 3

27

x 12 x 6

2

x

2

2

6

2x

3 r 3 3

x 6

x

3 r 3 3 2

x 70. x 2 8 x 14

x 3 r x 3 2

x 3 or or

x 8x 2

x 8x 4

2

x

2

2

x 7

x 3

2x

4

x

2

4

x 4

The only solution to the equation is x

2.

x

7

0

x 12 x

2x 3

7 z 3

8

0

25

2

r

x 7

4 or x

69. x 2 12 x 25

2x 3

2

x

2

x

1r3 2 2

3

r6 2 r 6

x 6x 3

1r3 2

2

92 .

0

x

x 6x 2

2x

64. 2 x 3 27

92

36

2

r 18

2

x

32 22

x2 6x 2

18

2x 1

9

30 68.

2

2x

32

x 2

r 30

2 x

x 4

x 5

8, 18

30

63. 2 x 1

x 7

or

x 5

The only solution to the equation is x

r 14

2

7

5 z 4

14

x

x

x 4 or

x 5

8

67. x 2 4 x 32 2

x 5

65.

x

13 r 5

x

x

x 4 2 r x 4

2

25

x 2

62.

or

x 13

2

5

x 5

12 r 4

13

x

r4

x

x 61.

66.

16

x 12

60.

Prerequisites requisites

25 62 11 r 11 6 r

11

0 14 14 16 2 r

2

4 r

2

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© Cengage Learning. All Rights Reserved.

NOT FOR SALE Chapter P.1

71.

8 4 x x2

0

5 x 2 15 x 7

0 0

x 4x 8

0

x 4x 8

0

7 x 2 3x 5

x2 4 x

8

x 2 3x

2

2

x 2 4 x 22

x

2

8 22

2

12

x 2

r 12 2r 2 3

x 9 x 2 12 x

72.

x2 x2

4 x 3

14

4x 3

x 23 x 23 23

x

73.

75.

14 9

2

14 9

2

2

x

2 3

2x2 5x 8

0

2x2 5x

8

5 x 2

4

5 §5· x ¨ ¸ 2 ©4¹

2

5· § ¨x ¸ 4¹ ©

2

x

x

r

x

3 2

r 17 2 5

x

3 17 r 2 2 5

x

3 85 r 2 10

x

15 r 85 10

2

89 16

0

5 3

0

x 2 3x

5 3

5 § 3· ¨ ¸ 3 © 2¹

5 9 3 4

§ 3· x 2 3x ¨ ¸ © 2¹

2

3· § ¨x ¸ 2¹ ©

2

3· § ¨x ¸ 2¹ ©

2

2

x

5 89 r 4 4

x

r

7 12

x

5 r 89 4

3 2

x

3 2

r

7 2 3

x

3 7 r 2 2 3

x

3 r 2

x

9 r 21 6

2

2

2

89 4

17 20

3x 2 9 x 5 x 2 3x

§5· 4¨ ¸ © 4¹

17 20

3 2

2 76.

2

x

4 9

r

7 § 3 · ¨ ¸ 5 © 2¹

r

x2 x

x 12 x 12

2

0

7 12

99 4

99 4

1 4

100 4

7

7 5

5 4

74. 4 x 2 4 x 99

x 2 x 12

3· § ¨x ¸ 2¹ ©

2 r 2

x2

2

18 9

2 3

x2

§ 3· x 2 3x ¨ ¸ © 2¹

Solving Solvin Equations Solvi

21 6

25

1 2

r

25

x

1 2

r5

11 , 2

92

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NOT FOR SALE

Chapter P

8

Prerequisites requisites

77. 2 x 2 x 1 b r

x

1 r

0

82.

x2 6x 4

b 2 4ac 2a

x

12 4 2 1 2 2

1 r 3 4

b r

x

0

20 2 25

0

x2 2x 2

0

x

4 25 3

10 r

84. 4 x 2 4 x 4 x x 1

1r

x

14 r

3

b r

1r

10 21

2

0

b 4ac 2a

1 2 21

5r

1 4 2

85. 9 x 2 24 x 16

x 3

41 1

1 5 r 2 2

41 22

100 88 2

7 3

2

1 r

b 2 4ac 2a

2 r 3 0

2

0

81. x 2 14 x 44

b r

122 4 9 3

12 r 6 7 18

x

10 r 2 3 2

0

2 9

2 4 1 2

10 r

3

b 2 4ac 2a

12 r

2 r

b r

b r

x

2 1

x

13

9 x 12 x 3

2

80. x 10 x 22

36 16 2

2

b 2 4ac 2a

2

62 41 4

12 x 9 x 2

83.

b r

2 r 2 3 2

6 r

3 r 2

400 300 50 20 r 10 3 1 , 50 5 5 2 2 x x2

b 4ac 2a

6 r 2 13 2

20 r

79.

b r

6 r

b 2 4ac 2a

20 r

0 2

21

1 , 1 2

78. 25 x 2 20 x 3

4 x2

6x

b r

0

b 2 4ac 2a

24 r

242 49 16 29

0

24 r 0 18 4 3

b 2 4ac 2a 142 41 44 21

14 r 2 5 2

7 r

5

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© Cengage Learning. All Rights Reserved.

NOT FOR SALE Chapter P.1

86. 16 x 2 40 x 5

x

b r

28 x 49 x 2

4

49 x 28 x 4

0

87.

2

b 4ac 2a 2

40 r 40 r

0

40 216

2

b r

x

416 5

Solving Solvin Equations Solvi

9

b 4ac 2a 2

282 4 49 4

28 r

2 49

1600 320 32

28 r 0 98

40 r 16 5 32 5 5 r 4 2

2 7

88. 3x x 2 1

0

x 3x 1

0

2

x

b r

b 2 4ac 2a

3 r

32 41 1 21

3 r 13 2 89.

2t 2 8t 5

b r

h

82 4 2 5

8 r

b 2 4ac 2a

2 2

0

8 r 2 6 4

2 r

§5 · ¨ x 14 ¸ ©7 ¹

2 25

6400 6100 50

x

2

y 2 12 y 25 y

b r

12 r 2 11 2

0 0

1372 2 25

2

4 25 9604

921,984 50

686 r 196 6 25

0

12 21

8x

b 2 4ac 2a

1372 r 2y

b 2 4ac 2a

12 r

b r

1372 r

8 3 r 5 5 5

8x

25 x 2 1372 x 9604

8 10 3 r 5 50

y

2

25 2 x 20 x 196 49 25 2 x 28 x 196 49

802 4 25 61

80 r

6 2

92.

b 2 4ac 2a

80 r

91.

13 2

0

90. 25h 2 80h 61

b r

3 r 2

5 2t 2

8t

t

93. 0.1x 2 0.2 x 0.5 2

6r

41 25 x 11

x

b r 0.2 r

0

b 2 4ac 2a

0.2 40.1 0.5 20.1 2

x | 1.449, 3.449

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© Cengage Learning. All Rights Reserved.

10

Chapter P

Prerequisites requisites

94. 0.005 x 2 0.101x 0.193 b r

x

100. x 2 14 x 49

0

x

b 4ac 2a 2

0.101 40.005 0.193 2 0.005

0.101 r 0.006341 0.01 | 2.137, 18.063

95. 422 x 2 506 x 347

7

11 4

12 x 12

x2 x

0

2

x

x | 1.687, 0.488 96. 3.22 x 2 0.08 x 28.651

b r

x

102. 3x 4

0

0.08 2 43.22 28.651 2 3.22

x2 2 x 1

0 Complete the square.

x 2x 2

1

x 2 2 x 12

x

1

98. 11x 33x

0

11 x 3 x

0

x x 3

0

0 or

x 3

2

2

x

99.

x

3

2

x 3

81

r

x

1 2

12 4

r

3

3 42 11 2 2 2

97 4 2x 8

4x 4

0

2

4 x 1

0

1 x 1

0

x 1

0

2

a 2 x2 b2

104.

ax

or

Factor.

x 1

0

x

1

1

x

Factor.

x

1 2

2

x

1r

2

12 4

103. 4 x 2 2 x 4

r 2

x

2

12

97

3 r 4

2

x 1

11 4

4

1 12

2

2

2 x 2 3x 11

3r

0.08 r 369.031 | 2.995, 2.971 6.44 97.

11 4

3 r

x

Complete the square.

2 x 2 7 Quadratic Formula

0

b 2 4ac 2a

0.08 r

0

x2 x

506 4422 347 2 422

506 r

0

x

x2 x

101.

Extract square roots.

0

x 7 2

0.101 r

x

7

0

2

b ax b

0 Factor. 0

ax b

0 x

ax b

0 x

b a

0 3

Extract square roots.

r9

2 x 4 50 x 2

0

2 x x 25

0

2 x 2 x 5 x 5

0

105.

2

x 3

9 or

x 3

x

6 or

x

9 12

2

2 x2

0 or

x 5

x

0 or

x

b a

or

x 5

0

5 or

x

5

0

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Chapter P.1

20 x 3 125 x

106.

5 x 4 x 25 2

x

0

5 x 2 x 5 2 x 5

9 x 3 x 3

0

2

0 No real solution

0 x

0

x 3

0 x

3

2x 5

0 x

52

x 3

0 x

3

2x 5

0 x

5 2

x 6 64

0

8 x 8

0

2 x 2 2 x 4 x 2 x 2 2 x 4

0

x

3

3

x 2 x 2x 4 2

x 2 x 2x 4 2

x3 216

0

x 6

3

0

6 x 6 x 36

0

109.

3

2

x 6

0 x

x 6 x 36

110. 9 x 4 24 x3 16 x 2 x 9 x 24 x 16 x 2 3 x 4

2

0 No real solution (by the Quadratic Formula) 0 x

2

0 No real solution (by the Quadratic Formula)

6

0

2

0 x

x3 3x 2 x 3

0

x x 3 x 3

0

111. 2

0

2

x 3 x x 3 x 1 x 2

0

x2

0 x

0

3x 4

0 x

4 3

1 1

0 x

3

x 1

0 x

1

x 1

0 x

1

0

x x 2 3 x 2

0

2 x 2 3

0

x

x 2 x2 3

0

x x 1 x 1

0

1 x 1

0

1 x 1 x 2 x 1

0

x x

3

0 x

2

0 No real solution

x3 1

x 4 x3 x 1 3

0

x 3

2

x4 x

0

x3 2 x 2 3x 6

112.

113.

11

0 No real solution (by the Quadratic Formula)

2

2

0

x 9

108.

x

2

0

5x

x

x 4 81

107.

0

Solving Solvin Equations Solvi

x 1

0 x

1

x 1

0 x

1

INSTRUCTOR USE ONLY x x 1 2

0 No real solution (by the Quadratic Formula)

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© Cengage Learning. All Rights Reserved.

12

NOT FOR SALE

Chapter P

Prerequisites requisites x 4 2 x3

114.

16 8 x 4 x3

x 4 2 x3 8 x 16

0

x x 2 8 x 2

0

3

x x

8 x 2

0

2 x 2 2 x 4 x 2

0

3

x 2

0 x

x 2x 4 x 2

0 x

x4 4x2 3

115.

x

x

3 x

0

3 x 1 x 1

0

2

2

36t

6t

x

3

0 x

x

3

0 x 0 x

1

x 1

0 x

1

x 1

3

2x

3

2

2 x 1

0

x 1

0

x

0 x

3

3

2x

2

2 3

2

x 1 x2 x 1 2 x 10

119.

2x 2x x

120. 7

x 6 7

x

0

7

0 t

7 6 7 6

0 No real solution

3 2

0 No real solution (by the Quadratic Formula) 0 x

1

0 No real solution (by the Quadratic Formula) 121.

10 100 50

0

6t

1

0

x2

0 t

0 No real solution (by the Quadratic Formula)

3

2

7

2

x 6 3x3 2

2 º¼» x 1 x

6t

t2 1

0 x

x x 1

2 ª« x 2 ¬

0

0 No real solution (by the Quadratic Formula)

2

3

3

0 x

x 2x 4

x

7 t 2 1

0

0

x 2

x

7 6t

0

0

2

3

7 t 1 2

0

x3 8 x3 1 x 2 x 2 2 x 4 x 1 x 2 x 1

118.

2

3

x 1

x6 7 x3 8

117.

36t 4 29t 2 7

116.

0

3 x 1

2

2

0 No real solution (by the Quadratic Formula)

2

122.

x 10 4

0

x 10

4

x 10

16

x

26

5 x 3

0

6

5 x

3

49 x

36

5 x

9

x

36 49

x

INSTRUCTOR USE ONLY 4

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.1

2x 5 3

123.

0

130.

x

3

2x 5

3

2x 5

x

3

3

9 4

x

3

3

2x x

2

125.

3

82 64 3

x

3

64

3 4

x

131.

23

0

3 2x

2

x 3

53 2

3 2x

4

x 3

r5 5

2 x

1

x

x

x2

3 r 5 5

x 22

43

16

x x 22 2

2x 1 8 3

132.

0

x 2 x 22

2x 1

8

2x 1

512

x

513 2

163 4 8

or

x 2 x 22

8 0

x x 30

0

x x 14

x

6 x 5

0

1r

x

6 or x

5

2

513

2x

1

5

3 2x 2

1 2

2

x

3

4x 3 2 3

1r

2

4x 3

8

4x

5

x

5 x 26 4

127.

x 1 5 x 2

0

6 or

134.

4 x 2 x 1

13

2 x ª2 x x 1 ¬

x 6 x 6 x

43

2 x x 1 ª¬2 x 3 x 1 º¼

0

13

x

7

2 x x 1

13

5 x

3

2, 5

0 x

1

extraneous

0 0

2x 5

32

0 x

43 3 x 1 º ¼

13

10

x

0 x 1

6 x x 1

2x 5

x 5

6

0

5x 2

x 8 x 16

x 5

ª¬3 x 2 x 1 º¼

x 1

5 x 26

0

0

12

2

x

32

12

x 4

6 x 7

2 x 1

x

0

x

x 1

12

5 x 26

x

129.

12

5 4

x 2 13 x 42

128.

133. 3x x 1

57 2

0

4x 3

1 41 14 2

21

x 126.

13

8

x 3

No solution. 124.

32

Solving Solvin Solvi Equations

0

2x

0 x

0

x 1

0 x

1

5x 3

0 x

3 5

8 82 3 4 10

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

14

NOT FOR SALE

Chapter P

Prerequisites requisites

x

135.

2 x x 2x2

136.

3 1 x 2 1 §3· 2 x ¨ ¸ 2 x §¨ ·¸ © x¹ © 2¹ 6 x

2x2 x 6

0

2 x 3 x 2

0

2x 3

0 x

x 2

0 x

3 2 2

13x 1

12 x

13 x 1

x

12 x

x2 x 3

x2 x 3

x2 3

0

x

r

1 x 2

0

1 x 3

3

x

x

0 x

1

x 3

0 x

3

20 x

x2 x 2 x 20

0

x

0 x

1

x 3

0 x

3

Only x

3 and x

3 and x

1 are extraneous

3 x 18

x2 6x

5 x 4

0 x

5

x 4

0 x

4

x 4 x

3 are solutions to the original

First equation:

x 5

138.

0

x 1

142. x 2 6 x

0

0

1 x 3

0

x 1

x2 x 3

x2 2 x 3

equation. x x

1

Second equation:

x 2 3x 2

x 2x 3

20 x x

11 13

12

13 x 1

x

1

2

3 x 18

x 3 x 18 2

x

0

3 x 6

0

x 3

0 x

3

x 6

0 x

6

4x 1

3 x

Second equation:

x 1

3 x x

x 2 6 x

3 x 18

4x2 x

3

0

x 2 9 x 18

4x2 x 3

0

0

x

0

0

x 3 x

3

0

x 6 x

6

4 x

139.

12

First equation:

4 x 8 3x 3

137.

13 x 1

141. x

4 3 x 1 x 2 4 x 2 3 x 1

x

140.

3 x 1 4x 3

0 x

x 1

0 x

2x 1

5

2x 1

5 x

3

2 x 1

5 x

2

3 4 1

3 x 6

The solutions to the original equation are x x 6.

r3 and

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

Section P.1 143. x 1

x2 5

x

x2 5

x2 x 6

0

3 x 2

0

x 3

0 x

3

x 2

0 x

2

x 14. To attempt to solve a quadratic equation by factoring, the equation should be in general form first. If it factors, then the zero-factor property can be employed to solve the equation.

Second equation:

4x2 4x

15

x 1

x 5

x 1

x2 5

2 x

0

2x 3

0 x

3 2

2x 5

0 x

52

2

x2 x 4

3 and x

Only x

4 x 4 x 15 2

1 r 17 2

x

1 17 are solutions to the 2

original equation. x

15

146. The zero-factor property states that if the product of two factors is zero, then one (or both) of the factors must be zero. To solve 4 x 2 4 x 15, a student factors 4x from the left side of the equation and sets each factor equal to 15 and 15. The resulting incorrect solutions are x 4

First equation: x 1

Solving Solvin Solvi Equations

2 and x

1 17 are 2

3 2 x 5

0 0

147. Equivalent equations are derived from the substitution principle and simplification techniques. They have the same solution(s).

2x 3

8 and 2 x

5 are equivalent equations.

extraneous. 144. x 10

148. Remove symbols of grouping, combine like terms, reduce fractions.

x 2 10 x

Add(or subtract) the same quantity to (from) both sides of the equation.

First equation: x 10

x 2 10 x

0

x 2 11x 10

Multiply (or divide) both sides of the equation by the same nonzero quantity.

0

x

Interchange the two sides of the equation.

0

x 1 x

0

x 10 x

1 x 10 1

0.432 x 10.44

149. Female: y 10

For y

Second equation: x 10

26.44

x 2 10 x

0

x 2 9 x 10

0

x

0

x 10 x

0

x 1 x

10 x 1 10

150. Male: y

1

For y 10 and

145. The student should have subtracted 15x from both sides so that the equation is equal to zero. By factoring out an x, there are two solutions. 0 or x

0.432 x 10.44 0.432 x

26.44 x 0.432 x | 61.2 inches

The solutions to the original equation are x x 1. x 1 is extraneous.

x

16

16:

0.449 x 12.15 19:

19 31.15

0.449 x 12.15 0.449 x

69.4 | x Yes, it is likely that both bones came from the same person because the estimated height of a male with a 19inch thigh bone is 69.4 inches.

6

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

16

NOT FOR SALE

Chapter P

151. (a) P

Prerequisites requisites

157. 1

200 million when 182.17 1.542t 1 0.018t

200

2001 0.018t

182.17 1.542t

200 3.6t

182.17 1.542t

2.058t

2 and 1

ªx 1 ¬

ª x 1 ¬

x

182.17 1.542t 1 0.018t 182.17 1.542t

241 4.338t

182.17 1.542t 58.83

158. x

0.2 x 26.25

x

3

2

0

3

159. 9 9 a

0

x 2 3 x 18

0

Any non-zero multiple of this equation would also have these solutions. 156.

x 4 x 11 x

0

4 x 11

0

x 15 x 44

0

2

0

x 6x 4

0

5

b b 9

b 9 OR 9 a

b 9

a

b 18

9 a

b 9

a

18 b

a

b

a So, a 18 b or a you know that b t 9.

b

b. From the original equation

Some possibilities are: b

0

3 x 6

0

2

9 a

0 has only one solution to check, 0.

5, so:

5 x 3

155. –3 and 6 One possible equation is:

x

0

x 2x 1

26,250 passengers

x 3 x 6

2

0

5, x

9 a

153. False—See Example 14 on page A58. 154. False. x

2

1

x 2x 1 2

x 3

0.2 x 1 0.2 x 1

0

x 3 5 x 3 5

2.5:

5.25

2 º¼

2

t | 21 years

The model predicts the total voting-age population will reach 241 million during 2011. This value is reasonable.

6.25

0

Any non-zero multiple of this equation would also have these solutions.

2411 0.018t

2.5

2

So, the total voting-age population reached 200 million during 1998.

152. When C

2º ¼

2 ºª ¼¬ x 1

17.83

2.796t

2 ºª x 1 ¼¬

t | 8.7 years

(b) 241

2

One possible equation is:

9, a

9

b

10, a

8 or a

10

b

11, a

7 or a

11

b

12, a

6 or a

12

b

13, a

5 or a

13

b

14, a

4 or a

14

160. Isolate the absolute value by subtracting x from both sides of the equation. The expression inside the absolute value signs can be positive or negative, so two separate equations must be solved. Each solution must be checked because extraneous solutions may be included. 161. (a)

ax 2 bx

0

x ax b

0

x

0

ax b

0 x

(b) ax 2 ax

0

ax x 1

0

ax

0 x

0

x 1

0 x

1

b a

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.2

Solving Inequalities

17

162. Sample answer:

(a) An identity is true for all real numbers, whereas a conditional equation is true for just some (or none) of the real numbers in the domain. (b) Sample answer: x 4

0

(c) The opposite of b plus or minus the square root of the quantity b 2 minus the product of 4, a, and c, all divided by the product of 2 and a. (d) No. For instance, x and x 2.

2 is not equivalent to x 2

4 because x 2

4 has solutions x

2

Section P.2 Solving Inequalities 13. Interval: f, 2

1. solution set

(a) Inequality: x 2

2. graph

(b) The interval is unbounded.

3. negative

14. Interval: f, 7@

4. union

(a) Inequality: f x d 7 or x d 7

5. key; test intervals

(b) The interval is unbounded. 6. zeros; undefined values

15. x 3

7. Interval: >0, 9

Matches (b).

(a) Inequality: 0 d x d 9

16. x t 5

(b) The interval is bounded.

Matches (h).

8. Interval: 7, 4

17. 3 x d 4

(a) Inequality: 7 d x d 4

Matches (e).

(b) The interval is bounded.

18. 0 d x d

9. Interval: >1, 5@

9 2

Matches (d).

(a) Inequality: 1 d x d 5

19. x 3 3 x 3

(b) The interval is bounded.

Matches (f).

10. Interval: 2, 10@

20. x ! 4 x ! 4 or x 4

(a) Inequality: 2 x d 10

Matches (a).

(b) The interval is bounded. 11. Interval: 11, f

21. 1 d x d

(a) Inequality: x ! 11

5 2

Matches (g).

(b) The interval is unbounded.

22. 1 x

12. Interval: >5, f

5 2

Matches (c).

(a) Inequality: 5 d x f or x t 5 (b) The interval is unbounded.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

18

Chapter P

Prerequisites requisites

23. 5 x 12 ! 0 (a) x

(b)

3 ?

3

x

(c) ?

53 12 ! 0

5 3 12 ! 0

3 ! 0

27 ! 0

Yes, x

3 is

52 12

! 0

1 2

! 0

?

5

5 2

Yes, x

a solution.

3 2

(d) x

3 is not

No, x

a solution.

5 2

x

32 12

! 0

92

! 0

?

5

is

3 2

No, x

a solution.

is not

a solution.

24. 2 x 1 3 (a) x

0

(b) ?

(c)

20 1 3

2 3

4 is

32 is not

No, x

a solution.

a solution.

a solution. 25. 0

?

2 32 1 3

7 3 Yes, x

14 is not

No, x

a solution.

32

(d) x ?

3

1 2

0 is not

4

x

2 4 1 3

?

2 14 1 3

1 3 No, x

14

x

x 2 2 4

(a) x

4

(b)

4 2 ? 2 4 1 0 2 2 Yes, x 4 is

x

(c)

10 ?

(d) x

?

0

No, x

0

0 2 ? 2 4 1 0 2 2 No, x 0 is not

10 2 ? 2 4 0 2 2

?

0

x 0

10 is not

a solution.

a solution.

?

0

7 2 7 2 2

4 3 2 0 8 7 is Yes, x 2 a solution.

a solution.

?

2

26. 5 2 x 1 d 1

12

(a) x

(b)

?

52

x

?

?

(c) ?

?

?

43 1 d 1

?

5 5 1 d 1

5

1d1

5 2 d 1

5 6 d 1

8 3

5

5 3

1

12 is

52 is not

No, x

a solution.

4 3

No, x

a solution.

?

5 20 1 d 1 5 d 1 d 1

?

5 1 1 d 1 Yes, x

?

?

5 2

?

0

(d) x

?

5 2 52 1 d 1

5 2 12 1 d 1

4 3

x

Yes, x

0 is

a solution.

is not

a solution.

27. x 10 t 3

(a) x

(b)

13 ?

13 10

t 3

x

1

1 10

3 t 3 Yes, x

13 is

a solution.

(c) ?

t 3

x

Yes, x

1 is

?

Yes, x

x

9

t 3

9 10

4 t 3

1

14 10

11 t 3 a solution.

(d)

14

14 is

a solution.

No, x

?

t 3 t

3

9 is not

a solution.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.2

Solving Inequalities

19

28. 2 x 3 15 6

(a) x

(b)

2 6 3

x

?

0

(c)

20 3

15

15 15 Yes, x

a solution.

1 4

4 x

1 4

12

No, x

a solution.

12

2

3

4

−6

−5

−4

−3

3

− 12 x −2

−1

0

1

−2

−1

0

1

2

42.

−3

11

12

13

43.

14

x d 5

x 3

5

4

6

x 0

2 x 4

1

2

3

44.

x −1

0

1

1 2

t 3x

5 2

7

8

9

x 0

1

2

3

4

5 2

9x 1

3 4

16 x

2

1 6 x −1

0

1

3

2

1 6

45. 3.6 x 11 t 3.4

2

3.6 x t 14.4

x −6

−5

−4

−3

−2

x t 4

37. 2 x 1 t 1 5 x

2 7

46. 15.6 1.3x 5.2

x −2

−1

0

1

2

x 14

1.3 x 20.8

2 7

15

16

17

18

x ! 16

38. 6 x 4 d 2 8 x

47.

x −5

−4

−2

−3

−1

1 2x 3 9 2 2 x 6

x t 3

x −1

0

1

2

3

1 x 3

39. 4 2 x 33 x

x 5

1 t 3 x

x !

1 2

1 2

4 2 x 9 3x

6

12 x 2

36. 3x 1 t 2 x

2 x d 6

8 x

36 x 4 48 x 6

4

x ! 2

7x t 2

x 5

x t 2

7

35. 2 x 7 3 4 x

2x t 1

1 2

4x

34. x 7 d 12

6

x 7

x 10

5

5 x ! 35

−1

−2

4

! x 2

2x 7

21 2 x ! 7 x 14

33. x 5 t 7 x t 12

3

x −4

3

x t 4

2

or x

x 2

3

3 2

52

6 d x 7 14 x d 1

x

12

−5

15 6

3 x 4

3 2

32. 6 x ! 15

x t

a solution.

x 12

41.

x

x t

7 is

−2

2 x ! 3

x

Yes, x

4x 4 2x 3

x

x 4

2 x

11 15

12 is not

40. 4 x 1 2 x 3

5

?

15

2 x 1

30. 10 x 40

12

7

27 3

15

a solution. x

1

x

21 15

0 is

x 3

31.

(d) ?

212 3

15

4 x 12

29.

x

3 15

6 is not

No, x

?

x 3

4

5

6

7

48. 8 d 3 x 5 13

8 d 3 x 5 13

x − 6 − 5 −4 −3 −2 −1

0

1

3 d 3 x 18 6 x d 1

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© Cengage Learning. All Rights Reserved.

20 49.

Chapter P

NOT FOR SALE

Prerequisites requisites

8 d 1 3 x 2 13

x −2 −1

8 d 1 3x 6 13

0

1

2

3

4

59.

5

8 d 7 3 x 13 5 t x ! 2 0 d 2 3 x 1 20

60.

1

−3

0 d 2 3 x 3 20

x −7 −6 −5 −4 −3 −2 −1

0 d 1 3x 20

0

51.

2x 3 4 3 12 2 x 3 12

0

2

4

8

6

x ! 3 5

x −10

10

20

62. No solution. The absolute value of a number cannot be less than a negative number. 63.

x 3 5 2 0 d x 3 10

−3

0 d

7

−2

0

2

4

6

! x 1 !

26

14

x 10

15

20

25

30

14 d x d 26

x −4

x 20 d 6 6 d x 20 d 6

8

64. x 8 t 0

3 d x 7 53.

3

No solution. The absolute value of a number cannot be less than a negative number.

x −6 −4 −2

9 15 x 2 2

3 4

0

2

61. x 5 1 15 2

− 29

9 2 x 15

52.

−20

1

t x ! 7

4

0

x x 3 or ! 3 5 5 x 15 x ! 15

1 d 3 x 21 13

x −3 −2 −1

x x 1 or !1 2 2 x 2 x ! 2

15 d 3 x 6

50.

x !1 2

x 8 t 0 or x 8 t 0 1 4

x t 8

− 41

− 43

x 8 t 0 x t 8

x

14 ! x ! 34

−1

0

1

x d 8

34 x 14

All real numbers x. x

x 1 3 3 6 x 3

−3 −2 −1

54. 1 2

1

3

5

7

9

11

4 x t 6

x 10

x d 32

13.5 12

11

13

14

− 23 x

10.5 d x d 13.5

−2 −1

1.5 x 6 ! 10.5 2 9 ! 1.5 x 6 ! 21

0

1

2

3

4

66. 1 2 x 5

56. 4.5 !

5 1 2 x 5

x −2

−1

0

1

2

3

6 2 x 4

3 ! 1.5 x ! 15

3 ! x ! 2

2 ! x ! 10

2 x 3

There is no solution.

58. x t 8

or 3 4 x t 9

x t 3 10.5

4.2 d 0.4 x d 5.4

5 x 5

3

4 x d 12

3.2 d 0.4 x 1 d 4.4

x 5

2

3 4 x d 9

3 x 9

57.

1

65. 3 4 x t 9

9 x 3

55.

0

x

−5

5 x

−6 −4 −2

0

2

4

6

x

INSTRUCTOR USE ONLY x t 8 or x d 8

−8

−4

0

4

8

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.2

67.

73. 6 x ! 12

x 3 t 4 2

10

x 3 t 4 2 x 3 t 8

x d 5

−10

74. 3x 1 d 5

5

10

10

3x d 6

x 0

10

−10

x t 11 11 15

x d 2 2x 1 3

68. 1

21

x ! 2

x 3 d 4 or 2 x 3 d 8

− 15 − 10 − 5

Solving Inequalities

−10

10

x −1

0

1

2

3

−10

4

75. 5 2 x t 1

2x 1 3 2x 2 0 3 3 ! x ! 0 1 1

10

2 x t 4 x d 2

−10

10

−10

0 x 3 76. 20 6 x 1

69. 9 2 x 2 1

x 3

9 2x 1

4

5

6

21 6 x 7 2

1 9 2 x 1

10

x

−10

10

10 2 x 8

−10

5 ! x ! 4 4 x 5

77. 4 x 3 d 8 x

70. x 14 3 ! 17

5 x d 20

x 14 ! 14

x 14 14 or

78. 3 x 1 x 7

10

3x 3 x 7

x 7

2x 4

x 10 t

or

29 2

− 29

9 2

x t

79. x 8 d 14

11 2

10

14 d x 8 d 14 −10

6 d x d 22

− 11

2

10

−10

9 2

x 10 d 92 x d

−10

x 2

71. 2 x 10 t 9

x 10 t

10

−10

x ! 0 0

−10

x d 4

x 14 ! 14

x 28 −35 −28 −21 −14 −7

10

4 x 12 d 8 x

2

24

x − 16

− 12

−8

−4

−10

72. 3 4 5 x d 9 4 5x d 3

80. 2 x 9 ! 13

7 5

1 5

x 0

1

2

2 x 9 13 or 2 x 9 ! 13

3 d 4 5 x d 3 7 d 5 x d 1 7 5

t x t

1 5

1 5

d x d

7 5

2 x 22

2x ! 4

x 11

x ! 2

10

−15

9

INSTRUCTOR USE ONLY −10

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22

Chapter P

NOT FOR SALE

Prerequisites requisites

81. 2 x 7 t 13

x 7 t

13 2 13 2 27 2

x 7 d x d

88.

4

6 x 15

6 x 15 t 0

or

x 7 t x t

13 2 12

6 x t 15

x t 52 ª 5 , f ¬ 2

10

− 15

89. x 10 8

1

All real numbers within 8 units of 10. −10

90. x 8 ! 4 82.

1 2

x 1 d 3

10

All real numbers more than 4 units from 8.

x 1 d 6 6 d x 1 d 6 7 d x d 5

10

−10

91. The midpoint of the interval >3, 3@ is 0. The interval

represents all real numbers x no more than 3 units from 0. x 0 d 3

83. x 5 t 0

x d 3

x t 5

>5, f 84.

−10

92. The graph shows all real numbers more than 3 units from 0. x 0 ! 3

x 10 x 10 t 0

x ! 3

x t 10

>10, f

93. The graph shows all real numbers at least 3 units from 7.

x 7 t 3

85. x 3 t 0 x t 3

>3, f 86.

x 1 d 4

3 x

95. All real numbers within 10 units of 12.

3 x t 0

x 12 10

3 t x

f, 3@

96. All real numbers at least 5 units from 8.

87. 7 2 x t 0

2 x t 7 x d

f, 72 º¼

94. The graph shows all real numbers no more than 4 units from 1.

7 2

x 8 t 5 97. All real numbers more than 4 units from 3.

x 3 ! 4 x 3 ! 4 98. All real numbers no more than 7 units from 6.

x 6 d 7

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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NOT FOR SALE Section P.2

Solving Inequalities

23

99. x 2 3 0

(a) x

(b)

3

3

2

?

x

0

0

3 0

2

3 2

x

32

?

3 0

6 0 No, x

(c)

2

?

3 0

3 0

3 is not

Yes, x

a solution.

5

(d) x

5

?

2

3 0 22 0

34 0

0 is

3 2

Yes, x

a solution.

is

5 is not

No, x

a solution.

a solution.

100. x 2 x 12 t 0

(a) x

(b)

5

5 2

?

5 12 t 0

x

0

0 2

(c) ?

0 12 t 0

5 is

No, x

a solution.

4 12 t 0

3 12 t 0

?

?

9 3 12 t 0 0 t 0

8 t 0

a solution.

4 is

Yes, x

a solution.

3 is

a solution.

x 2 t 3 x 4

(a) x

(b)

5

Yes, x

x

4

(c)

4 2 ? t 3 4 4 6 is undefined. 0 No, x 4 is not

5 2 ? t 3 5 4 7 t 3 5 is

a solution.

x

9 2

9 2

(d) x

9 2 ? 2 t 3 9 4 2 13 t 3

9 2 ? 2 t 3 9 4 2 5 t 3 17 9 is not No, x 2 a solution.

a solution.

102.

?

3 2

16 4 12 t 0

0 is not

Yes, x

101.

3

(d) x ?

4 2

12 t 0

8 t 0 Yes, x

4

x

9 is 2 a solution. Yes, x

3x 2 1 x 4 2

2

(a) x

3 2

2

2

2

4

(b)

1

x

3 1

?

1

1

2

(c)

2

?

4

12 1 8 2 is not No, x

Yes, x

a solution.

a solution.

103. 3x 2 x 2 3x 2 x 1

3 x

2 x 1

0 x 0 x

23 1

The key numbers are 23 and 1.

1

3 1 5 1 is

x

0 30

0

2

(d) x

2

33

?

1

4

3

0 1 Yes, x

3 2

2

4

?

1

27 1 13 No, x 3 is not

0 is

a solution.

a solution. 104. 9 x3 25 x 2

0

x 9 x 25

0

2

x2

0 x

0

9 x 25

0 x

25 9

The key numbers are 0 and

25 . 9

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24

Chapter P

105.

1 1 x 5

NOT FOR SALE

Prerequisites requisites 1 1 x 5

x 4

x 5 x 4 x 5 0 x 4

Key numbers: x

x 5

0 x

Test intervals: f, 4 , 4, 4 , 4, f

x 2 16 d 0

x

5

x x 1 2 x 2

2 x x 2 x 1

x

x

x

4 x 1

r4

x-Value Value of x 2 16

Interval

2 x 1

x2 x 2x 4 x 2 x 1

x x

4 x 4 d 0

Test: Is x 4 x 4 d 0?

The key numbers are 4 and 5. 106.

x 2 d 16

108.

4 x 1

2 x 1

f, 4

–5

9

Positive

4, 4

0

16

Negative

4, f

5

9

Positive

Solution set: > 4, 4@

0 x

x 4

0 x

4

x 1

0 x

1

2 x 1

−6 −4 −2

0 x

2

x 1

0 x

1

0

x

109.

0

x 2

2

4

6

2 d 25 2

x 2 4 x 4 d 25 x 2 4 x 21 d 0

x

The key numbers are 2, 1, 1, and 4.

7 x 3 d 0 7, x

Key numbers: x x2 9

107.

Test: Is x 7 x 3 d 0?

3 x 3 0

Test intervals: f, 3 , 3, 3 , 3, f Test: Is x 3 x 3 0?

x-Value

Value of x 2 9

f, 3

–4

7

Positive

3, 3

0

–9

Negative

3, f

4

7

Positive

Interval

Conclusion

−4 −3 −2 −1

0

1

2

3

x

Value of 7 x 3

f, 7

–8

1 11

7, 3

0

7 3

3, f

4

11 1

11

Conclusion

Positive

21

Negative

11

Positive

Solution set: >7, 3@ −7

3 x

−8 −6 −4 −2

Solution set: 3, 3 x

x-Value

Interval

r3

Key numbers: x

3

Test intervals: f, 7 , 7, 3 , 3, f

x2 9 0

x

Conclusion

x

110.

0

2

4

6

3 t 1 2

x2 6x 8 t 0

4

x

2 x 4 t 0

Key numbers: x

2, x

4

Test intervals: f, 2 x 2 x 4 ! 0

2, 4 x 2 x 4 0 4, f x 2 x 4 ! 0 Solution set: f, 2@ >4, f x

INSTRUCTOR USE ONLY 1

2

3

4

5

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© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.2

111.

x2 4x 4 t 9

x

x2 2 x 3 ! 0

x

5 x 1 t 0 5, x

Key numbers: x

3 x 1 ! 0

f, 5

x

Value of 5 x 1

1 7

6

5, 1

0

5 1

1, f

2

7 1

7

3, 1 x 3 x 1 0 1, f x 3 x 1 ! 0

Conclusion

Solution set: f, 3 1, f Positive

x − 4 −3 −2 −1

5

Negative

7

Positive

115.

112.

1

Interval

1 x 7 0 1, x

1, 7 x 1 x 7 0 7, f x 1 x 7 ! 0

4

6

Conclusion

1 5

5

Positive

3, 1

0

3 1

3

Negative

1, f

2

5 1

5

Positive

x −3

x 2

Value of 3 x 1

x

4

−2

−1

7 0

1

Solution set: 3, 1

Solution set: 1, 7 −1

3, x

x-Value

f, 3

7

Test intervals: f, 1 x 1 x 7 ! 0

−2

3 x 1 0

Test: Is x 3 x 1 0?

x2 6 x 7 0

Key numbers: x

2

Test intervals: f, 3 , 3, 1 , 1, f

2

x 2 6 x 9 16

x

1

Key numbers: x

x 0

0

x2 2 x 3 0

x

Solution set: f, 5@ >1, f −6 −5 −4 −3 −2 −1

1

Test intervals: f, 3 x 3 x 1 ! 0

Test: Is x 5 x 1 t 0?

x-Value

3, x

Key numbers: x

1

Test intervals: f, 5 , 5, 1 , 1, f

Interval

25

x2 2 x ! 3

114.

x2 4 x 5 t 0

Solving Inequalities

1

x2 ! 2 x 8

116.

8

0

x2 2x 8 ! 0 x2 x 6

113.

x

x2 x 6 0

x

3, x

Test: Is x 4 x 2 ! 0?

f, 3

x

x-Value

Interval

Test: Is x 3 x 2 0?

x-Value

4

Test intervals: f, 2 , 2, 4 , 4, f

2

Test intervals: f, 3 , 3, 2 , 2, f

Interval

2, x

Key numbers: x

3 x 2 0

Key numbers: x

4 x 2 ! 0

Value of 3 x 2

Conclusion

4

1 6

6

Positive

3, 2

0

3 2

6

Negative

2, f

3

6 1

6

Positive

x

Value of 4 x 2

Conclusion

f, 2

–3

7 1

7

Positive

2, 4

0

4 2

8

Negative

4, f

5

1 7

7

Positive

Solution set: f, 2 4, f x

Solution set: 3, 2

−3 −2 −1

0 1 2

3 4

5

x

INSTRUCTOR USE ONLY −3

−2

−1

0

1

2

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26

NOT FOR SALE

Chapter P

Prerequisites requisites

3 x 2 11x ! 20

117.

119.

x

3 x 2 11x 20 ! 0

3x 4 x 5 ! 0 Key numbers: x

x 2 3x 18 ! 0 3 x 6 ! 0 3, x

Key numbers: x

5, x

6

Test intervals: f, 3 , 3, 6 , 6, f

43

Test intervals: f, 54 , 34 , 5 , 5, f

Test: Is x 3 x 6 ! 0?

Test: Is 3x 4 x 5 ! 0?

Interval

Interval

x-Value

f, 43 43 , 5

3 x

Value of 4 x 5

Conclusion

–3

5 8

40

Positive

0

4 –5

20

Negative

6

22 1

22

Positive

5, f

3

4

5

–4

1 10

3, 6

0

3 –6

6, f

7

10 1

120.

0

2

4

6

2 x 2 6 x 15 t 0

x

2 x 2 d 0

10

Positive

6 42 15 2 2 2

2

2, x

Key numbers: x

2

Test intervals: f, 2 x 3 2 x 2 4 x 8 0

2, 2 x3 2 x 2 4 x 8 0 2, f x3 2 x 2 4 x 8 ! 0

156 4

6 r 2 39 4 3 r 2

Negative

x 2 x 2 4 x 2 d 0

6

2 x 2 4 d 0

6r

18

x3 2 x 2 4 x 8 d 0

x

6 r

Positive

8

118. 2 x 2 6 x 15 d 0

x

10

Conclusion

x

x 2

f, 3

−4 −2

3 1

Value of 3 x 6

−3

−4 0

x

Solution set: f, 3 6, f

Solution set: f, 43 5, f

−2 −1

x-Value

Solution set: f, 2@

39 2

x 0

3 2

Key numbers: x

39 ,x 2

3 2

1

2

3

4

39 2

Test intervals: § 3 ¨¨ f, 2 ©

39 · 2 ¸ 2 x 6 x 15 0 2 ¸¹

§3 ¨¨ ©2

39 3 , 2 2

39 · 2 ¸ 2 x 6 x 15 ! 0 2 ¸¹

§3 ¨¨ ©2

39 · , f ¸¸ 2 x 2 6 x 15 0 2 ¹

§ 3 Solution set: ¨¨ f, 2 © 3 − 2

39 2

3 + 2

ª3 39 º » « 2 ¼» ¬« 2

39 · , f ¸¸ 2 ¹

39 2 x

−2 −1

0

1

2

3

4

5

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© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.2

Solving Inequalities

27

x3 3 x 2 x ! 3

121.

x3 3x 2 x 3 ! 0 x 2 x 3 x 3 ! 0

x 3 x 2 x 3 x 1 x

1 ! 0 1 ! 0

Key numbers: x

1, x

1, x

3

Test intervals: f, 1 , 1, 1 , 1, 3 , 3, f Test: Is x 3 x 1 x 1 ! 0? Interval

x-Value

Value of x 3 x 1 x 1

Conclusion Negative

f, 1

–2

5 1 3

1, 1

0

3 1 1

3

Positive

1, 3

2

1 3 1

3

Negative

3, f

4

1 5 3

15

15

Positive

Solution set: 1, 1 3, f x −2 −1

122.

0

1

2

3

4

5

2 x3 13 x 2 8 x 46 t 6

123. 4 x 2 4 x 1 d 0

2 x3 13 x 2 8 x 52 t 0

2 x

x 2 2 x 13 4 2 x 13 t 0

2 x 13 x 13 x 2 x 2

2 x

f, 2 x 132 , 2 2 x 2, 2 2 x 2, f 2 x3

2, x

, 12 , f

Test: Is 2 x 1 d 0? x-Value Value of 2 x 1

Interval 3

13 x 8 x 52 0

3

13 x 2 8 x 52 ! 0

2

2

13 x 2 8 x 52 ! 0

, 2º¼, >2, f Solution set: ª¬ 13 2

1 2

2

2

13 x 8 x 52 0

3

1 2

Test intervals: f,

2 t 0

Test intervals: 13 2

2

Key number: x

4 t 0

13 ,x 2

Key numbers: x

1 d 0

f, 12 12 , f

0

1

1

1 2

2

1 1

2

Conclusion Positive Positive

1 2

Solution set: x 1 2

x −2

−1

0

1

2

− 13 2

x −8 − 6 −4 −2

0

2

4

124. x 2 3 x 8 ! 0

The key numbers are imaginary:

3 i 23 r 2 2

So the set of real numbers is the solution set. x −3 −2 −1

0

1

2

3

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28

125.

NOT FOR SALE

Chapter P

Prerequisites requisites

130. x 4 x 3 d 0

4 x3 6 x 2 0 2 x 2 2 x 3 0

Key numbers: x

Key numbers: x 0, x

0, 32 2 2 x 2 x 3 0 32 , f 2 x 2 x 3 ! 0 Solution set: f, 0 0, 32

0, 3 x 4 x 3 0 3, f x 4 x 3 ! 0

2

Solution set: f, 3@

2

126. 4 x3 12 x 2 ! 0

131.

4x 1 ! 0 x

Key numbers: x

4 x x 3 ! 0 2

0, x

3

Test intervals: f, 0 x 4 x 3 0

3 2

Test intervals: f, 0 2 x 2 2 x 3 0

Key numbers: x

0, x

0, x

1 4

14 , 14 , f

Test intervals: f, 0 , 0,

3

Test intervals: f, 0 4 x 2 x 3 0

Test: Is

0, 3 4 x 2 x 3 0 3, f 4 x 2 x 3 ! 0 Solution set: 3, f

4x 1 ! 0? x

Interval

x-Value

f, 0

–1

§ 1· ¨ 0, ¸ © 4¹

1 8

§1 · ¨ , f¸ ©4 ¹

1

Value of

x 4x t 0

4x 1 x

Conclusion

5 1

5

Positive

1 2 1 8

4

Negative

3 1

3

Positive

3

127.

x x 2 x 2 t 0

Key numbers: x

0, x

r2

Test intervals: f, 2 x x 2 x 2 0

2, 0 x x 2 x 2 ! 0 0, 2 x x 2 x 2 0 2, f x x 2 x 2 ! 0

§1 · Solution set: f, 0 ¨ , f ¸ ©4 ¹

Solution set: >2, 0@ >2, f

1 4 x −2

−1

0

1

2

128. 2 x x d 0 3

4

x3 2 x d 0

Key numbers: x

0, x

2

Test intervals: f, 0 x 3 2 x 0

0, 2 x3 2 x ! 0 2, f x3 2 x 0 Solution set: f, 0@ >2, f 129.

x

1 x 2 t 0 2

3

Key numbers: x

1, x

2

Test intervals: f, 2 x 1 x 2 0 2

2, 1 1, f

3

x 1 x 2 ! 0 2

3

x 1 x 2 ! 0 2

3

INSTRUCTOR USE ONLY Solution So u o sset: se : >2,, f

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.2

132.

x2 1 0 x x 1 x 1 0 x

134.

1 2x 1, x

Key numbers: x

0, x

x-Value

f, 1

–2

1, 0

1 2

3 1

1, f

2

§ 3 ·§ 1 · ¨ ¸¨ ¸ © 2 ¹© 2 ¹ 1 2 § 1 ·§ 3 · ¨ ¸¨ ¸ © 2 ¹© 2 ¹ 1 2

1 2

0, 1

Value of x 1 x 1 x

1 3

2

3 2

Positive

3 2

2

1 x d 0? 1 2x

Test: Is Interval

Negative

Positive

x-Value

1· § ¨ f, ¸ 2¹ ©

–1

2 1

§ 1 · ¨ , 1¸ © 2 ¹

0

1, f

2

133.

0

1

135.

5 ,x 3

5

5· §5 · § Test intervals: ¨ f, ¸, ¨ , 5¸, 5, f 3¹ © 3 ¹ © 3x 5 t 0? x 5

Interval § ¨ f, ©

x-Value

5· ¸ 3¹

§5 · ¨ , 5¸ ©3 ¹

5, f

2

Negative

1 1

1

Positive

1 5

1 5

Negative

x −2

3x 5 t 0 x 5

Test: Is

Conclusion

2

2

Key numbers: x

1 x 1 2x

−1

x −1

Value of

1· § Solution set: ¨ f, ¸ >1, f 2¹ ©

Solution set: f, 1 0, 1 −2

1

1· § 1 · § Test intervals: ¨ f, ¸, ¨ , 1¸, 1, f 2¹ © 2 ¹ © Negative

3 2

d 0

1 ,x 2

Key numbers: x

Conclusion

3 2

29

1 x d 0 1 2x

1

Test intervals: f, 1 , 1, 0 , 0, 1 , 1, f Interval

5 7x d 4 1 2x 5 7 x 41 2 x

Solving Inequalities

−1

0

1

2

x 6 2 0 x 1 x 6 2 x 1 0 x 1 4 x 0 x 1 1, x

Key numbers: x Value of

3x 5 x 5

Conclusion

0

5 5

2

65 2 5

1 3

Negative

6

18 5 65

13

Positive

4 x 4 1, 4 x 4 4, f x

Test intervals: f, 1

Positive

1

4

x 0 1 x ! 0 1 x 0 1

Solution set: f, 1 4, f x −2 −1

0

1

2

3

4

5

5º § Solution set: ¨ f, » 5, f 3¼ © 5 3 x 0

1

2

3

4

5

6

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30

136.

Chapter P

NOT FOR SALE

Prerequisites requisites

x 12 3 t 0 x 2 x 12 3 x 2 t 0 x 2 6 2x t 0 x 2

138.

x

2, x

Key numbers: x

5 3 ! x 6 x 2 5 x 2 3 x 6 ! 0 x 6 x 2

3

2 x 28 ! 0 6 x 2

14, 2

2, 6

6, f

x

137.

−1

0

1

2

3

−14

5, x

3, x

Test intervals: f, 5

5, 3

3, 11

11, f

x

−2

139.

11

x 11 0 x 5 x 3 x 11 ! 0 5 x 3

−5

5

3

6

2 x 28 ! 0 6 x 2

x

2 x 28 0 6 x 2

x

2 x 28 ! 0 6 x 2

10

x

30 5 x d 0 3 4 x 3

Key numbers: x

3, x

3 ,x 4

6

x 11 0 x 5 x 3

3· 30 5 x § ! 0 Test intervals: ¨ f, ¸ x 4 3 4 x 3 © ¹

x 11 ! 0 5 x 3

30 5 x § 3 · 0 ¨ , 3¸ x 3 4 x 3 © 4 ¹

x

3, 6

6, f

11 0

x

1 9 d x 3 4x 3 1 9 d 0 x 3 4x 3 4 x 3 9 x 3 d 0 x 3 4 x 3

x −9 −6 −3

2 x 28 0 6 x 2

6 0

Solution set: 5, 3 11, f −5

x

x −15 −10

x 11 ! 0 5 x 3

Key numbers: x

6

Solution intervals: 14, 2 6, f

2 1 ! x 5 x 3 2 1 ! 0 x 5 x 3 2 x 3 1 x 5 ! 0 x 5 x 3

x

2, x

Test intervals: f, 14

Solution interval: 2, 3@ −2

14, x

Key numbers: x

6 2x Test intervals: f, 2 0 x 2 6 2x ! 0 2, 3 x 2 6 2x 0 3, f x 2

9 12 15

x

30 5 x ! 0 3 4 x 3

x

30 5 x 0 3 4 x 3

§ 3 · Solution set: ¨ , 3¸ >6, f © 4 ¹ −3

3

4

x −4 −2

0

2

4

6

8

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.2

140.

1 1 t x x 3 1 x 3 1 x t 0 x x 3

142.

3, x

Key numbers: x

3, x

3, 0

3 ! 0 x x 3

Test intervals: f, 3

0, x

Test intervals: f, 3

0

3, 0

3 0 x x 3

0, f

3 ! 0 x x 3

31

x2 x 6 t 0 x x 3 x 2 t 0 x Key numbers: x

3 t 0 x x 3

Solving Inequalities

0, 2

2, f

x

2

3 x 2

x x 3 x 2

x x 3 x 2 x 3 x x 2 x

0

! 0 0 ! 0

Solution set: >3, 0 >2, f

Solution intervals: f, 3 0, f

x −3

x − 4 −3

141.

−2

−1

0

−2 −1

0

1

2

3

1

x2 2x d 0 x2 9 x x 2 d 0 x 3 x 3 Key numbers: x

143.

2, x

0, x

Test intervals: f, 3

3, 2 2, 0

0, 3 3, f

x x

x

x x

3x 2 x 2 ! 0 x 1 x 1

r3

x x 2

3 x 3

x x 2

3 x 3

x x 2

3 x 3

x x 2

3 x 3 x x 2

3 x 3

Solution set: 3, 2@ >0, 3

3 2x ! 1 x 1 x 1 3 x 1 2 x x 1 1 x 1 x 1 ! 0 x 1 x 1

! 0

Key numbers: x

0

Test intervals: f, 1

! 0 0 ! 0

1, x

1

3x 2 x 2 ! 0 x 1 x 1

1, 1

3x 2 x 2 0 x 1 x 1

1, f

3x 2 x 2 ! 0 x 1 x 1

Solution set: f, 1 1, f x −4 −3 −2 −1

0

1

2

3

4

x −3 −2 −1

0

1

2

3

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© Cengage Learning. All Rights Reserved.

32

144.

NOT FOR SALE

Chapter P

Prerequisites requisites

3x x d 3 x 1 x 4 3x x 4 x x 1 3 x 4 x 1 d 0 x 1 x 4

147.

x

x

4, x

Key numbers: x

Test: intervals f, 4

1 x 4

2, x

1, x

6

0

x 1 x 4 x 6 x 2 ! 0 4, 2 x 1 x 4 x 6 x 2 0 2, 1 x 1 x 4 x 6 x 2 ! 0 1, 6 x 1 x 4 x 6 x 2 0 6, f x 1 x 4

2

4

2

r2

x2 4 t 0 2, x

4, 5

9 2

12 12

14

Negative

5, f

6

2 1

2

Positive

2 x 9 2 x t 0 9 2

r

x-Value

9

Value of 2 x 9 2 x

–5

19 1

0

9 9

5

1 19

Conclusion

19

Negative

81

Positive

19

Negative

ª 9 9º Domain: « , » ¬ 2 2¼ 149.

x t 0 x 2 2 x 35 x t 0 x 5 x 7 5, x

0, x

x

2

2, 2 x 2 x 2 0 2, f x 2 x 2 ! 0

Conclusion

81 4 x 2 t 0

Test intervals: f, 5

Test intervals: f, 2 x 2 x 2 ! 0

Domain: f, 2@ >2, f

Positive

Key numbers: x

2 x 2 t 0

Key numbers: x

20

§9 · ¨ , f¸ ©2 ¹

Domain: >2, 2@

x

4 5

9· § ¨ f, ¸ 2¹ © § 9 9· ¨ , ¸ © 2 2¹

Test intervals: f, 2 4 x 2 0

146.

0

Interval

6

2, 2 4 x 2 ! 0 2, f 4 x 2 0

Value of 4 x 5

x

9· § 9 9· §9 · § Test intervals: ¨ f, ¸, ¨ , ¸, ¨ , f ¸ 2¹ © 2 2¹ © 2 ¹ ©

x 2 x t 0

Key numbers: x

f, 4

Key numbers: x

4 x2 t 0

145.

x-Value

9

x 0

Interval

148.

1 −2

5

Domain: f, 4@ >5, f

Solution set: f, 4 >2, 1 >6, f

−4

4, x

Test intervals: f, 4 , 4, 5 , 5, f

d 0

x 6 x 2

4 x 5 t 0

Key numbers: x

x 2 4 x 12 d 0 x 1 x 4 x 6 x 2

x 2 9 x 20 t 0

5, 0

0, 7

7, f

x x

7

x 0 5 x 7

x ! 0 5 x 7 x

5 x 7

0

x ! 0 5 x x 7

Domain: 5, 0@ 7, f

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© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.2

x t 0 x 9

150.

x

x t 0 3 x 3

Key numbers: x

3, x

0, x

Test intervals: f, 3

3, 0

0, 3

3, f

x

x

x 0 3 x 3

x ! 0 3 x 3

Key numbers: x | 2.39, x | 2.26 Test intervals: f, 2.26 , 2.26, 2.39 , 2.39, f

0

x

3 x 3

x

x ! 0 3 x 3

Domain: 3, 0@ 3, f 151.

1 3.4 ! 0 2.3 x 5.2 1 3.4 2.3x 5.2 ! 0 2.3 x 5.2 7.82 x 18.68 ! 0 2.3x 5.2

3

x

0.4 x 2 5.26 10.2 0.4 x 2 4.94 0

Solution set: 2.26, 2.39 156.

2 ! 5.8 3.1x 3.7 2 5.83.1x 3.7 ! 0 3.1x 3.7 23.46 17.98 x ! 0 3.1x 3.7

Key numbers: x | 1.19, x | 1.30

0.4 x 2 12.35 0

23.46 17.98 x 0 3.1x 3.7 23.46 17.98 x ! 0 1.19, 1.30 3.1x 3.7 23.46 17.98 x 0 1.30, f 3.1x 3.7 Solution set: 1.19, 1.30 Test intervals:

Key numbers: x | r 3.51 Test intervals: f, 3.51 , 3.51, 3.51 , 3.51, f Solution set: 3.51, 3.51 152. 1.3x 2 3.78 ! 2.12 1.3 x 2 1.66 ! 0

f, 1.19

157. x a t 2

Key numbers: x | r1.13 Test intervals: f, 1.13 , 1.13, 1.13 , 1.13, f Solution set: 1.13, 1.13

x a d 2 or

x a t 2

x d a 2

x t a 2

or

Matches graph (b) 158. x b 4

153. 0.5 x 2 12.5 x 1.6 ! 0

Key numbers: x | 0.13, x | 25.13

4 x b 4

Test intervals: f, 0.13 , 0.13, 25.13 , 25.13, f

b 4 x b 4

Solution set: 0.13, 25.13

Matches graph (b) 159. ax b d c c must be greater than or equal to zero.

c d ax b d c

154. 1.2 x 2 4.8 x 3.1 5.3

b c d ax d b c

1.2 x 2 4.8 x 2.2 0

1, then b c

Key numbers: x | 4.42, x | 0.42

Let a

Test intervals: f, 4.42 , 4.42, 0.42 , 0.42, f

This is true when b

Solution set: 4.42, 0.42

33

1 ! 3.4 2.3x 5.2

155.

2

Solving Inequalities

0 and b c c

10.

5.

One set of values is: a

1, b

5, c

5.

(Note: This solution is not unique. The following are also solutions. a

2, b

c

10

a

3, b

c

15.)

In general, a

k, b

c

5k , k t 0 or

INSTRUCTOR USE ONLY a

k, b k,

5k , c

5k , k 0

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© Cengage Learning. All Rights Reserved.

34

Chapter P

160.

x

NOT FOR SALE

Prerequisites requisites

a x b

166. Let x

(a) The polynomial is zero when x (b)

x

x a : x b : a x b :

a or x

−

+

+

−

−

+

+

−

+

number of dozens of doughnuts sold per day.

Revenue: R

b.

Cost:

C

2.75 x 220

Profit:

P

R C 4.50 x 2.75 x 220

x a

4.50 x

b

1.75 x 220

(c) A polynomial changes signs at its zeros.

60 d 1.75 x 220 d 270 280 d 1.75 x d 490

161. 9.00 0.75 x ! 13.50

160 d x d 280

0.75 x ! 4.50 x ! 6 You must produce at least 6 units each hour in order to yield a greater hourly wage at the second job. 162. Let x gross sales per month

The daily sales vary between 160 and 280 dozen doughnuts per day. h 68.5 d1 2.7

167.

1000 0.04 x ! 3000

h 68.5 d1 2.7 2.7 d h 68.5 d 2.7 1 d

0.04 x ! 2000 x ! $50,000

65.8 inches d h d 71.2 inches

You must earn at least $50,000 each month in order to earn a greater monthly wage at the second job.

65.8

163. 10001 r 2 ! 1062.50

71.2 h

65 66 67 68 69 70 71 72

1 2r ! 1.0625

h 50 d 30

168.

2r ! 0.0625

30 d h 50 d 30

r ! 0.03125

20 d h d 80

r ! 3.125%

The minimum relative humidity is 20 and the maximum is 80.

164. 825 7501 r 2 825 7501 2r

169. 2 L 2W

100 W

50 L

LW t 500

825 750 1500r

L50 L t 500

75 1500r 0.05 r

L 50 L 500 t 0

The rate must be more than 5%.

By the Quadratic Formula you have:

165. E

(a)

2

25 r 5 5

1.52t 68.0

Key numbers: L

70 d 1.52t 68.0 d 80

Test: Is L 50 L 500 t 0?

2.0 d 1.52t d 12.0

Solution set: 25 5 5 d L d 25 5 5

2

1.32 d t d 7.89

13.8 meters d L d 36.2 meters

The annual egg production was between 70 and 80 billion eggs between 1991 and 1997. (b) 1.52t 68.0 ! 100

170. 2 L 2W

440 W 220 L LW t 8000

L 220 L t 8000

1.52t ! 32.0

L2 220 L 8000 t 0

t ! 21.05 The annual egg production will exceed 100 billion eggs sometime during 2011.

By the Quadratic Formula we have: Key numbers: L

110 r 10 41

Test: Is L 220 L 8000 t 0? 2

Solution set: 110 10 41 d L d 110 10 41 45.97 feet d L d 174.03 feet

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.2

10001 r ! 1100 2

171.

1 r 2

174. s

! 1.1

16t 2 v0t s0

(a) 16t 2 128t

0

16t t 8

0

1 2r r 2 1.1 ! 0

By the Quadratic Formula we have: 1 r

16t

0 t

0

0 t

8

It will be back on the ground in 8 seconds.

1.1 1 r

Since r cannot be negative, r

16t 2 128t 128

(b)

1.1 | 0.0488

16t 2 128t 128 0

4.88%.

16t 2 8t 8 0

Thus, r ! 4.88%. 172. R

P

t 2 8t 8 ! 0

x50 0.0002 x and C

12 x 150,000

Key numbers: t

R C

Test intervals:

50 x 0.0002 x 2 12 x 150,000 P t 1,650,000 0.0002 x 2 38 x 150,000 t 1,650,000 100,000

The solution set is >90,000, 100,000@ or 90,000 d x d 100,000. The price per unit is R p 50 0.0002 x. x 90,000, p $32. For x 100,000, For x p $30. So, for 90,000 d x d 100,000, $30 d p d $32. 173. s

16t v0t s0

16t 160t 2

175.

1 R

1 1 R1 2

2 R1

2 R RR1

2 R1

R 2 R1

2 R1 2 R1

R

Because R t 1, 2 R1 t1 2 R1 2 R1 1t 0 2 R1 R1 2 t 0. 2 R1

(a) 16t 2 160t

0

16t t 10

0

t

10

0, t

2, 4 2 2 ,

and 4 2 2 seconds t d 8 seconds

Test intervals: 0, 90,000 , 90,000, 100,000 , 100,000, f

2

4 2 2

Solution set: 0 seconds d t 4 2 2 seconds

0.0002 x 2 38 x 1,800,000 t 0 90,000 and x

4 2 2, t

f, 4 2 2 , 4 2 4 2 2, f

0.0002 x 2 38 x 150,000

Key numbers: x

35

16t 2 128t

t 8

r 2 2r 0.1 ! 0

Critical Numbers: r

Solving Inequalities

Because R1 ! 0, the only key number is R1 2. The inequality is satisfied when R1 t 2 ohms.

It will be back on the ground in 10 seconds. (b)

16t 2 160t ! 384 16t 2 160t 384 ! 0 16t 2 10t 24 ! 0 t 2 10t 24 0

t

4 t 6 0

Key numbers: t

4, t

6

Test intervals: f, 4 , 4, 6 , 6, f Solution set: 4 seconds t 6 seconds

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE

Chapter P

36

176. (a)

Prerequisites requisites

4

d

6

8

10

(b)

12

2472.1 d 168.5d 2

Load 2223.9 5593.9 10,312 16,378 23,792

14.67 d d 2 3.83 d d

L

Maximum safe load

2000 d 168.5d 2 472.1

25,000

The minimum depth is 3.83 inches.

20,000 15,000 10,000 5,000 d 4

6

8 10 12

Depth of the beam

177. False. If c is negative, then ac t bc. 178. False. If 10 d x d 8, then 10 t x and x t 8. 179. True

The y-values are greater than zero for all values of x. 180. When each side of an inequality is multiplied or divided by a negative number the direction of the inequality symbol must be reversed.

Section P.3 Graphical Representation of Data 1. (a) v horizontal real number line

9.

y

(b) vi vertical real number line

8 6

(c) i point of intersection of vertical axis and horizontal axis

4 2

(d) iv four regions of the coordinate plane (e) iii directed distance from the y-axis

2

4

6

8

−4 −6

(f) ii directed distance from the x-axis 2. Cartesian

x

−6 − 4 −2 −2

10.

y 4

3. Distance Formula

3

4. Midpoint Formula

2 1

5. A: 2, 6 , B : 6, 2 , C : 4, 4 , D : 3, 2 6. A:

32 , 4 ; B : 0, 2 ; C : 3, 52 , D: 6, 0

7.

x 2

3

−2

11. 3, 4

y

12. 4, 8

6 4 2 x

−6 − 4 −2 −2

2

4

6

13. 5, 5 14. 12, 0

−4 −6

15. x ! 0 and y 0 in Quadrant IV.

y

8.

−3 −2 −1 −1

16. x 0 and y 0 in Quadrant III.

4 3

17. x

2 1 x

4 and y ! 0 in Quadrant II.

18. x ! 2 and y

3 in Quadrant I.

INSTRUCTOR S USE ONLY −3 −2 −1 −1 −2

1

2

3

1 y 5 in Quadrant III or IV. 19.

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.3

20. x ! 4 in Quadrant I or IV. 21.

x, y

31. d

is in the second Quadrant means that x, y is

5

22. If x, y is in Quadrant IV, then x, y must be in

x1 y2 y1 2

6 6

12

2

2

2

2

13 units

! 0 means x and y have the same signs.

This occurs in Quadrant I or III.

32. d

x2

x1 y2 y1 2

24. If xy 0, then x and y have opposite signs. This happens in Quadrant II or IV.

0 8

25.

8

y

2

20 5

2

15

2

2

2

64 225

7500

Number of stores

2

37

25 144

Quadrant III.

x, y , xy

x2

3 2

in Quadrant III.

23.

Graphical Representa Representation of Data Represent

7000

289

6500 6000

17 units

5500 5000 4500

33. d

4000

x2

x1 y2 y1 2

2

x 1

2

3

4

5

6

7

5 1

Year (0 ↔ 2000)

26.

6 2

Temperature (in °F)

y

Month, x Temperature, y

1

– 39

2

– 39

40 30 20 10 0 − 10 − 20 − 30 − 40

5

6

8

10 12

34. d

x2

x1 y2 y1 2

4

–5

3 1 2

5

17

2

6

27

4 25

7

35

8

32

9

22

10

8

11

– 23

12

– 34

28. d

18

29. d

2 3

30. d

4 6

2

61 units

x 2

– 29

5 3

2

36 25

3

27. d

1 4

2

Month (1 ↔ January)

2 3

5

2

2

2

2

29 units

35. d

x2

x1 y2 y1 2

2

1· 4· § § ¨ 2 ¸ ¨ 1 ¸ 2¹ 3¹ © © 2

§ 3· § 7· ¨ ¸ ¨ ¸ 2 © ¹ © 3¹

2

277 36

7

277 units 6

5

10

2

9 49 4 9

8

7

2

10

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

38

NOT FOR SALE

Chapter P

36. d

x2

Prerequisites requisites

x1 y2 y1 2

41. (a) The distance between 1, 1 and 9, 1 is 10.

2

2

§ § 2 ·· §5 · ¨ 1 ¨ ¸ ¸ ¨ 3¸ 3 4 © ¹ © ¹ © ¹ 2

§ 1· § 7· ¨ ¸ ¨ ¸ 3 © ¹ © 4¹

The distance between 9, 1 and 9, 4 is 3.

2

The distance between 1, 1 and 9, 4 is

2

9 1 (b) 102 32

1 49 9 16

4 2

x1 y2 y1 2

12.5 4.2 8.3 2

100 9

109

1 5

Distance

457 units 12

x2

109

2

109.

2

42. (a) 1, 5 , 5, 2

457 144

37. d

4 1

2

1.7

7

2

2

16 49

65

1, 5 , 1, 2

2

4.8 3.1

2

5 2

2

5 2

Distance

2

5 2

7

7

1, 2 , 5, 2

2

15

Distance

68.89 2.89

(b) 42 7 2

71.78

4

16 49

4

65

65

2

| 8.47 units

38. d

x2

x1 y2 y1

3.9

2

2

9.5 8.2 2.6 2

13.4 10.8 2

2

2

43. d1

4

2 0 1

2

41

d2

4

1 0 5

2

25 25

d3

2

1 1 5

2

2

2

5 2

179.56 116.64 296.2

2

3 1

44. d1

| 17.21 units

45

d2

5 3

The distance between 4, 2 and 4, 5 is 3.

d3

5 1

The distance between 0, 2 and 4, 5 is

39. (a) The distance between 0, 2 and 4, 2 is 4.

4 0 2 (b) 42 32

5 2

2

16 9

16 9

25

52

25

40. (a) 1, 0 , 13, 5

13 1 5 0 2

Distance

122 52

2

169

13, 5 , 13, 0 Distance

50

1 13

2

5

5

12

12

20

2

2

40

16 4 4 16

1 3

2

45

2

5 3

1 5 2

9 36

50

2

20 20

36 4

40

2

5.

45. d1

1 3 2

3 2

d2

3 2 2

2 4

d3

1 2 2

3 4

4 2 2

9 3

d1

2

4 25

2

25 4 2

9 49

29 29 58

d2

13 46. d1

1, 0 , 13, 0 Distance

20

2

2

50

2

5

d2

2 4 2

d3

2 2

d1

2

7 9 2

4 36 2

3 7

36 4 2

16 16

40 40 32

d2

INSTRUCTOR USE ONLY (b) 52 122 (b

25 144

169

132

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.3

47. (a)

Graphical Representa Representation of Data Represent

51. (a)

y

39

y

12

5

10

(5, 4)

4

(9, 7)

8

3

6

(− 1, 2)

4 2

(1, 1) −2

2

4

9

(b) d

−1

x 6

8

10

1 7 1 2

§ 9 1 7 1· , (c) ¨ ¸ 2 ¹ © 2

2

64 36

10

5, 4

12

(b) d

(1, 12)

52. (a)

3

4

5

5 1 2

4 2

36 4

2 10

2

2, 3

y

(2, 10)

10 10

8

8

6

6

4 2 −2

2

§ 1 5 2 4 · , (c) ¨ ¸ 2 ¹ © 2

y

48. (a)

x 1

−1

4

(6, 0) 2

4

6

x

8

2

10

(10, 2) x

1 6

(b) d

12 0

2

§ 1 6 12 0 · (c) ¨ , ¸ 2 ¹ © 2 49. (a)

2

25 144

2

4

13

2

(b) d

§7 · ¨ , 6¸ ©2 ¹

6

8

10

10 10 2 2

64 64

8 2

§ 2 10 10 2 · , (c) ¨ ¸ 2 ¹ © 2

y

(− 4, 10)

2

6, 6

10 8

53. (a)

6

y 5 2

2 x

− 8 −6 − 4 − 2

4

−4

4

(− 25 , 34 )

8

3 2

(4, − 5)

−6

(b) d

6

2

1 2

4 5 10 2

64 225

2

5 2

3

1

(b) d

5· 4· §1 § ¨ ¸ ¨1 ¸ 2 2 3¹ © ¹ © 9

(2, 8) 8

2

(−7, − 4)

(b) d

−2

1 9

2

82 3

§ 5 2 1 2 4 3 1 · (c) ¨ , ¸ 2 2 © ¹

6

−6

1 2

−2 − 2 −1 − 2

2

§ 5· ¨ 0, ¸ © 2¹

y

50. (a)

x −

17

§ 4 4 5 10 · (c) ¨ , ¸ 2 © 2 ¹

−10 −8

( 21 , 1)

§ ¨ 1, ©

7· ¸ 6¹

x 2

−2 −4

7

2 4 8 2

81 144

§ 7 2 4 8 · , (c) ¨ ¸ 2 ¹ © 2

2

15

§ 5 · ¨ , 2¸ © 2 ¹

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© Cengage Learning. All Rights Reserved.

40

Chapter P

Prerequisites requisites

54. (a)

42

y

58. d x − 63

− 26

− 16

2

2020

− 26

2 505 | 45

− 36

(− 61 , − 21)

2

242 382

− 16

(− 31 , − 31 )

18 50 12

The pass is about 45 yards. 2

§ 1 1· § 1 1· ¨ ¸ ¨ ¸ © 3 6¹ © 3 2¹

(b) d

1 1 36 36

59. midpoint

2 6

§ 1 § 1· 1 § 1·· ¨ ¨ ¸ ¨ ¸ ¸ 3 © 6¹ 3 © 2¹¸ (c) ¨ , 2 2 © ¹ 55. (a)

2

§ 2003 2007 4174 4656 · , ¨ ¸ 2 2 © ¹

5· § 1 ¨ , ¸ © 4 12 ¹

2005, 4415 In 2005, the sales for Big Lots were about $4415 million. 60. midpoint

y

(− 3.7, 1.8)

(6.2, 5.4)

4

2005, 3521.50

2 −4

In 2005, the sales for the Dollar Tree were about $3521.50 million.

x

−2

2

4

6

−2

(b) d

6.2

3.7 5.4 1.8 2

98.01 12.96

56. (a)

61. 2 2, 4 5

0, 1 2 2, 3 5 4, 2 1 2, 1 5 1, 4

2

110.97

§ 6.2 3.7 5.4 1.8 · , (c) ¨ ¸ 2 2 © ¹

1.25, 3.6

62. 3 6, 6 3

5 6, 3 3 3 6, 0 3 1 6, 3 3

y 20

(−16.8, 12.3)

15 10 5 x

−5

5 −5

(b) d

16.8 5.6

2

12.3 4.9

501.76 54.76

§ 16.8 5.6 12.3 4.9 · (c) ¨ , ¸ 2 2 © ¹ 57. d

3, 3 1, 0 3, 3 5, 0

63. 7 4, 2 8

(5.6, 4.9) −20 −15 −10

§ x1 x2 y1 y2 · , ¨ ¸ 2 ¹ © 2 § 2003 2007 2800 4243 · , ¨ ¸ 2 2 © ¹

8 6

§ x1 x2 y1 y2 · , ¨ ¸ 2 ¹ © 2

2

556.52

5.6, 8.6

1202 1502 36,900

2 2 7

3, 6 4, 2 8 2, 10 4, 4 8 2, 4 4, 4 8 3, 4

64. 5 10, 8 6

3 10, 6 6 7 10, 6 6 5 10, 2 6

5, 2 7, 0 3, 0 5, 4

30 41 | 192.09

The plane flies about 192 kilometers.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.3

65. To reflect the vertices in the y-axis, negate each xcoordinate.

Original Point

1, 5

5, 4

5, 4

2, 2

2, 2

Original Point

66. Negate each x-coordinate.

Original Point

4, 5

2, 3

2, 3

5, 1

5, 1

69.

Reflected Point

0, 3

0, 3

3, 2

3, 2

6, 3

6, 3

3, 8

3, 8

68. Negate each x-coordinate.

Reflected Point

4, 5

41

67. Negate each x-coordinate.

Reflected Point

1, 5

Graphical Representa Representation of Data Represent

Original Point

Reflected Point

7, 1

7, 1

5, 4

5, 4

1, 4

1, 4

3, 1

3, 1

y 6

(−2, 3) 4 d1

(2, 1) x

−6 −4 −2

2

d3

4

6

d2

(−1, −5)

−6

d1

2 2

2

d2

1 2

5 1

d3

2 1

2

Since d12 d 2 2 70.

1 3

2

4

2

3

3 5

2

20 45

65

2

2

2

2

2

6

1

2

20 2

2 5

45

8

2

3 5 65

d32 , the triangle is a right triangle.

y

(4, −1)

2 −2 −2

x 2

6

d1

−4 −6

(2, −5)

d3

8

d2

10

(10, −4)

−8 −10

5 1

d1

2 4 2

d2

10

4 4 1

d3

10

2 4 5

2

Since d12 d 2 2

2

20 45

71. On the x-axis, y

0

On the y-axis, x

0

2

2

2

65

2 2

4

6

2

3

8

2

1

2

2

2

20

2 5

45

3 5

65

d32 , the triangle is a right triangle.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

42

NOT FOR SALE

Chapter P

Prerequisites requisites

y

72.

(a) The point is reflected through the y-axis.

8 6

(−3, 5)

(−2, 1) 2 −8 −6 − 4 −2

(−7, −3)

(b) The point is reflected through the x-axis.

(3, 5)

4

(c) The point is reflected through the origin.

(2, 1) x 2

4

−4 −6

6

8

(7, −3)

−8

73. The highest price of milk is approximately $3.87. This occurred in 2007. 74. Price of milk in 1996 | $2.73 Highest price of milk Percent change

$3.87 in 2007

3.87 2.73 | 41.8% 2.73

75. (a) Cost during Super Bowl XXXVIII 2004 | $2,302,000

Cost during Super Bowl XXXIV 2000 | $2,100,000 Increase

$2,302,000 $2,100,000

Percent increase

$202,000

$202,000 | 0.096 or 9.6% $2,100,000

(b) Cost during Super Bowl XLII 2008

$2,700,000

Cost during Super Bowl XXXIV 2000 Increase

$2,700,000 $2,100,000

Percent increase

Cost during 1996 awards

$795,000

(b) Minimum wage in 1990: $3.80

$495,000 | 0.623 or 62.3% $795,000

Minimum wage in 1995: $4.25

Cost during 1996 awards

$1,700,000

§ $4.25 $3.80 · Percent increase: ¨ ¸100 | 11.8% $3.80 © ¹

$795,000

Minimum wage in 1995: $4.25

$1,700,000 $795,000

Percent increase

78. (a) The minimum wage had the greatest increase in the 2000s.

$495,000

(b) Cost during 2007 awards Increase

$1,290,000

$1,290,000 $795,000

Percent increase

$600,000

$600,000 | 0.286 or 28.6% $2,100,000

76. (a) Cost during 2002 awards

Increase

$2,100,000

$905,000

$905,000 | 1.138 or 113.8% $795,000

77. The number of performers elected each year seems to be nearly steady except for the middle years. Five performers will be elected in 2010.

Minimum wage in 2009: $7.25 § $7.25 $4.25 · Percent increase: ¨ ¸100 | 70.6% $4.25 © ¹

(c) $7.25 0.706$7.25 | $12.37

The minimum wage will be approximately $12.37 in the year 2013. (d) Answers will vary. Sample answer: No, the prediction is too high because it is likely that the percent increase over a 4-year period (2009 –2013) will be less than the percent increase over a 14-year period (1995 –2009).

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.3

§ x1 x2 y1 y2 · , ¨ ¸ 2 ¹ © 2

81. (a) Pieces of mail (in billions)

§ 1999 2007 19,805 28,857 · , ¨ ¸ 2 2 © ¹

2003, 24,331 In 2003, the sales for the Coca-Cola Company were about $24,331 million. y

22

53

29

74

35

57

44

79

48

90

53

76

58

93

65

83

76

99

205 200 195 190 185 180 x 8

10 12 14 16 18

(b) The greatest decrease occurred in 2008.

100 90 80 70 60 50 40 30 20 10

(c) Answers will vary. Sample answer: Technology now enables us to transport information in ways other than by mail. The Internet is one example. 82. (a) 10 20 30 40 50 60 70 80

66

210

Year (6 ↔ 1996)

x

40

215

6

y

x

Final exam score

80. (a)

43

y

Math entrance test score

Number of basketball teams

79. midpoint

Graphical Representa Representation of Data Represent

y

Men Women

1050 1000 950 900 850

x 4

6

8 10 12 14 16

Year (4 ↔ 1994)

(b) In 1994, the number of men's and women's teams were nearly equal. (c) In 2005, the difference between the number of men's and women's teams was the greatest: 1036 983 53 teams.

(b) The point 65, 83 represents an entrance exam score of 65. (c) No. There are many variables that will affect the final exam score. x1 x2 and ym 2

83. Because xm

2 xm 2 xm x1

x1 x2

y1 y2 we have: 2 y1 y2

2 ym 2 ym y1

x2

y2

So, x2 , y2

2 xm

x1 , 2 ym y1 .

84. (a)

x2 , y2

2 xm

x1 , 2 ym y1

2 4 1, 21 2

7, 0

(b)

x2 , y2

2 xm

x1 , 2 ym y1

2 2 5 , 2 4 11

9, 3

§ x x2 y1 y2 · 85. The midpoint of the given line segment is ¨ 1 , ¸. 2 ¹ © 2 x x2 y y2 · § § x x2 y1 y2 · x1 1 y1 1 ¸ , The midpoint between x1 , y1 and ¨ 1 2 , 2 ¸ is ¨¨ ¸ 2 ¹ © 2 © ¹ 2 2 y y2 § x1 x2 · § x x2 y1 y2 · x2 1 y2 ¸ ¨ The midpoint between ¨ 1 , x , y is and 2 2 2 2 ¸ , ¨ ¸ 2 ¹ © 2 © ¹ 2 2

§ 3x1 x2 3 y1 y2 · , ¨ ¸. 4 4 © ¹ § x1 3 x2 y1 3 y2 · , ¨ ¸. 4 4 © ¹

So, the three points are § 3x1 x2 3 y1 y2 · § x1 x2 y1 y2 · § x1 3x2 y1 3 y2 · , , , ¨ ¸, ¨ ¸, and ¨ ¸. 4 4 2 ¹ 4 4 © ¹ © 2 © ¹

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44

NOT FOR SALE

Chapter P

Prerequisites requisites

§ 3 1 4 3 2 1 · , ¨ ¸ 4 4 © ¹

§ 3x x2 3 y1 y2 · , 86. (a) ¨ 1 ¸ 4 4 © ¹ § x1 x2 y1 y2 · , ¨ ¸ 2 ¹ © 2

§ 1 4 2 , ¨ 2 © 2 § x1 3 x2 y1 3 y2 · § 1 3 4 , , ¨ ¸ ¨ 4 4 4 © ¹ ©

§ 5 3· ¨ , ¸ © 2 2¹ 3 1 · § 13 5 · ¸ ¨ , ¸ 4 4¹ ©4 ¹

§ 3 2 0 3 3 0 · , ¨ ¸ 4 4 © ¹

§ 3x x2 3 y1 y2 · (b) ¨ 1 , ¸ 4 4 © ¹ § x1 ¨ © § x1 ¨ ©

1· ¸ ¹ 2

§7 7· ¨ , ¸ © 4 4¹

§ 3 9· ¨ , ¸ © 2 4¹

x2 y1 y2 · 3· § 2 0 3 0 · § , , ¸ ¨ ¸ ¨ 1, ¸ 2 2 ¹ 2 2 2¹ © ¹ © 3 x2 y1 3 y2 · § 2 0 3 0 · § 1 3 · , , ¸ ¨ ¸ ¨ , ¸ 4 4 4 ¹ © 2 4¹ ¹ © 4

87. No. It depends on the magnitude of the quantities measured. 88. (a)

First Set d A, B

2 2

2

3 6

2

d B, C

2 6 2

6 3

2

d A, C

2 6

3 3

Because 32 42 of a right triangle.

2

2

89. False, you would have to use the Midpoint Formula 15 times. 9

3

16 9 16

90. True. Two sides of the triangle have lengths 5

4

52 , A, B, and C are the vertices

d A, B

8 5 2

3 2

d B, C

5 2 2

2 1

d A, C

8 2

3 1

2

2

2

10

§a b c· , ¸ ¨ 2¹ © 2

§a b 0 c 0· , ¨ ¸ 2 2 ¹ ©

§a b c· , ¸ ¨ 2¹ © 2

10

92. (a) Because x0 , y0 lies in Quadrant II, x0 , y0

40

must lie in Quadrant III. Matches (ii).

A, B, and C are the vertices of an isosceles triangle or are collinear: 10 10 2 10 40. (b)

18.

91. Use the Midpoint Formula to prove the diagonals of the parallelogram bisect each other.

§b a c 0· , ¨ ¸ 2 ¹ © 2

Second Set

2

the third side has a length of

149 and

y

(b) Because x0 , y0 lies in Quadrant II, 2 x0 , y0 must lie in Quadrant I. Matches (iii).

(c) Because x0 , y0 lies in Quadrant II, x0 , 12 y0

must lie in Quadrant II. Matches (iv).

8

(d) Because x0 , y0 lies in Quadrant II, x0 , y0

6 4

must lie in Quadrant IV. Matches (i).

2 x

−2

2

4

6

8

−2

First set: Not collinear Second set: The points are collinear. (c) If A, B, and C are collinear, then two of the distances will add up to the third distance.

Section P.4 Graphs of Equations 1. solution or solution point

4. y-axis

2. graph

5. circle; h, k ; r

INSTRUCTOR USE ONLY 3. intercepts

6. numerical

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.4

7. y

(a)

x 4

0, 2 :

11. (a) ?

2

Yes, the point is on the graph. (b)

5, 3 :

?

2 1 2

3

1 2

3

3

(b)

1, 0 :

?

?

9

3

0

?

2 2

0 z 4

3

No, the point is not on the graph.

Yes, the point is on the graph. 8. (a) 1, 2 : 2

1 1 2

0

?

3

4

No, the point is not on the graph.

2

(b)

5, 0 :

55

0 0

0

(a)

2, 0 : 2

46 2

2, 8 : 2

(b)

5

2

16 4

8

20

14. y

(a) 4 1 2

1 x3 3

6, 0 :

0

2, 163 :

1 3

2 3

?

2 2

8 3 8 3

4 6 2 4 4

0

0

Yes, the point is on the graph.

20 ?

20 20

?

2

16 3

?

8 24

1 3

4 1

0

?

2x2

No, the point is not on the graph. (b)

20

Yes, the point is on the graph.

5 z 3 ?

2

8 ?

4 x 2

?

4, 2 : 4 2

?

No, the point is not on the graph.

(a) 1, 5 : 5

20 ?

No, the point is not on the graph.

0

3 2 2

?

13 z 20

0

12 z 8

?

2

9 4

0

4 6 2

10. y

2

?

Yes, the point is on the graph. (b)

3, 2 : 3 2

?

0

0

Yes, the point is on the graph. 13. (a)

3 2 2

2

0

0

x 2 3x 2 2

0 ?

213

Yes, the point is on the graph. 9. y

?

(b) 1, 1 : 21 1 3

Yes, the point is on the graph. ?

0

3 z 0

?

2

?

12. (a) 1, 2 : 21 2 3

51

2

45

Yes, the point is on the graph.

5 4

3

?

3 ?

0 4

2 2

2, 3 :

Graphs of Equations

16 3

?

8

16 3

?

24 3 16 3

16 3 16 3

Yes, the point is on the graph. (b)

3, 9 : 13 3 3 1 3

2 3

27

2

29

9 18

?

9 ?

9 ?

9

27 z 9

INSTRUCTOR USE ONLY No, the point is not on the graph.

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© Cengage Learning. All Rights Reserved.

46

NOT FOR SALE

Chapter P

15. y

Prerequisites requisites

2 x 5

18. y

5 x2

x

1

0

1

2

5 2

x

2

–1

0

1

2

y

7

5

3

1

0

1

4

5

4

1

(x, y)

0, 5

1, 3

2, 1

0

y

1, 7

x, y

2, 1

1, 4

0, 5

1, 4

2, 1

5 , 2

y

y

7

6

5

4

4

3

3

2

2

1

1

x –4 –3

x

−3 −2 −1 −1

1

2

4

5

1

3 x 4

x

2

0

1

y

52

–1

14

19. y

2, 52

1

3

x

3

x

2

x-intercept: 0

0

1 2

0

x 3

3

x

y

3 2 1 x 3

2

3, 0 0 3 2

y-intercept: y

2

3

4 3

1, 14 43 , 0 2, 12

0, 1

2

4

–4 –3 –2 –1

4

–2

16. y

x, y

–1

y

3

y

9

2

0, 9

4

–2 –3 –4

17. y

20. y

16 4 x 2 0

16 4 x 2

4x

2

16

x

2

4

x

r2

x-intercepts:

x 2 3x

x

1

0

1

2

3

y

4

0

–2

–2

0

(x, y)

1, 4 0, 0 1, 2 2, 2 3, 0

2, 0 , 2, 0 y-intercept: y

y

16 40

2

16

0, 16

5 4 3

21. y −2 −1

x −1 −2

1

2

4

5

x 2 x 2

x-intercept: 0 0

x 2

x

2

2, 0 0 2

y-intercept: y y

2

y

2

0, 2

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.4

22. y 2

4 x

27. y 4 x

x-intercept: 0 x

x-intercept: 0

3x 7

0

3x 7

7 3

0

4, 0

73 , 0

40

y-intercepts: y y

2

4

y

r2

28. y

5x 6

6

5x

6 5

x

10

8

x

8 3

29. y

2 x3 4 x 2

8 30

8

2 x 2 x 2

x

0

or

x

2

20 40 3

y-intercept: y y

2

0

0, 0

x 4 x 4

x-intercept: 0

30. y

x 4

0 4

x

x 4 25

x4

25 r 4 52

x

0 4

y-intercept: y

x 4 25

x-intercept: 0

4, 0

2

0, 2

r 5

5, 0 , 5, 0

0

y-intercept: y

4

25

25

0, 25

2x 1 2x 1

0 2x 1

0

x

31. y 2

6 x 6 x

x-intercept: 0 x

12 , 0 y-intercept: y

0

0, 0 , 2, 0

0, 8

x-intercept:

2 x3 4 x 2

x-intercepts: 0

0

26. y

10

0, 10

8 3x

3x

0 10

y-intercept: y

6

8 , 3

25. y

10

10, 0

8 3x

y-intercept: y

0

x

50 6

x-intercept: 0

x 10

0 x 10

0, 6 24. y

x 10

x-intercept:

56 , 0 y-intercept: y

7

0, 7

5x 6

x-intercept: 0

30 7

y-intercept: y

0, 2 , 0, 2 23. y

47

3x 7

4

2

Graphs of Equations

6

6, 0 20 1 1

There is no real solution.

y-intercepts: y 2

6 0

y

r 6

INSTRUCTOR USE ONLY There is no y-intercept. y-intercept.

0,

6 , 0,

6

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© Cengage Learning. All Rights Reserved.

48

NOT FOR SALE

Chapter P

Prerequisites requisites

x 1

32. y 2

x 1

x-intercept: 0

1

x

1, 0 01

y-intercepts: y 2

r1

y

0, 1 , 0, 1 33. x 2 y

x 2

0

y

0 x2 y

0 y -axis symmetry

x y

0 x y

0 No x-axis symmetry

2

x

2

2

y

x y2

34.

x y

0 x2 y

0 No origin symmetry

0

2

0

x y2

0

x-axis symmetry 35. y

y

x3

x 3

y

y

x3 y

y

x

36. y

y

3

x3 No y -axis symmetry x3 No x-axis symmetry

y

x3 y

x4 x2 3

x

4

x 3 y 2

y

x4 x2 3 y

y

x

37. y

y y y

38. y

y y y

x3 Origin symmetry

4

x 4 x 2 3 No x-axis symmetry

x 3 y 2

x x2 1 x

x 1 2

y

x 4 x 2 3 y -axis symmetry

x 4 x 2 3 No origin symmetry

x No y -axis symmetry x2 1

x x y No x-axis symmetry 2 x 1 x 1 x x x y y Origin symmetry 2 x2 1 x2 1 x 1 2

1 1 x2 1 1 x

2

y

1 y -axis symmetry 1 x2

1 1 y No x-axis symmetry 1 x2 1 x2 1 1 y No origin symmetry 2 1 x2 1 x

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.4

39. xy 2 10

x y

0 xy 2 10

x y 10 2

x y 40. xy

49

0

10

2

Graphs of Equations

2

0 No y -axis symmetry

0 xy 10

0 x-axis symmetry

2

10

0 xy 2 10

0 No origin symmetry

4

x y 4 x y 4 x y

x3 3

45. y xy

4 No y -axis symmetry

xy

4 No x-axis symmetry

4 xy

x-intercept:

3, 0

3

y-intercept: 0, 3

4 Origin symmetry

No symmetry 3 x 1

41. y

x-intercept:

y

13 , 0

5

x

–2

–1

0

1

2

y

–5

2

3

4

11

4

y-intercept: 0, 1

( (

1 1 (0, 1) ,0 3

y

x

−4 −3 −2 −1 −1

No symmetry

1

2

3

4

7

−2

6

−3

5 4

2x 3

42. y

x-intercept:

y

32 , 0

2

( 3 −3, 0)

2

−4 −3 −2

( 32 , 0 )

1

(0, 3)

1 x 1

−1

2

3

4

x

y-intercept: 0, 3

–3

–2

–1 –1

2

3

–3

x3 1

46. y

–2

No symmetry

y 2

x-intercept: 1, 0

(0, −3)

1

(1, 0)

y-intercept: 0, 1 x2 2x

43. y

–3

–1

2

x 3

(0, −1) –2

No symmetry

x-intercepts: 0, 0 , 2, 0

–2

–3 –4

y-intercept: 0, 0 x 3

47. y

No symmetry x

–1

0

1

2

3

y

3

0

–1

0

3

x-intercept: 3, 0 y-intercept: none No symmetry

y

x

3

4

7

12

y

0

1

2

3

4 3 2

(0, 0) −2

−1

y

(2, 0) 1

−1

2

3

x

5

4

4

−2

3 2

44. y

x 2x 2

x-intercept: 2, 0 , 0, 0 y-intercept: 0, 0 No symmetry

1

y

(−2, 0) −5 − 4 −3

1

2

3

4

5

6

–1

(0, 0) −1 −1

(3, 0) x

2 x 1

2

3

−2 −3 −4 −5

INSTRUCTOR T USE ONLY −6

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50

NOT FOR SALE

Chapter P

Prerequisites requisites

y

1 x

48. y

x-intercept: 1, 0

5

y-intercept: 0, 1

3

52. x

x-intercept: 5, 0

4

(0, 1)

No symmetry

(1, 0) –2 –1

1

5

x-axis symmetry

x

2

–1

y 4

x 6

49. y

5 , 0,

y-intercepts: 0,

2

–4 –3

y2 5

3

x-intercept: 6, 0

(−5, 0) −6

y-intercept: 0, 6

( 0,

5)

1 x

−3 −2 −1

1

( 0, −

2

5)

−3

No symmetry

−4

x

–2

0

2

4

6

8

10

y

8

6

4

2

0

2

4

53. y

3

1x 2 10

y −10

12

10

10 8 −10

6 (0, 6) 4

−2

Intercepts: 6, 0 , 0, 3

(6, 0)

2

x −2

2

4

6

8

10 12

54. y

1 x

50. y

y

−3

−2

1

−10

(0, 1) (1, 0)

10

x −1

1

2

3 −10

−2 −3

y2 1

51. x

10

2

(−1, 0)

y-axis symmetry

1

3

x-intercepts: 1, 0 , 1, 0 y-intercept: 0, 1

2x 3

Intercepts: 0, 1 , 55. y

x-intercept: 1, 0

x2 4 x 3 10

y-intercepts: 0, 1 , 0, 1 −10

x-axis symmetry x

–1

0

3

y

0

±1

±2

32 , 0

10

−10

Intercepts: 3, 0 , 1, 0 , 0, 3

y

56. y

3

x2 x 2

2 10

(0, 1)

(−1, 0)

x –2

1

(0, −1)

2

3

4 −10

10

–2 –3 −10

Intercepts: 2, 0 , 1, 0 , 0, 2

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.4

57. y

2x x 1

62. y

6

x

Graphs of Equations

51

x

10

10

−10

−10

10

10

−10

−10

Intercepts: 0, 0 , 6, 0

Intercept: 0, 0 58. y

x 3

63. y

4 x2 1

10

10 −10 −10

10

10 −10

Intercepts: 3, 0 , 0, 3

−10

Intercept: 0, 4 59. y

3

64. y

2 x

x 2

10

10 −10

− 10

10

10 −10

Intercepts: r2, 0 , 0, 2

−10

Intercepts: 8, 0 , 0, 2 60. y

3

65. Center: 0, 0 ; Radius: 4

x 1

x

0 y 0 2

10

−10

10

66.

x

2

42

x2 y2

16

0 y 0 2

2

52

x2 y2

25

−10

Intercepts: 1, 0 , 0, 1 61. y

x

67. Center: 2, 1 ; Radius: 4

x

x 6

2 y 1

2

x

2

2

2 y 1 2

42 16

10

68.

−10

10

−10

Intercepts: 0, 0 , 6, 0

x 7 x

y 4

2

7 y 4

2

2

2

72 49

69. Center: 1, 2 ; Solution point: 0, 0

x 1

2

y 2

0

1 0 2

x

1 y 2

2

2

2 2

2

r2 r2 5

r2

5

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© Cengage Learning. All Rights Reserved.

52

NOT FOR SALE

Chapter P

Prerequisites requisites

3 1

70. r

42 3

x

2

2 1

2

25 2

x 3 y 2

2

2

2

52

1 x

−3 −2

25

§ 0 6 0 8· Center: ¨ , ¸ 2 ¹ © 2

−3

3 0 4

2

r 2 25

x

3 y 4

2

8 2

−5

76. x 2 y 1

0

4 4 2

2

1

Center: 0, 1 , Radius: 1

r2

y

25 3

1 1

2

2

(0, 1)

1

2

x –2

–1

1

2

–1

64 4

68

Center: 12 , 12 ,

77. x

§1· ¨ ¸ 2 17 © 2¹

17

1 2

Midpoint of diameter (center of circle):

x

5

−7

r2

1 2 1 2 1 2 1 2

4

−6

2

72. r

2

(1, −3)

−4

3, 4

3 y 4

2

1

−1 −2

x

2

9

Center: 1, 3 , Radius: 3

5

71. Endpoints of a diameter: 0, 0 , 6, 8

2

2

y

3 y 2 2

75. x 1 y 3

2

§ 4 4 1 1 · , ¨ ¸ 2 ¹ © 2

0, 0

0 y 0

2

2

x2 y2

73. x 2 y 2

17

2

y

1 2

2

9 4

Radius:

y 3

2 1

( 12 , 12 ) x

17

–1

1

2

3

25

Center: 0, 0 , Radius: 5

78.

x

2 y 3 2

2

16 9

Center: 2, 3 , Radius:

y 6

1

(0, 0) x

−3 −2 −1 −2 −3 −4

1 2 3 4

6

x −1

1

2

−4

16

Center: 0, 0 , Radius: 4

79. y-axis symmetry

y

y

5

−2 −3

4

(2, −3)

−3

74. x 2 y 2

−3 −2 −1

3

−1 −2

−6

3 2 1

4 3

y

4 3 2 1

−5

3 2

4 3 2

(0, 0)

1 x

1 2 3

5

x –4 –3

–1

1

3

4

–2

INSTRUCTOR S USE S ONLY −5 − 5

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.4

80.

Graphs of Equations

53

85. (a)

y 4

y

3 2 1

x

x –1

1

2

3

6

7

8

–2

(b) 2 x 2 y

1040 3

2y

1040 3

–3 –4

81. Origin symmetry y

y

520 3

A

xy

4 3

(c)

2

2x x x

5203 x

8000

1 x –4 –3 –2

1

2

3

4

–2 –3

0

82.

180 0

–4

(d) When x

y

86 23 yards, the area is a maximum

of 751119 square yards.

4 3 2

(e) A regulation NFL playing field is 120 yards long and 53 13 yards wide. The actual area is 6400 square

1

x – 4 – 3 – 2 –1

y

1

2

3

4

yards.

–2 –3 –4

86. (a)

83. Answers will vary.

y

One possible equation with x-intercepts at x 2, x 4, and x 6 is: y

x x

x

2 x 4 x 6 2 x 2 10 x 24

x3 10 x 2 24 x 2 x 2 20 x 48

(b) P 360 meters so: 2x 2 y 360 w y 180 x A

x3 8 x 2 4 x 48 Any non-zero multiple of the right side of this equation, y k x3 8 x 2 4 x 48 , would also have these

(c)

lw

x180 x

9000

x-intercepts. 0

84. Answers will vary. One possible equation with x-intercepts at 3 x 52 , x 2, and x is: 2 y

2 x 2 x

180 0

5 x 2 2 x 3 5 2 x 2 7 x 6

4 x3 14 x 2 12 x 10 x 2 35 x 30 4 x3 4 x 2 23x 30

90 (d) x 90 and y A square will give the maximum area of 8100 square meters. (e) Answers will vary. Sample answer: The dimensions of a Major League Soccer field can vary between 110 and 120 yards in length and between 70 and 80 yards in width. A field of length 115 yards and width 75 yards would have an area of 8625 square yards.

Any non-zero multiple of the right side of this equation, y k 4 x3 4 x 2 23 x 30 , would also have these x-intercepts.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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54

Chapter P

87. (a)

NOT FOR SALE

Prerequisites requisites

88. (a)

100

The model fits the data very well.

x

5

10

20

30

40

y

430.43

107.33

26.56

11.60

6.36

x

50

60

70

80

90

100

y

3.94

2.62

1.83

1.31

0.96

0.71

Answers will vary. 0

100 0

(b) graphically: 75.4 years algebraically: 2

75.66 years

y

(c) 76

0.0025t 2 0.574t 44.25

0

0.0025t 2 0.574t 31.75

0.574 r

t

0.574 40.0025 31.75 2 0.0025 2

0.574 r 0.011976 0.005 t | 92.9, or about 93 years

115

0.0025115 0.574115 44.25 2

y

x 20

40

60

80 100

Diameter of wire (in mils)

(c) When x

In the year 1900 93 1993 the life expectancy was approximately 76 years. 2015 o t

y 450 400 350 300 250 200 150 100 50

When x

t

(d) Year

(b) Resistance (in ohms)

0.002590 0.57490 44.25

y

y

85.5, the resistance is 1.1 ohms. 85.5,

10,770 0.37 85.52

1.10327.

(d) As the diameter of the copper wire increases, the resistance decreases.

y | 77.2

The model predicts the life expectancy to be 77.2 years in 2015. (e) Answers will vary. Sample answer: In 50 years the tvalue will exceed 100 (since 1920), so the model will no longer provide accurate predictions. 89. y

ax 2 bx3

a x b x 2

(a) y

3

ax 2 bx3 To be symmetric with respect to the y-axis; a can be any non-zero real number, b must be zero. (b) y

y y

a x b x 2

3

ax 2 bx3 ax 2 bx3

To be symmetric with respect to the origin; a must be zero, b can be any non-zero real number. 90. (a) (v)

(b) (i) (c) None (d) (ii), (vi) (e) (i), (iv) (f) (i), (iv) 91. Assuming that the graph does not go beyond the vertical limits of the display, you will see the graph for the larger values of x.

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© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.5

Linear Equations in Two Tw T Variables

55

Section P.5 Linear Equations in Two Variables 14. The line appears to go through 0, 7 and 7, 0 .

1. linear 2. slope

y2 y1 x2 x1

Slope

3. parallel 15. y

4. perpendicular

5

y-intercept: 0, 3

6. linear extrapolation

y 5

7. general

4

0

(iii) general form

(b) x

a

(i) vertical line

(c) y

b

(v) horizontal line

(d) y

mx b

(ii) slope-intercept form

m x x1

(e) y y1

(0, 3)

3

8. (a) Ax By C

(iv) point-slope form

x

−4 −3 −2 −1

16. y

1

y

2

3

x 10

Slope: m 9.

1

5x 3

Slope: m 5. rate or rate of change

0 7 7 0

y

1

y-intercept: 0, 10

(2, 3)

2

m= 0

x –2

2

4

6

10 12

–4 –6

2

m= 1

–8

m = −3

1

(0, −10)

–10

m= 2 x 1

10.

2

17. y

m= −3

m=3

y-intercept: 0, 4

4

2

11. Two points on the line: 0, 0 and 4, 6 y2 y1 x2 x1

6 4

32 x 6

Slope

8 2

4 3 1

5 2

4

(0, 6)

5

2 x

−1

19. 5 x 2 x

13. Two points on the line: 0, 8 and 2, 0 y2 y1 x2 x1

6

y-intercept: 0, 6

3 2

5 0 31

y

32

Slope: m

12. The line appears to go through 1, 0 and 3, 5 . y2 y1 x2 x1

1 2 3 4 5 6 7 8

−2

x

−2

18. y

Slope

x

−1

−2

Slope

(0, 4)

3 2 1

m = 21

(−4, 1) −6

y 7 6 5

12

Slope: m

y

undefined

12 x 4

1

2

4

5

6

7

y

0 2, 5

3

2

vertical line

1

Slope: undefined No y-intercept

x –1

1

2

3

–1 –2

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56

Chapter P

20. 3 y 5

Prerequisites requisites y

0

3y

5

y

53

−2

x

−1

1

2

−1

Slope: m

26. x 2

0

x

2

1

0

y-intercept: 0, 53

No y-intercept

−2

Slope: undefined (vertical line)

(

0, − 35

)

y 4

−3

3 2

21. 7 x 6 y

y

30 76 x 5

y

Slope: m

x

−4 −3 −2 −1 −1

(0, 5)

5

76

1

4

−2

3

−3

2

−4

1

3

4

1

y-intercept: 0, 5

09 60

x

−1

1

3

2

4

6

7

27. m

−2

9 6

3 2

y

22. 2 x 3 y

y

9

(0, 9)

3y

2 x 9

y

23 x 3

6

4

(0, 3)

23

Slope: m

8

5

4

2

2

1

−2 −2

x

y-intercept: 0, 3

–1

1

2

3

(6, 0) 2

y

4

Slope: m

y-intercept: 0, 3

8 12

2

(0, 3)

0

x 10

2 3

y

5

3, horizontal line

8

8 0 0 12

y

0

6

4

28. m

23. y 3

4

(12, 0)

−2 −2

2

2

4

6

8

x

12

−4

1 x

−3 −2 −1 −1

1

2

−6

3 − 10

(0, − 8)

− 12

24. y 4

y

0

4

y Slope: m

2

0

−4 − 3 −2 −1 −1

y-intercept: 0, 4

6 2

29. m

1 1

2

3

2

4 y

−2 −3

8 4

1 3

x

(1, 6)

6

(0, − 4)

5

−5

4

−6 2

25. x 5 x

1

0

x –5 – 4 –3

5

–1

1

2

3

(−3, −2)

Slope: undefined (vertical line)

4 4 4 2

30. m

No y-intercept y

4

y

4

(2, 4)

4

3 2

2

1 x −7 −6

−4 −3 −2

−1 −2

1

x –2

2

6

–2

−3 −4

4

–4

(4, −4)

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NOT FOR SALE Section P.5

7 7

31. m

0 3

85

Linear Equations in Two Tw T Variables 1 3 4 4 5 7 4 8

0

36. Slope

y

1 3 8

57

8 3

4 y

2 x

−4 −2 −2

2

4

6

8

( 78 , 43 )

3 4

10

1 2

−4

(5, − 7)

−6

1 4

(8, −7)

−8

x

−10

32. m

1 4

− 14

3 4

3 1.6 3.1 5.2 4.8

37. m

y 2

(− 2, 1)

1.5 10

0.15

y

1

8

x

−1 −1

1

2

6

−2 −3

(− 4, − 5)

(4.8, 3.1)

4

(−5.2, 1.6)

−4

33. m

1

( 45 , − 41 )

− 12

5 1 4 2

− 6 − 5 − 4 −3

1 2

−5

−6

−6

−4

−2

x 2

4

6

−2 −4

4 1

5 0

6 6

2.6 8.3

38. Slope

1.425

2.25 1.75

m is undefined.

y

y 1

6

x −5 − 4 −3 −2 −1

(−6, 4)

2

(−6, −1) –2

−8 −9

–2

34. m

0 10

4 0

39. (a) m

5 2 x

2

(c) m

4

–2 –4

40. (a) m

(b) m –10

35. m

the slope is positive, the line rises.

(b) m is undefined. The line is vertical. Matches L3.

(− 4, 0) –2

2 . Since 3

Matches L2 .

y

–6

(2.25, − 2.6)

(−1.75, −8.3)

x –8

1 2 3 4 5

−2 −3 −4 −5

4

(0, −10)

1 § 4· ¨ ¸ 3 © 3¹ 3 11 2 2

2. The line falls. Matches L1. 0. The line is horizontal. Matches L 2 . 34 . Because the slope is negative, the line

falls. Matches L1. (c) m 1. Because the slope is positive, the line rises. Matches L 3.

y

1 7

3 2 3 1 − 2, − 3 1

(

)

x −1 −2 −3 −4 −5 −6

41. Point: 2, 1 , Slope: m

0

3 4 5 6

( 112, − 34 )

Because m

0, 1 , 3, 1 ,

0, y does not change. Three points are

and 1, 1 .

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58

Chapter P

NOT FOR SALE

Prerequisites requisites

42. Point: 3, 2 , Slope: m

Because m

51. Point 0, 2 ; m

0

0, y does not change. Three other points

are: 4, 2 , 0, 2 , 5, 2 43. Point: 5, 6 , Slope: m

Because m

y

2

3 x 0

1 x

3x 2

–2

–1

1

2

3

4

–1 –2

(0, −2)

1

1, y increases by 1 for every one unit

increase in x. Three points are 6, 5 , 7, 4 , and

8, 3 . 44. Point: 10, 6 , Slope: m

Because m

y 2

y

3

1

1, y decreases by 1 for every one unit

52. Point 0, 10 ; m

1

y 10

1 x 0

y 10

x

y

y

(0, 10)

10

5

x 10

increase in x. Three other points are: 0, 4 , 9, 5

x 5

11, 7 . 53. Point 3, 6 ; m

45. Point: 8, 1 , Slope is undefined.

Because m is undefined, x does not change. Three points are 8, 0 , 8, 2 , and 8, 3 .

y 6 y

2

y 6

(−3, 6)

2 x 3

4

2 x

x –6

–4

–2

46. Point: 1, 5 , Slope is undefined.

Because m

2, 1

2

2

54. Point 0, 0 ; m

y 0

y increases by 2 for every one

y

y

4

4 x 0

5

4x

3

4

2 1

and 2, 10 .

48. Point: 0, 9 , Slope: m

Because m

6

–6

unit increase in x. Three additional points are 4, 6 ,

3, 8 ,

4

–4

Because m is undefined, x does not change. Three other points are: 1, 3 , 1, 1 , 1, 7 47. Point: 5, 4 , Slope: m

2

−3

2

2, y decreases by 2 for every one unit

increase in x. Three other points are: 2, 5 ,

1, 11 , 3, 15 .

55. Point 4, 0 ; m

y 0 y

−2

13

(0, 0)

−1

x

1

3

2

y 4

13 x 4

3

13 x

1

2

4 3

(4, 0) x

–1

49. Point: 7, 2 , Slope: m

Because m

1, 2

1 2

3

4

–2

y increases by 1 unit for every two

and 13, 1 .

50. Point: 1, 6 , Slope: m

Because m

2

–1

unit increase in x. Three additional points are 9, 1 ,

11, 0 ,

1

12

12 , y decreases by 1 for every 2 unit

56. Point 8, 2 ; m

x

y 2

1 4

y 2

1 x 4

y

1x 4

1 4

8 2

y 8 6 4

(8, 2)

2

x −2

2

4

6

8

10

−4 −6

increase in x. Three other points are: 3, 5 ,

1, 7 , 5, 9 .

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

Section P.5 57. Point 2, 3 ; m

1 2

y 3 y

58. Point 2, 5 ; m

y 5

3 4

4 y 20

x

62. Point 12 ,

y 4 3

1 x 2 2 1 x 1 2 1 x 2 2

y 3

2 1 −5 −4

x

−1 −1

1

2

3

(2, −3)

−3

3 2

; m

y

3 2

0 x

y

3 2

0

y

3 2

1 2

4

3

(− 12 , 32 ) 2 1 x − 3 − 2 −1

63. Point 5.1, 1.8 ; m

–2

2

2

y 3

5 x 27.3

y

(−5.1, 1.8)

2 1 x

−7 −6 −5 −4 −3 −2 −1

–2

3 x 4

1

−2 −3

3x 14

y

3

5

5 x 5.1

y 1.8 x

2

1

−1 −2

y

3 4

59

y

0

−4

3x 6

4y

−4 −5

(−2, −5)

7 2

64. Point 2.3, 8.5 ; m

59. Point 6, 1 ; m is undefined.

Because the slope is undefined, the line is a vertical line. x

Linear Equations in Tw Two Variables T

6

2.5

y 8.5

2.5 x 2.3

y 8.5

2.5 x 5.75

y

2.5 x 2.75

y

y 2 −4

x

−2

2

4

6

8

−4 −6

(2.3, −8.5)

−8

6

−10

4 2 –4

–2

2

(6, −1)

4

–2

65. 5, 1 , 5, 5

x

–4

y 1

–6

60. Point 10, 4 ; m is undefined.

y

Because the slope is undefined, the line is a vertical line. x

y

10

y 3

6

(−10, 4)

4 2

−8 −6 −4 −2 −2

y 3

x 2

−4

y 3

−6

52 ; m

61. Point 4,

51 x 5 5 5 3 x 5 1 5 3 x 2 5

(−5, 5) 6 4

x –6

–4

–2

(5, −1)

–2 –4

66. 4, 3 , 4, 4

y 8

−12

y 8

y

0

y

5 2

0 x 4

y

5 2

0

y

5 2

y

5

4 3 x 4 4 4 7 x 4 8 7 7 x 8 2 7 1 x 8 2

y 6 4

(4, 3)

2

x –6

–4

–2

2

4

6

(− 4, − 4) –4 –6

4

)4, 52 )

3 2 1

x −1

−1

1

2

3

4

5

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60

Chapter P

Prerequisites requisites

67. 8, 1 , 8, 7

3· § 9 9· § 1 71. ¨ , ¸, ¨ , ¸ © 10 5 ¹ © 10 5 ¹

Because both points have x 8, the slope is undefined, and the line is vertical. x

9 § 3· ¨ ¸ § 1 ·· 5 © 5¹ § ¨ x ¨ ¸¸ 9 § 1 ·© © 10 ¹ ¹ ¨ ¸ 10 © 10 ¹

§ 3· y ¨ ¸ © 5¹

8 y 8

6§ 1· 3 ¨x ¸ 5© 10 ¹ 5 6 18 x 5 28

(−8, 7) 6

y

4 2

y

(−8, 1) x –10

–6

–4

–2 y

–2 2

68. 1, 4 , 6, 4

y 4

4 4 x 1 6 1

y 4

0 x 1

y 4

0

y 1

8 x −2

6

(−1, 4)

(6, 4)

(

§ 69. ¨ 2, ©

1 y 2 y y

2

5· ¸ 4¹

2· § 70. 1, 1 , ¨ 6, ¸ 3¹ ©

4

6

y 1 y 1 y

( 109 , − 95)

8

§3 72. ¨ , ©4 3

y

2

( 12 , 45 )

2 1 3 x 1 6 1 1 x 1 3 1 1 x 3 3 1 4 x 3 3

3· § 4 ¸, ¨ , 2¹ © 3

y

7· ¸ 4¹

3

7 3 4 2 §x 3· ¸ 4 3¨ 4¹ © 3 4 1 4 §x 3· ¸ 25 ¨ 4¹ © 12 3§ 3· ¨x ¸ 25 © 4¹

3 2

(2, 12 )

1

x

−1

1

2

3

−1

y

y

3 2

y

3 2

y

3 2

2

y x

−1 −2

1

2

3

4

(

6, − 23

)

2 1

( 34 , 32 ) x

−2

−1

1

2

−1

3 9 x 25 100 3 159 x 25 100

(1, 1)

1

(− 43 , 74 )

4 3

y 1

)

y

5 1 4 2 x 2 1 2 2 1 1 x 2 2 2 1 3 x 2 2

2

x –2

4

1· §1 ¸, ¨ , 2¹ © 2

1

−2

2

–2

y

−1 1 , −3 − 10 5

73. 1, 0.6 , 2, 0.6

y

0.6 0.6 x 1 2 1 0.4 x 1 0.6

y

0.4 x 0.2

y 0.6

y 3 2 1

(1, 0.6) x

–3

1

2

3

(−2, −0.6) –2 –3

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.5

74. 8, 0.6 , 2, 2.4

10 y 6

3 x 8 10 3 x 8

10 y 6

3 x 24

y 0.6

y

6 4

(−8, 0.6)

The slope is undefined. The line is vertical. x

–6

3 9 x 10 5

or

y 3

0.3x 1.8

y

2 1

y

−3 − 2 − 1

3

y 2 y

1 −2

)

1 , −1 3

2

)

3

4

5

79. L1 : y

(2, −1)

m1

−3

x

3

(− 6, − 2)

−2 −3 −4 −5 −6 −7 −8

−3 −4 −5

2

1x 3

3

1 3

1 −6 −5 −4 − 3 − 2 − 1 −1

1x 3

The lines are parallel.

2 x 1

2

(15, − 2)

4x 1

80. L1 : y

m1

4 4x 7

L2 : y

m2

4

The lines are parallel.

m1 9 and is undefined. 0

1x 2 1 2

12 x 1

L2 : y m2

3

12

The lines are neither parallel nor perpendicular. 54 x 5

82. L1 : y

m1

y

−1

(1.5, −2)

1 3

81. L1 : y

The line is vertical. 2 1

m2

y

§7 · §7 · 77. ¨ , 8¸, ¨ , 1¸ 3 © ¹ ©3 ¹ m

3

x

−1

2

1 8 7 7 3 3 7 3

2

1

§1 · 76. ¨ , 2 ¸, 6, 2 ©5 ¹

y 2

1

−3

L2 : y

2 2 x 6 1 6 5 0 x 6 1 6 5 0

(1.5, 0.2) x

−1 −2

2

The line is horizontal.

y 2

1.5

–8

1

y

(2, −2.4)

x

–4

1 1 x 2 1 2 3 0

y 1

y 2

2

–10 –8 –6

§1 · 75. 2, 1 , ¨ , 1¸ ©3 ¹ y 1

2 0.2 x 1.5 1.5 1.5 2 0.2 x 1.5 0

y 2

y

3 x 18

10 y

61

78. 1.5, 2 , 1.5, 0.2

2.4 0.6 x 8 2 8

y 0.6

Linear Equations in Two Tw T Variables

) 73 , 1) x

54 5 x 4

L2 : y

1 2 3 4 5 6 7 8

m2

1

5 4

The lines are perpendicular.

)

7 , −8 3

)

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62

Chapter P

NOT FOR SALE

Prerequisites requisites

83. L1 : 0, 1 , 5, 9 m1

88. x y

x 7

y

9 1 50

2

1, 3, 2

(a) m

13 40

1 2

y 2

1 x 3

y 2

x 3

y

x 1

The lines are perpendicular. 84. L1 : 2, 1 , 1, 5 m1

5 1

6 3

1 2

8 4

§ L2 : 0, 1 , ¨ 5, ©

m2

34 x

23 , 78 , m

(a)

y

7· ¸ 3¹

y

2 3

7 8

34 x 32

y

34 x

m1

28 4 4

6 8

y

4x 3

m2

1 5 3 1 3

y

Slope: m (a)

2, 1 , m y 1 y

(b)

2, 1 , m y 1

y

53 x

(a) m 16 3 4

53 ,

78 , 34

53 x

3 4

24 y

40 x 53 53 x

2x 3

40 y 30

3 x 78 5 24 x 78

12

40 y 30

24 x 21

12 x 2

y

3 4

40 y

40 x 35

(b) m

2 x 2

24 y 18 y

2

7 8

40 x

3 2

2

127 72

24 y 18

3 2x

53

y

4 3

The lines are perpendicular. 87. 4 x 2 y

5 x

Slope: m

1· § L2 : 3, 5 , ¨ 1, ¸ 3¹ ©

2 3

0

3y 3 4

3 8

x

4 3

90. 5 x 3 y

4 3

7 8

The lines are parallel. 86. L1 : 4, 8 , 4, 2

34

23 , 78 , m

(b)

7 4

34

Slope: m

2 3

7 1 3 50

7

y

2

85. L1 : 3, 6 , 6, 0 0 6 6 3

x 5

y 89. 3x 4 y

5 3 51

1 x 3

y 2

2

The lines are neither parallel nor perpendicular.

m1

1, 3, 2

(b) m

L2 : 1, 3 , 5, 5 m2

1

Slope: m

L2 : 0, 3 , 4, 1 m2

7

7 8

53 24

3 7 3 , , 5 8 4

24 x 9 3 x 5

9 40

INSTRUCTOR USE ONLY y

12 x 2

y

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© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.5

91. y 3

3

y

(a)

1, 0 , m

0

y

2

Slope: m (a)

(b)

is undefined.

(b) 0

4, 1 , m

0 x 4

y 1

0

y

1

x

93. x 4

97.

is undefined.

4

3, 2 , m x

(b)

94. x 2

3

2

99.

(a)

5.

(b)

1 3

1 3

x 3.9

y 1.4

1x 3

1.3

y

1x 3

0.1

x y 2 3 3x 2 y 6

1 0

x y 1 6 2 3

3x y 2 2 3x y 2

1

x y c c x y

1 x 2.5

1 2

101.

1

2.5, 6.8 , m

x 4.3

2.5, 6.8 , m y 6.8

3.9, 1.4 , m

x y 2 3 2

x 4

y

3x 13.1

1

12 1 0

1

1

1 x

2.5

1 0

§2 · 100. ¨ , 0 ¸, 0, 2 ©3 ¹

4

y 6.8

y

6x

vertical line x

Slope: m

3 x 11.7

3 y 2 12 x 3 y 2

0

(b) A perpendicular to a vertical line is a horizontal line, whose slope is 0. The horizontal line containing 5, 1 is the line y 1.

y

y 1.4

0

x 2 Slope: m is undefined. (a) The original line is the vertical line through x 2. The line parallel to this line containing 5, 1 is the

95. x y

3 x 3.9

x y 3 4 x y 12 12 4 3 4 x 3 y 12

is undefined.

3, 2 , m y

3

98. 3, 0 , 0, 4

0

x 4 Slope: m is undefined.

(a)

3.9, 1.4 , m

y 1.4

0

y 1

4, 1 , m

9 2

3

y 1.4

1

92. y 2

3x

Slope: m

0

x

6 x 9

y 0

63

9

2y 0

1, 0 , m y

(b)

96. 6 x 2 y

0

Slope: m (a)

Linear Equations in Two Tw T Variables

1 1 0 1, c z 0 c c

3

c

x y

3

x y 3

0

INSTRUCTOR USE ONLY y

x 9.3

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64

NOT FOR SALE

Chapter P

Prerequisites requisites

102. d , 0 , 0, d , 3, 4

x y d d x y

1

(b) y

d

(c) y

2x 3

d

1

d

x y

1

x y 1

0

9

(b)

2 −6

105. (a) y

12 x

(b) y

12 x

4

2 x

(a)

(b)

(c) y

(c)

−6

1x 2

(c) y

(a) −9

(a) is parallel to (c). (b) is perpendicular to (a) and (c).

2x

(b) y

(c)

32 x

3 4

103. (a) y

6

2x 3

104. (a) y

(c)

3

−10

−4

(b)

2x 4

106. (a) y

x 8

(b) y

x 1

(c) y

x 3

14

(a) −8

(a) and (b) are parallel. (c) is perpendicular to (a) and (b).

6

(b) and (c) are perpendicular.

8

10

(b) (a) −14

16

(c)

(a) is parallel to (b). (c) is perpendicular to (a) and (b).

−10

107. Set the distance between 4, 1 and x, y equal to the distance between 2, 3 and x, y .

x

4 ª¬ y 1 º¼ 2

x 4 y 1 2

ª¬ x 2 º¼ y 3

2

2

x 2 y 3

2

2

x 2 8 x 16 y 2 2 y 1

y

2

(−2, 3)

2

x2 4 x 4 y2 6 y 9

8 x 2 y 17

4 x 6 y 13

0

12 x 8 y 4

0

43 x 2 y 1

0

3x 2 y 1

4

(1, 1) −4

x

−2

2

(4, −1) −4

This line is the perpendicular bisector of the line segment connecting 4, 1 and 2, 3 . 108. Set the distance between 6, 5 and x, y equal to the distance between 1, 8 and x, y .

x

6 y 5

x

6 y 5

2

2

x

2

x

2

x 12 x 36 y 10 y 25 2

2

x y 12 x 10 y 61 2

2

12 x 10 y 61

1 y 8 2

1 y 8 2

2 y 8

2

6

2

x y 2 x 16 y 65 2

2

2 x 16 y 65

10 x 26 y 4

0

25 x 13 y 2

0

5 x 13 y 2

0

(6, 5)

4

x 2 x 1 y 16 y 64 2

2 −6 −4 − 2 −2 −4 −6 −8

x 6

8 10

( 72, − 32 ) (1, − 8)

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

Section P.5

Linear Equations in Tw Two Variables T

65

52 and x, y equal to the distance between 7, 1 and x, y .

109. Set the distance between 3,

52 y 52

x

3 y

2

x

3

2

2

2

ª¬ x 7 º¼ y 1 2

x

7 y 1 2

y

2 8 6

2

x2 6x 9 y 2 5 y

25 4

x 2 14 x 49 y 2 2 y 1

6 x 5 y

61 4

14 x 2 y 50

24 x 20 y 61

56 x 8 y 200

80 x 12 y 139

0

(3, 52)

4

7 − 2, 4 ) (− 7, 1) (

x

−8 − 6 − 4

2

4

6

−8

52 and 7, 1 .

This line is the perpendicular bisector of the line segment connecting 3,

110. Set the distance between 12 , 4 and x, y equal to the distance between

x 1 2

2

y 4

x 12 x2 x

1 4

2

2

y 4

2

y 2 8 y 16

x 72 y 54 x 72 y 54 2

2

49 4

y2

x2 y 2 x 8 y

65 4

x2 y 2 7 x

x 8y

65 4

7 x

39 16

0

128 x 168 y 39

0

8x

111. (a) m

21 y 2

2 y

5 y 2

5 y 2

(72, 54)

2

2

x2 7 x

72 , 54 and x, y .

1

5 y 2

25 16

221 16

221 16

−2 −1 −1

x 1

−2

(

3 , 2

3

4

)

− 11 8

(− 12, − 4)

135. The sales are increasing 135 units per year.

(b) m

0. There is no change in sales during the year.

(c) m

40. The sales are decreasing 40 units per year.

112. (a) m

400. The revenues are increasing $400 per day.

(b) m

100. The revenues are increasing $100 per day.

(c) m

0. There is no change in revenue during the day. (Revenue remains constant.)

113. (a) greatest increase

largest slope

18, 97,486 , 16, 90,260 m1

97,486 90,260 18 16

3,613

The salary increased the greatest amount between 2006 and 2008. least increase

smallest slope

14, 86,160 , 12, 83,944 m2

86,160 83,944 14 12

1108

The salary increased the least amount between 2002 and 2004. (b) m

97,486 69,277 18 6

9403 12

or

2350.75

(c) The average salary of a senior high school principal increased $2350.75 per year between 1996 and 2008.

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66

Chapter P

Prerequisites requisites

114. (a) greatest increase

largest slope

5, 13.93 , 4, 8.28 13.93 8.28 5.65 5 4 Sales showed the greatest increase between 2004 and 2005. least increase smallest slope m

2, 5.74 , 1, 5.36 5.74 5.36 0.38 2 1 Sales showed the least increase between 2001 and 2002. m

24.01 5.36 | 3.11 7 1

(b) m

(c) The average sales increased about 3.11 billion dollars per year between 2001 and 2007. 115. y

y

6 x 100 6 100

200

12 feet

116. (a) and (b)

x

300

600

900

1200

1500

1800

2100

y

–25

–50

–75

–100

–125

–150

–175

Horizontal measurements Vertical measurements

600 1200 1800 2400 x −50 −100 −150 −200 y

50 25

(c) m

600 300

y 50 y 50 y

25 300

1 12

1 x 600 12 1 x 50 12 1 x 12

1 , for every change in the horizontal measurement of 12 feet, the vertical 12 measurement decreases by 1 foot.

(d) Because m

(e)

1 | 0.083 12

117. 10, 2540 , m

8.3% grade

V – 2540

125t 10

V 2540

125t 1250

V

118. 10, 156 , m

125

125t 3790, 5 d t d 10

4.50

V – 156

4.50t 10

V 156

4.50t 45

V

4.5t 111, 5 d t d 10

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.5

119. The V-intercept measures the value of the molding machine at the time of purchase (when t 0 ). The slope measures the amount the value of the machine depreciates per year. 120. The C-intercept measures the fixed costs of manufacturing when zero bags are produced. The slope measures the cost to produce one laptop bag. 121. Using the points 0, 875 and 5, 0 , where the first

coordinate represents the year t and the second coordinate represents the value V, you have m V

V

0 875 175 50 175t 875, 0 d t d 5.

128. 3, 1078 , 7, 1067

1067 1078 11 7 3 4 11 y 1078 t 3 4 11 4345 y t or y 4 4 For 2012, use t 12 m

y

2.7512 1086.25

y

1053.25

For 2014, use t

2000 25000 2300 10 0 2300t 25,000, 0 d t d 10

123. Sales price

L 0.20 L

S

0.80 L

2.7514 1086.25

y

1047.75

y | 1047 stores

Answers will vary. 129. (a)

m

125. W

0.07 S 2500

126. C

0.55 x 120

1.46 1.21 17 9

y

y

0.03125

0.03125 22 0.92875

24

0.03125 24 0.92875

y | $1.68

442.62510 40,571

y

44997.25

15

y

442.62515 40,571

y

47,210.375

y | 47,210 students

(c) m 442.625; Each year, enrollment increases by about 443 students.

22

y | $1.62

y

10

y

For 2015, use t

0.03125t 0.92875

For 2014, use t

442.625t 40,571

y | 44,997 students

0.03125t 17

For 2012, use t

442.625

442.625t 0

(b) For 2010, use t

127. 17, 1.46 , 9, 1.21

y 1.46

44,112 40,571 80 y

0.75 x 12.25

m

0, 40,571 , 8, 44,112

y 40,571

124. W

2.75t 1086.25

14

y

List price 20% discount

S

67

y | 1053 stores

122. 0, 25,000 and 10, 2000

m

Linear Equations in Tw Two Variables T

130. (a)

2000, 46,107 , 2008, 51,413 Average annual change: 51,413 46,107 5306 663.25 2008 2000 8 So the average annual change in enrollment from 2000 to 2008 was 663.25.

(b) 2002: 46,107 2663.25 | 47,434 students 2004: 46,107 4663.25

48,760 students

2006: 46,107 6663.25 | 50,087 students (c) m

663.25, so N t

663.25 46,107

Each year, enrollment increases by about 663 students. (d) Answers will vary.

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68

Chapter P

NOT FOR SALE

Prerequisites requisites

cost of operation per hr cost of operator per hour cost of machine

131. (a) Cost

C

6.50t 11.50t 42,000

C

18t 42,000

(b) Revenue

charge per hour

R 30t R C

(c) P P

30t 18t 42,000

P

12t 42,000

(d)

0 42,000 12 t

12t 42,000 12t 12 3500.

To break even the company must use the equipment for 3500 hours. 132. 580, 50 and 625, 47

135. (a) and (b) Doctors (in thousands)

47 50 3 1 625 580 45 15 1 x 50 p 580 15 1 116 x 50 p 15 3 1 266 x p 15 3

(a) m

1 266 655 15 3 1 266 595 15 3

(b) x (c) x

2

3

4

5

6

7

8

49 units

Using 2 points to estimate:

3, 51.7 , 4, 54.1 54.1 51.7 43

m x x1

y 51.7

2.4 x 3

y

(d)

8 x 50

Because m 8, each 1-meter increase in x will increase y by 8 meters.

2.4

y y1

x

2.4 x 44.5

(d) Answers will vary. Sample answer: The slope describes the change (increase or decrease) in the number of osteopathic doctors per year. The y-intercept describes the number of doctors in the year 2000 (when x 0). (e) The model is a good fit to the data.

10 1

134. Answers will vary. Sample answer: Choosing the points 1, 2.1 and 7, 2.8

2.8 2.1 | 0.1167. 7 1 y y1 mt t1

m

y

35

Sample answer: Using technology: y 2.3726 x 44.658

0

y 2.1

40

45 units

215 2 x 210 2 x

0

45

(c) Answers will vary.

x

150

50

Year (0 ↔ 2000)

10 m

(c)

55

x

m

(b) y

60

1

133. (a)

15 m

y 65

(f) Sample answer: For 2012, use x y

12

2.372612 44.658

y | 73.1 In the year 2012, there will be approximately 73.1 thousand osteopathic doctors.

0.1167t 1 0.1167t 1.9833

INSTRUCTOR USE ONLY Using a calculator the regression line is y 0.1167 1167t 1.9667

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section P.5

y

136. (a) and (b)

Linear Equations in Tw Two Variables T

69

140. On a vertical line, all the points have the same x-value, y2 y1 , you would have a so when you evaluate m x2 x1 zero in the denominator, and division by zero is undefined.

100 90 80 70 60 50

141. No. The slope cannot be determined without knowing the scale on the y-axis. The slopes will be the same if the scale on the y-axis of (a) is 2 12 and the scale on the

x 10 12 14 16 18 20

(c) Two approximate points on the line are 10, 55 and

19, 96 . m y 55 y

(d) y

y-axis of (b) is 1. Then the slope of both is 54 .

96 55 41 19 10 9 41 x 10 9 41 85 x 9 9

142. Because 4 !

143. Both lines have positive slope, but their y-intercepts differ in sign. Matches (c). 144. The lines intersect in the first quadrant at a point x , y where x y. Matches (a).

(e) Each point will shift four units upward, so the bestfitting line will move four units upward. The slope remains the same, as the new line is parallel to the old, but the y-intercept becomes

145. The line y

−15

41 121 x . 9 9

138. False, the lines are not parallel. 4 2 8, 2 and 1, 4 : m1 1 8

−10

−15

11 7

d2

x2

x1 y2 y1 2

2

2

15

−10

y = −2x

y = −0.5x

The greater the magnitude of the slope (the absolute value of the slope), the faster the line rises or falls.

2

2

y = −4x 10

40 , the triangle is a right triangle.

x1 y2 y1

4 x falls most quickly.

The line y y = −x

7 0

x2

y = 4x

2 7

7 4

146. d1

15

y =x

139. Using the Distance Formula, we have AB 6, BC 40, and AC 2. Since

62 2 2

y = 0.5x

10

137. False. The slope with the greatest magnitude corresponds to the steepest line.

and 7, 7 : m2

4 x rises most quickly. y = 2x

§ 121 · ¨ 0, ¸ 9 ¹ ©

so the new equation is y

0, 4

, the steeper line is the one with a

slope of – 4. The slope with the greatest magnitude corresponds to the steepest line.

41 85 | 87 17 9 9

§ 85 · 4¸ ¨ 0, © 9 ¹

5 2

1 0 2

m1 0

1 0 2

m2 0

1 m 1

2

2

1 m 2

2

2

Using the Pythagorean Theorem:

d1

distance between 1, m1 , and 1, m2

d 2

2

§ 1 m 2· § 1 m 2· 1 ¸ ¨ 2 ¸ ¨ © ¹ © ¹

2

§ ¨ ©

1 m 1 1 m 2 2

2

m 1 2

m 2 2

m2 m1 m 2 2 2m1m2

2

2

2

2 1 m2

1 1 2

2 m2 m1 ·¸ ¹

2

2

2

m1

2

2m1m2

INSTRUCTOR USE ONLY m1

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE

Chapter P

70

Prerequisites requisites

147. No, the slopes of two perpendicular lines have opposite signs. (Assume that neither line is vertical or horizontal.) 148. (a) Matches graph (ii).

The slope is –20, which represents the decrease in the amount of the loan each week. The y-intercept is 0, 200 , which represents the original amount of the loan. (b) Matches graph (iii). The slope is 2, which represents the increase in the hourly wage for each unit produced. The y-intercept is 0, 8.5 , which represents the hourly rate if the employee produces no units. (c) Matches graph (i). The slope is 0.32, which represents the increase in travel cost for each mile driven. The y-intercept is 0, 30 , which represents the fixed cost of $30 per day for meals. This amount does not depend on the number of miles driven. (d) Matches graph (iv). The slope is –100, which represents the amount by which the computer depreciates each year. The y-intercept is 0, 750 , which represents the original purchase price.

Review Exercises for Chapter P 1. 6 x 2

6 x 2 4 x 4

2

7. 4 x 3 3

2 4x x2

3x 6 2 x 3x

8 6x 4

4x 9

The equation is an identity. 2. 3 x 2 2 x

2 4 3 x 4

4 x 12 3

2 x 3

10 x

5

x

12

2x 6 12

8.

x

1 2

Conditional equation

6 x 4

3 2 x 1

5

x 3 4x 4

10 17

x 3 x7 x 3

x3 x 2 7 x 3

3 x

x x 2 x 7 x 1 4

x3 x 2 7 x 3

x

3.

The equation is an identity.

9.

2 x 2 x 28

0 0

4. 3 x 4 x 8

10 x 2 3x 6

2 x 7 x 4

3x 2 12 x 24

10 x 20 3 x 2 6

2x 7

2

2

6 x 2 2 x 38

x

0

Conditional equation 5. 3x 2 x 5

10

3 x 2 x 10

10

x

20

6. 4 x 27 x

5

4 x 14 2 x

5

2x

9

x

92

10.

0

or

x 4

0

72

or

x

4

15 x 2 x 2

0

2 x x 15

0

2

2 x

5 x 3

x

52 , 3

11. 16 x 2

25

2

25 16

x

r

x

12.

17 3

6 x2

25 16

0

r 54

3x 2 2

INSTRUCTOR USE ONLY x

r

2

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE

Review Exercises fo ffor Chapter P

13.

x

8

2

15

x 8

x

4

8r

2

0

x

0,

0

9 x x 3 4 x x 3

0

3

9 x 4 x x 3 x9 x 2 4 x 3

4 r 3 2

15. x 2 6 x 3

1, b

12 5

21. 9 x 4 27 x3 4 x 2 12 x

r3 2

x

3

0

3

6, c

0 0

22. x 4 5 x 2 6

0 12 r

x

6 r

3, c

3r

41 30

x x

20 2

249

1 r 2

6

18. 2 x 2 5 x 27

24.

5 r 5 r

5 2 42 27 2 2 25.

241 4

x 2 4 x 6

0

x2

0 x

0

4x 6

0 x

3 2

2

26.

x 5 2

0 x

2 3

x 3

0 x

3

0

u 5u 6

0

2 u 3

0

2, 3 2

2, 3 r

2, r

3

x 2 8

0

x 2

8

x 2

64

x

66

x 4

3

x 4

9

x

5

3x 2

4 x

3x 2

4 x 2

3x 2

16 8 x x 2

0

18 11x x 2

0

x

0

x 9 x

9, extraneous

0

x 2 x

2

9 x 2

x 5

x

x 2 10 x 25

x 6 x 25 2

b 4ac

3x 2

x

4x 2

23.

249 6

0

x

0

23

2

u

6

3 43 20 23

19. 4 x3 6 x 2

0 x

x2.

Let u

0

3 r

x

2

0

3x 2 3 x 20 3, b

12 21

u

17. 20 3x 3x 2

a

3x 2

3 r 2 3

16. x 2 12 x 30

0

x

21

6 r 48 2

0

x3x 2 3 x 2 x 3

62 41 3

6 r

x

x 5 x 12

15

18

x 4

a

0

3

r 15

x 14.

20. 5 x 4 12 x3

71

0

6a 41 25 2

64 0 no real solutions

INSTRUCTOR USE ONLY Original equation has no real solutions.

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72

27.

28.

NOT FOR SALE

Chapter P

x

x

2

Prerequisites requisites

34

27

x 2

27

x 2

81

x

79

1

23

30. 43

25

0

x 1

253 2

x 1

r125

13

x

33. x 2 6

0

x

4 ª¬5 x 20 x 1º¼

0

x

x 2 4

43

13

2

13

12

2

12

0 4

4

3 x

x

0

4

0

2 3 x 2

0

13

9 x 2

0

10 r 5

4,

2x

2

x 2

0 x

2

3x 2

0 x

2 3

3x 2

0 x

23

95

95

7

2x 3

0 x

10 r 5

25

31. 2 x 3

x 2

0

202 45 1

20 r

x

7 or 2 x 3

7

4 or

2x

10

x

5

2

32. x 5

10

First Equation

Second Equation

x 5

10

x 5

x

15

x

x 2 6

x2 6

x

x x 6

0

3 x 2

x2 x 6

0

0

x 3 x 2

0

10 5

or

x

x 3

0 x

3

x 2

0 x

2

x 2

0 x

2, extraneous

x 3

0 x

3, extraneous

34. x 2 3

2x

First Equation x 3 2

Second Equation

x 2

2x

x 2x 3

0

x 3 x 1

0

2

3

x 2x 3

x

2x

2

0

3 x 1

0

x 3

0 x

3

x 3

0 x

3, not a solution

x 1

0 x

1, not a solution

x 1

0 x

1

35. Let x

4

x

2

x

4 ª¬1 5 x x 4 º¼

0

12

5 x 20 x 1

4 ª¬8 x 2 x 2 4º¼

x2 2

x

32

2

126, 124 13

x

5 x x 4

12

x

8 x 2 x 2 4

x2

4

x

x

29.

x

the number of liters of pure antifreeze.

30% of 10 x 100% of x 0.3010 x 1.00 x

50% of 10 0.5010

3 0.30 x 1.00 x

5

0.70 x

2

x

2 0.70

20 7

6 2 liters 7

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

Review Exercises fo for f Chapter P 36. 42

0.001x 2 42 29.95

12.05 x

2

42.

29.95 0.001x 2

3 x 33 x

1 2

3 x ! 5 x ! 53

Number of units demanded per day: 143,203

53 , f

37. 6 x 17 ! 0

3 ?

63 17 ! 0

3x 17 d 34 2 38 3x 17 d 68

43. 19

1 ! 0 Yes, x

21 3x d 85

3 is a solution.

4

(b) x

!

7 x d ?

6 4 17 ! 0 4 is not a solution.

x 3 2 5 (a) x 3

2x 5 5 3 9 d 2 x 5 15 4 d 2 x 20 2 d x 10

33 2 5 3 d 0 2 ?

3 d

Yes, x

3 is a solution.

12

(b) x

12 is a solution.

39. 9 x 8 d 7 x 16

2 x d 24 x d 12

1 2

8 x

20 8x d 4 12 x x d 16 15 2 x t

15 x 2

>6, 4

6 d x d 4 or 46. x 2 1

1 x 2 1 1 x 3

1, 3 x 3 ! 4 or

40. 45 2x d

41.

x 1 d 5

45.

47. x 3 ! 4

f, 12@

ª 32 , f ¬15

>2, 10 5 d x 1 d 5

12 3 ? 3 d 2 5 3 d 3 2 ?

Yes, x

1º § or ¨ 7, 28 » 3¼ ©

44. 3 d

38. 3 d

?

85 3

7 x d 28

41 ! 0

No, x

2 3 x 2 2 3 x 1 3

9 3x ! 4 6 x

0.001x 2

143,202.5

(a) x

!

73

32 15

4 ! 3x 5

x 3 4

x ! 7

x 1

f, 1 7, f 48. x

x

3 2 3 2

t

3 2

d 32

x

x d 0

3 2

t

3 2

x t 3

f, 0@, >3, f

15 x 8 ! 6 x 10 9 x ! 18 x ! 2

INSTRUCTOR USE ONLY 2, f@

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74

NOT FOR SALE

Chapter P

Prerequisites requisites

49. R ! C 125.33 x ! 92 x 1200

2 x 2 x 15 t 0

33.33 x ! 1200

2 x

1200 x ! | 36 units 33.33

19.3 r 0.5 cm

A

19.3 2

3, 52 2 x 5 x 3 0 52 , f 2 x 5 x 3 ! 0 Solution intervals: f, 3@ ¬ª 52 , f

| 372 cm 2

18.8 2

Smallest area: A Largest area: A

19.8 2

| 353 cm 2 | 392 cm 2 55.

Interval containing area of square: 353 cm 2 A 392 cm 2 51. x 2 6 x 27 0 3, x

Critical numbers: x

9

x

Test intervals: f, 3 , 3, 9 , 9, f Test: Is x 3 x 9 0?

x 5

1 x 1

Test: Is

x 2x t 3 x2 2x 3 t 0

Critical numbers: 1, 3

56.

Test intervals: f, 1 x 3 x 1 ! 0

x

x 3 x 3, 5 3 x 5, f 3

Test intervals: f, 3

6 x2 5 4 6 x2 5x 4 0

43 , x

Critical numbers: x

Test intervals: f, 43 , 43 ,

1 2 1 2

Test: Is 3x 4 2 x 1 0?

5 0 x 5 ! 0 x 5 0 x

Solution intervals: f, 3 5, f

4 2 x 1 0

1 2

d 0?

1 x 1

x 5 0 3 x

Solution intervals: f, 1@ >3, f

x 5

Critical numbers: 3, 5

1, 3 x 3 x 1 0 3, f x 3 x 1 ! 0

Solution set: 43 ,

r1

Solution set: >5, 1 1, f

3 x 1 t 0

3 x

5, x

Test intervals: f, 5 , 5, 1 , 1, 1 , 1, f

2

53.

d 0

Critical numbers: x

Solution set: 3, 9

x

2 3 d x 1 x 1 2 x 1 3 x 1 d 0 x 1 x 1 2 x 2 3x 3 d 0 x 1 x 1

3 x 9 0

52.

5 2

Test intervals: f, 3 2 x 5 x 3 ! 0

s

x

5 x 3 t 0

Critical numbers: 3,

s2

50. A

2 x 2 x t 15

54.

, 12 , f

57.

x 2 7 x 12 t 0 x x 4 x 3 t 0 x Critical numbers: x

4, x

3, x

0

Test intervals: f, 4 , 4, 3 , 3, 0 , 0, f Test: Is

x

4 x 3 x

t 0?

Solution set: > 4, 3@ 0, f

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE

Review Exercises for fo f Chapter P

58.

y

62.

1 1 ! x 2 x 1 1 ! 0 x 2 x 2 ! 0 x x 2

6 4 2 x

−4 −2 −2

2

4

6

8

−4 −6

Critical numbers: 0, 2

63. x ! 0 and y

Test intervals: f, 0

2 ! 0 x x 2

0, 2

2 0 x x 2

2, f

2 ! 0 x x 2

Solution intervals: f, 0 2, f

2 in Quadrant IV.

64. Because the products positive, x and y must have the same sign.

x ! 0 and y ! 0 in Quadrant I. x 0 and y 0 in Quadrant III. y

65. (a) (−3, 8)

8

59. 50001 r ! 5500 2

1 r

2

(1, 5) 4

! 1.1

2

1 r ! 1.0488

−4

r ! 4.9%

10001 3t

2

4

3 1 2

(b) d

8 5

2

16 9

§ 3 1 8 5 · (c) Midpoint: ¨ , ¸ 2 ¹ © 2

t 2000

5t 10001 3t 2000 t 0 5t 1000t 9000 t 0 t 5

66. (a)

5

13 · § ¨ 1, ¸ 2¹ ©

y

(−2, 6)

6

2

t must be greater than zero, so the critical value to check is t 9. 1000t 9000 0 0, 9 t 5 1000t 9000 ! 0 9, f t 5 t t 9

−4

x

−2

2

4

−2

2

(b) d

6

(4, −3)

−4

4 6 3 2

36 81

So, the required time is at least 9 days. 61.

x

−2

r ! 0.0488

60. P

75

117

§ 2 4 6 , (c) Midpoint: ¨ 2 © 2

2

3 13

3· ¸ ¹

§ ¨1, ©

3· ¸ 2¹

y

67. (a)

y

6

(0, 8.2) 8

4 2

6 x

−6 −4 −2 −2

2

4

6

8

4

−4

2

−6

(5.6, 0)

−8

−2

(b) d

x 2

5.6

4

6

0 0 8.2 2

31.36 67.24

2

98.6 | 9.9

§ 0 5.6 8.2 0 · ((c)) Midpoint: p , ¨ ¸ 2 ¹ © 2

INSTRUCTOR USE ONLY 2.8, 4.1

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76

NOT FOR SALE

Chapter P

Prerequisites requisites y

68. (a) 4 −3 −2 −1

3x 5

73. y (1.8, 7.4)

8

x

–2

1

0

1

2

y

–11

–8

–5

–2

1

x −4

1

2

3

−8 y

(− 0.6, −14.5)

1 x –3

1.8 0.6

(b) d d

2.4 2

d

485.37

2

21.9

7.4 14.5

–2

–1

1

2

3

–1

2

–2 –3

2

–4 –5

d | 22.03 units

12 x 2

74. y

(c) Midpoint: § 1.8 0.6 7.4 14.5 · , ¨ ¸ 2 2 © ¹

0.6, 3.55

x

–4

2

0

2

4

y

4

3

2

1

0

y

69. New vertices:

4 6 4 6

4, 8 8 4, 8 8 4, 3 8 4, 3 8

10

0, 0 2, 0 0, 5 2, 5

8 6 4

x −6

70. New vertices:

0 2, 1 3 3 2, 3 3 0 2, 5 3 3 2, 3 3

−4

−2

4

x 2 3x

75. y

2, 4 1, 6 2, 8 5, 6

2

−2

x

–1

0

1

2

3

4

y

4

0

–2

–2

0

4

y 5 4

71. Midpoint: § 2000 2008 2.17 10.38 · , ¨ ¸ 2 2 © ¹

x –3 –2 –1

2004, 6.275

1

2

4

5

–2 –3

In 2004, the sales were $6.275 billion. 72. (a)

76. y

Apparent temperature (in °F)

y 150 140 130 120 110 100 90 80 70

2x2 x 9

x

–2

–1

0

1

2

3

y

1

–6

–9

–8

–3

6

1

3 4 5

y 1

x 65 70 75 80 85 90 95 100

−5 −4 −3

Actual temperature (in °F)

(b) Change in apparent temperature

150qF 70qF

−1

x

−2 −3 −4

80qF −9

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE

Review Exercises for fo f Chapter P

77. y 2 x 3

y

0

77

x 2 4 x is a parabola.

82. y

2x 3

Line with x-intercept 32 , 0 and y-intercept 0, 3

x

–1

0

1

2

3

4

y

5

0

–3

–4

–3

0

y y

6 5 4 3 x

1

–2 –1 –1

x –5 –4 –3

–1

1

2

3

1

2

3

5

6

–2 –3

–2

–4

78. 3x 2 y 6

0

2y

3 x 6

y

32 x 3

83. y

2x 7

x-intercept: Let y

Line with x-intercept 2, 0 and y-intercept 0, 3

0.

0

2x 7

x

72

72 , 0

y 2 1

y -intercept: Let x

x –5 –4 –3

1

–1

2

3

–3 –4 –6

5 x 6 4

5

4

1

4

3

y

0

1

2

3

1

x –2 –1

1

2

3

4

5

6

x 2

x 1 3

0

x 1 3

y

x 1 3, or 2

For x 1 0, 0

x 1 3, or 4

y-intercept: Let x

5

0

2

2

7

2

3

3 2

x

−3 − 2 − 1

1

2

3

4

y

x 1 3

y

0 1 3 or y

0

y

0 0

r1

–2

r2

–8

x

3 4 2

x-intercepts: 0

y

2 x 2 is a parabola.

2

0, 2 85. y

x

3 4 x 3 2

x 1

2

3

–2

4 r2

x

3 r2

x

5 or x

1

5, 0 , 1, 0

–3 –4 –5

2

x 3

1 –3 – 2 – 1

x

x.

5

−2

81. y 2 x 2

x.

0.

y

4

0

0.

For x 1 ! 0, 0

6

Domain: [2, f)

y

7

2, 0 , 4, 0

–2

–2

y

y

5

x

x

20 7

84. x-intercept: Let y

y

Domain: (f, 5]

80. y

y

0, 7

–5

79. y

0.

y -intercept: y y

9 4

y

5

0

3 4 2

INSTRUCTOR USE ONLY 0, 5

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78

NOT FOR SALE

Chapter P

86. y

Prerequisites requisites

x 4 x2 x 4 x2

x-intercepts: 0

4 x2

0

x

4 x

0

2

0 r2

x

0, 0 , 2, 0 , 2, 0 0

y -intercept: y

40

0

0, 0 87. y

4 x 1

Intercepts:

y

14 , 0 , 0, 1

y

4 x 1 y

y

4 x 1 y

y

4 x 1 y

4 x 1 No y -axis symmetry

1

4 x 1 No x-axis symmetry

x −4 −3 −2 −1

4 x 1 No origin symmetry

1

2

3

2

3

4

4

−2 −3 −4

88. y

5x 6

Intercepts:

y

56 , 0 , 0, 6

1 x

−2 −1 −1

2

3

4

5

6

−2

No symmetry

−3 −4 −5 −6 −7

89. y

5 x2

y

5, 0 , 0, 5

Intercepts: r

4

y

5 x y

y

5 x2 y

y 90. y

6

2

5 x 2 y -axis symmetry

3 2

5 x 2 No x-axis symmetry

5 x y 2

1 −4 −3

5 x No origin symmetry 2

x 2 10

−1

x 1

−2

y

Intercepts: r 10, 0 , 0, 10 y-axis symmetry

2 − 6 −4

x −2

2

4

6

8

−4

−12

91. y

x3 3

Intercepts: y

x

3

y

3

3, 0 , 0, 3

3 y

y

x3 3 y

y

x

3

7

3 y

6 5 4

x 3 No y -axis symmetry 3

2

x3 3 No x-axis symmetry x 3 No origin symmetry 3

1 −4 −3 −2

x −1

1

2

3

4

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE

Review Exercises fo for f Chapter P

92. y

6 x3

Intercepts:

3

79

y

2

6, 0 , 0, 6

x

−6 −4

2

4

6

No symmetry −8 −10

93. y

y

x 5 7

Domain: [ 5, f)

6

Intercepts: 5, 0 , 0, y

5 4 3

x 5 No y -axis symmetry

y

x 5 y

y

x 5 y

94. y

5

1

x 5 No x-axis symmetry

−6 −5 − 4 −3 −2 −1

x 1

2

x 5 No origin symmetry 98. x 2 y 8

x 9

Intercepts: 0, 9

2

y

81 18

Center: 0, 8

y-axis symmetry

14 12 10 8 6 4 2

Radius: 9

y 15

(0, 8)

x

−8 −6 −4 −2

4 6 8 10

9 6

99.

3 x

−9 −6 −3 −3

3

95. x 2 y 2

6

9

2

2

y

9

Center: 0, 0

y

36 8

2

6

62

2 x −8

−4

4 2 1 –4

–2

–1

1

4

100.

32 x 4 y 32 Center: 4, 32 x

4 y 2

2

y

Center: 0, 0

8

(

x 2

–4

4

(

2 4 1 , −1 2

−8

(0, 0)

–2

96. x 2 y 2

−2 −4

Radius: 6

Radius: 3

3

y

2

100 15

2

100

9

(− 4, )

6

3 2

3 x

−15

− 9 −6 −3

Radius: 10

3

9

−6

1

Radius: 2

(0, 0) −3

−1

x

1

−1

3

101. Endpoints of a diameter: 0, 0 and 4, 6

§ 0 4 0 6 · , Center: ¨ ¸ 2 © 2 ¹

−3

x

97.

x 12 y 1 x 12 y 1 Center: 12 , 1

x 2

2

2 y 2 2

y 0

Center: 2, 0

2

y

16 4

Radius:

6

2

r (−2, 0)

2 x

–8

–4

–2

4 –2

2, 3

2

0 3 0 2

2

4 9

Standard form: x 2 y 3

Radius: 4 –6

2

x

2 y 3 2

2

2

13

13

2

13

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© Cengage Learning. All Rights Reserved.

80

NOT FOR SALE

Chapter P

Prerequisites requisites

102. Endpoints of a diameter: 2, 3 and 4, 10

§ 2 4 3 10 · , Center: ¨ ¸ 2 © 2 ¹

1 2

Radius: r

13 · § ¨1, ¸ 2¹ ©

§ 13 · ¨ 3 ¸ © 2 ¹

2

§ 2 § 13 · · Standard form: x 1 ¨ y ¨ ¸ ¸ © 2 ¹¹ ©

2

13 · 2 § 1 ¨ y ¸ 2¹ ©

2

x 103. (a)

2

9

§ ¨¨ ©

85 · ¸ 4 ¹¸

85 4

2

85 4 107. y

N

Number of Walgreen stores

49 4

7000 6500 6000 5500 5000 4500 4000 3500 3000 2500

3 x 13

Slope: m

x −9 −6 −3 −3

3

4

5

6

7

108. y

0 d x d 20

10 x 9

3

6

9

y

10

12

y-intercept: 0, 9

9 6

x

0

4

8

12

16

20

F

0

5

10

15

20

25

(b)

3 −9

F

109. (a) m Force (in pounds)

30 25

(b) m

15 10

−3

x −3

9

! 0 The line rises.

0 The line is horizontal.

Matches L 3.

5 4

8

(c) m

12 16 20 24

Length (in inches)

(c) When x

10, F

3 0 The line falls.

Matches L1. 50 4

12.5 pounds.

(d) m

15 0 The line gradually falls.

Matches L 4 .

y

6

Slope: m

3 2

−6

Matches L 2 .

20

x

8

0

110. (a) m

y-intercept: 0, 6

4 2 −4

x

−2

2

4

6

−2

106. x

6

8

Slope: m 5 x, 4

3

−6

t 2

Year (0 ↔ 2000)

105. y

6 3

(b) 2008

(a)

12

3

y-intercept: 0, 13

1

104. F

y

3

y

Slope: m is undefined.

4

y-intercept: none

2

52 The line falls. Matches L3.

(b) m is undefined. The line is vertical. Matches L1. (c) m

0 The line is horizontal. Matches L 4 .

(d) m

1 2

The line rises. Matches L 2 .

3 1 −4

−2 −1

x 1

2

3

4

−2

INSTRUCTOR UC USE ONLY −3

−4 − 4

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE

Review Exercises fo ffor Chapter P

111. 6, 4 , 3, 4

m

116. 10, 3 , m

y

4 4

4 4 63

6 3

5 4

8 9

(6, 4)

y 3

3 2 1 x

−3 −2 −1

12 x 5

y

12 x 2

Slope: m

5

(5, 52)

3 2

(32, 1) 2

3

4

y 8

0

y 6

0 x 2

y 6

0

(−2, 6) 4 2

4

y

0

y 1

0 x 3

y 1

0 or y

4 3 2

1

(− 3, 1) x 1

2

−1 −2

119. Point: 8, 5 y

Slope: Undefined

8

x

(−4.5, 6)

4 2

0

x

2

5 11

−6

−4

−2

x −6 −4 − 2

2

−2

4

120. 12, 6 , m is undefined.

y 6

The line is vertical. x

6

0

−2

6

y

4

12

2 x

(− 3, 2)

4

−2

(8, 2)

–2

2

4

6

−6

y

m

2 1

(3, 0) −2

−1

4

6

8

10

(12, −6)

121. 0, 0 , 0, 10

–4

2 3

2

−4

8

–2

2 x 3 3 2 x 2 3

6

(− 8, 5)

8

x 8

(2.1, 3)

8

0 11

y

x

y

2

−4 −3 −2 −1

–4

y 0

−2

5

118. 3, 1 , m

114. 3, 2 , 8, 2

115. 3, 0 , m

(10, −3)

x

1

30 66

2 2 3 8

x

10 12

−6

−4

−4

m

8

4

−4

x

36 2.1 4.5 3 6.6

2

1

113. 4.5, 6 , 2.1, 3 m

−2

117. Point: 2, 6

4

m

4

−8

y

5 1 2 3 5 2 5 2 2 2 10 3 2 2 3 2 7 2 6 3 or 14 7

6

y 3

−3 −4

§ 3 · § 5· 112. ¨ , 1¸, ¨ 5, ¸ © 2 ¹ © 2¹

y

12 12 x 10

2 3 4 5 6

(−3, −4)

81

1

3

10 0 0 0

10 , undefined 0

The line is vertical. x

x

0

−1

122. 2, 1 , 4, 1 −3

m

1 1 4 2

y 1 y 1

0 2

0

0 x 2 0

INSTRUCTOR USE ONLY y

1

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82

Chapter P

NOT FOR SALE

Prerequisites requisites

129. Point: 3, 2

123. 1, 0 , 6, 2 2 0

m

5x 4 y

2 7

6 1

y

y y

125. x 5

(b)

126. x 4

(a)

127. y 6

128. y 1

(a)

(b)

y

5 x 4

23 4

y 2

54 x

12 5

y

54 x

2 5

0 x

4 and m is undefined.

3 or x 3

0

2 or y 2 0 y

2x 3

23 x 8 2 x 16 2 x 7

1 or y 1

0

0

2 or x 2

1 and m

0

3 2

x

3 x 24

2y

3x 30 3 x 2

131. 10, 12,500 , m

15 850

V 12,500

850t 10

V 12,500

850t 8500

V 72.95 V 72.95 V

3 2

8

2y 6

850t 21,000

132. 10, 72.95 , m

0

7 3

(b) Perpendicular slope: m

V

0

4 or y 4

23 x

y

6 and m

23

(a) Parallel slope: m

0

0

0 y

5 3

y 3

0

5

5 2x

y

0

54

54 x 3

3y

1 or y 1

3, 4 , m y

15 4

3y 9

0

2, 1 , m is undefined. x

y 3

0

2, 1 , m y

(b)

5 and m is undefined.

2 or x 2

3, 2 , m y

(a)

0 x

y

3, 2 , m is undefined. x

(b)

3y

1 1 x 5 5

2, 1 , m y

3

130. Point: 8, 3 , 2 x 3 y

x 1

2, 1 , m is undefined. x

x

5 x 4

y 2

1 5

1 x 11 5 x 11

y

5 4

y 2

5y

(a)

6 11

5 y 10

5 4

(b) Perpendicular slope: m

1 2

y 2

2

y 2

124. 11, 2 , 6, 1

m

8

(a) Parallel slope: m

2 x 1 7 2 x 1 7 2 2 x 7 7

y 0

5 x 4

5.15 5.15t 10 5.15t 51.5 5.15t 21.45, 10 d t d 15

0

3, 4 , m is undefined. x

3 or x 3

0

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

Chapter Test ffo for Chapter P 133. Sample answer:

When x

20

x

x a

b

20 a

b

20 If a

134. Isolate the radical by subtracting x from both sides of the equation. Then square both sides to eliminate the radical, and rewrite the equation in standard form. The solutions can now be found by using the Quadratic Formula. Each solution must be checked because extraneous solutions may be included.

20,

20 So, a

0

b

b

20

20, b

83

20

Chapter Test for Chapter P 2 3

1.

x

1

12 ª¬ 23 x 1

1x 4

1210

1 xº 4 ¼

8 x 1 3 x

120

8 x 8 3x

120

11x

128

3.

3x 1

3 x 2

14

x2 x 6

14

x 2 x 20

0

4 x 5

0

x

x2

5. x

3x

8

3x

6

x

8 3

x

2

11 d x 3 2 11

−2

3 x

−6

−4

−2

0

2

4

8. x 15 t 5

x 15 d 5 or

x 15 t 5

x d 10

x t 20 x

or x

5

5

10

15

20

2 x 2 5 x ! 12

9. 4.

7

11 d 2 x 6

0

4

x

7 or 3x 1

3 d 2 x 8 14

x 2 4 0 4 x 2 x 2 x 2 4 x 2 1 z 4 No solution

x

7

7. 3 d 2 x 4 14

128 11

x 2.

6. 3 x 1

10

x4 x2 6

0

2 x 2 3

0

x2

2

x

r

2

Critical numbers: x

x2

3 x

r

3i

Test intervals: f, 4 , 4,

2x 1

x 2x 1 x x 4

0

x

0

x 4

3 , 2

Solution set: f, 4

2x 1 0

3 x 4 ! 0

x

4

3 2

, 32 , f

Test: Is 2 x 3 x 4 ! 0? 2x 1

x 4x 2

2 x

1

x 1 2

2 x 2 5 x 12 ! 0

32 , f

In inequality notation: x 4 or x !

3 2

3 2 x

0 x

−5 −4 −3 −2 −1 0

1 2

3

4

4 is a solution to the original equation. Only x x 0 is extraneous.

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84

NOT FOR SALE

Chapter P

Prerequisites requisites 2 5 ! x x 6

10.

15. y

3 2

y-intercept: 0, 1

16. Center: 1, 3

! 0 4, x

6

0, x

x

6 5 4

1 y 3 2

2

(− 3, 3)

16

! 0?

2 6

2

89

y 3

2 x 4

y

2 x 1

m y 6

(− 2, 5)

y 6

5

1.7 x 11.9 1.7 x 5.9

y

1

(6, 0)

−2 −1

1

2

3

4

5

x

6

19. 5 x 2 y

3 5 x 3

2y

−2

4, 2 2, 4 3, 1

5 3 x 2 2

y

(a) m

3 5x 4

y-intercept: 0, 3

1

3

(0, 3)

y 4

2

( ( 3, 0 5

−4 −3 −2 − 1

No axis or origin symmetry

5 , 0, 4 2

y 4

y

53 , 0

1

2

x 3

y

4

−2 −3 −4

4 x

y-intercept: 0, 4

6 0.8 1.7 7 3 1.7 x 7

2

y

(b) m

y 4

6 5 4

(0, 4)

y 4

3

(− 4, 0)

2

(4, 0)

1

y-axis symmetry

x

5 x 0 2 5 x 2 5 x 4 or 5 x 2 y 8 2

0

2 , 0, 4 5

y

x-intercepts: r 4, 0

2

18. 3, 0.8 and 7, 6

3 2

14. y

4 2 2 x 2

y 3

§ 5· ¨ 2, ¸ © 2¹ 5 0

9 3

m

6

64 25

x-intercept:

4 5 6

17. 2, 3 and 4, 9

§ 2 6 5 0 · , 11. Midpoint: ¨ ¸ 2 ¹ © 2

2 2, 1 3 4 2, 1 3 5 2, 2 3

x

−4 −3 −2

x

Distance: d

(5, 3)

−2

x x 6

4

r=4 (1, 3)

1

In inequality notation: x 6 or 0 x 4

13. y

x 4

y

Radius: 4

Solution set: f, 6 0, 4

12.

3

−4

Standard form:

2

2

(0, −1)

8

3 x 4

0

1

−2

Test intervals: f, 6 , 6, 0 , 0, 4 , 4, f

−8 −6 −4 −2

(1, 0)

−4 −3 −2 − 1 −3

Critical numbers: x

Test: Is

1

(− 1, 0)

y-axis symmetry

3 x 12 ! 0 x x 6 x x 6

y 4

x-intercepts: r1, 0

2 5 ! 0 x x 6 2 x 6 5 x ! 0 x x 6

3 x 4

x2 1

y

2 x 0 5 2 x 5 2 x 4 or 5

2 x 5 y 20

0

INSTRUCTOR C USE ONLY − 4 − 3 −2

−1

1

2

3

4

−2 −2

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE

Problem Solving ffo for Chapter P

0.067 x 5.638

20. (a) y

21. r

220 A

220 20

85

200 beats per minute

0.50 200 d r d 0.85 200

5

100 d r d 170 The target heartrate is at least 100 beats per minute and at most 170 beats per minute. 75

150 0

(b) From the graph you see that y t 3 when x t 129. Algebraically:

3 d 0.067 x 5.638 8.638 d 0.067 x x t 129

IQ scores are not a good predictor of GPAs. Other factors include study habits, class attendance, and attitude.

Problem Solving for Chapter P 1. (a) 3 x 4 x 4 2 2

x 4.

Let u

3u 2 u 2

3u

2. (a)

0

0, a z 0, b z 0

x ax b

0

0 or ax b

x

0

2 u 1

ax 2 bx

0

b

x

1

u

2, 3

x

u 4

, 5 10 3

(b)

0

ax

b a

ax 2 a b x b

0, a z 0, b z 0

(b) 3 x 2 8 x 16 x 4 2

0

ax ax bx b

0

3 x 2 24 x 48 x 4 2

0

ax x 1 b x 1

0

3 x 2 25 x 50

0

3 x

0

x

10 x 5

103 , 5

2

ax ax b

b x 1 0

or

ax

b

x

0

x 1

0

x

1

b a

3. (a) x 2 bx 4 0 To have at least one real solution,

b 2 41 4 t 0 b 2 16 t 0. Critical numbers: b

r4

Test intervals: f, 4 , 4, 4 , 4, f Test: Is b 2 16 t 0? By testing values in each test interval, we see that b 2 16 is greater than or equal to zero on the intervals f, 4@ >4, f . (b) x 2 bx 4 0 To have at least one real solution, b 2 41 4 t 0 b 2 16 t 0. This is true for all values of b: f b f

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86

Chapter P

NOT FOR SALE

Prerequisites requisites

(c) 3x 2 bx 10

0

To have at least one real solution, b 2 43 10 t 0 b 2 120 t 0.

Critical numbers: b

r 120

r 2 30

Test intervals: f, 2 30 , 2 30, 2 30 , 2 30, f

Test: Is b 2 120 t 0? By testing values in each test interval, we see that b 2 120 is greater than or equal to zero on the intervals

f, 2

30 ª¬2 30, f .

(d) 2 x 2 bx 5

0

To have at least one real solution, b 2 4 2 5 t 0 b 2 40 t 0.

This is true for b d 2 10 or b t 2 10, f, 2 10 º¼ ª¬2 10, f . (e) If a ! 0 and c d 0, then b can be any real number since b 2 4ac would always be positive. If a ! 0 and c ! 0, then b d 2 ac or b t 2 ac , as in parts (a), (e), and (d). (f ) Since the intervals for b are symmetric about b

0, the center of the interval is b

4. (a) Estimate from the graph: when the plate thickness is 2 millimeters, the frequency is approximately 330 vibrations per second.

(b) Estimate from the graph: when the frequency is 600, the plate thickness is approximately 3.6 millimeters. (c) Estimate from the graph: when the frequency is between 200 and 400 vibrations per second, the plate thickness is between 1.2 and 2.4 millimeter. (d) Estimate from the graph: when the plate thickness is less than 3 millimeters, the frequency is less than 500 vibrations per second. 5. (a) Since there are three solutions, the equation is not linear nor is it quadratic. Neither

(b) Since there is only one solution, the equation could have been either linear or quadratic. Both (c) Since there are two solutions, the equation is not linear but could be quadratic. Quadratic (d) Since there are four solutions, the equation is not linear nor is it quadratic. Neither

0. x2 6x y2 8 y

6. (a)

x

2

6 x 9 y 8 y 16 2

x

3 y 4

Center: 3, 4

2

2

0 9 16 25

Radius: 5

(b) Slope of line from 0, 0 to 3, 4 is

4 . Slope of 3

3 tangent line is . Hence, 4 y 0

3 x 0 y 4

3 x. Tangent line 4

(c) Slope of line from 6, 0 to 3, 4 is Slope of tangent line is y 0 3 (d) x 4 3 x 2 x

40 36

4 . 3

3 . Hence, 4

3 x 6 y 4

3 9 x . Tangent line 4 2

3 9 x 4 2 9 2 3

9· § Intersection: ¨ 3, ¸ 4¹ ©

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE

Problem Solving ffo for Chapter P

7.

d1d 2

1

ª x 1 y 2 ºª x 1 y 2 º ¬ ¼¬ ¼

1

2

x

1 x 1 y 2

2

x

x ¬

2ª

2

1 x 1 º y ¼ 2

2

1

x 2x 1 2x y 2 y y

1

2x2 y2 y 4 2x2 2 y 2

0

2

x4

2

2 2

2

x2 2 x2 x

0. Then x 4

y2

0 or x 2

1

(

2

2, 0)

(0, 0) x

−1

4

2

2, 0)

1

1 y ª¬2 x 2º¼ y 2

2

y

(−

4

4

Let y

4

87

1

−1

2 x 2 y 2

2. Thus, 0, 0 ,

8. The original point is x, y .

2, 0 and

2, 0 are on the curve.

11. (a) and (b)

(a) The transformed point x, y is a reflection through the y-axis. (b) The transformed point x, y is a reflection through

x

165

184

150

210

196

240

y

170

185

200

255

205

295

1.3 x 36

179

203

159

237

219

276

x

202

170

185

190

230

160

y

190

175

195

185

250

155

1.3 x 36

227

185

205

211

263

172

the x-axis. (c) The transformed point x, y is a reflection through the origin. 7000 5500 9. (a) 10

150 students per year

(b) 2003: 5500 3150

5950 students

2007: 5500 7150

6550 students

2009: 5500 9150

6850 students

(c) Equation: y 150x 5500 where x 0 represents 2000. Slope: m

150

This means that enrollments increase by approximately 150 students per year. 0.23t 22.3

10. Milk: M

Bottled Water: B (a) B

1.87t 16.1

M

1.87t 16.1 2.1t

0.23t 22.3

300

140 150

250

(c) One estimate is x t 182 pounds. (d) 1.3 x 36 t 200 1.3x t 236 x t 181.5385 | 181.54 pounds (e) An athlete's weight is not a particularly good indicator of the athlete's maximum bench-press weight. Other factors, such as muscle tone and exercise habits, influence maximum bench press weight.

6.2

t | 2.952 Point of intersection: 2.952, 21.621 (b) and (c)

25

(2.952, 21.621) 0

7 15

(d) Per capita consumption of milk was equal to per capita consumption of bottled water in 2002.

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88

Chapter P

NOT FOR SALE

Prerequisites requisites

12. S | 22.6t 94

t

s

2

140.8

3

158.7

4

182.1

5

207.9

6

233.0

7

255.4

8

270.3

400

0

s

10 50

300 when t | 9, which corresponds to 2009.

13. (a) Choice 1: W

3000 0.07s

Choice 2: W

3400 0.05s

(b) 3000 0.07s 0.02s s

3400 0.05s 400 $20,000

The salaries are the same ($4400 per month) when sales equal $20,000. (c) An ambitious salesperson who call sell more than $20,000 per month would be wise to select choice 1. A more conservative choice for a salesperson who is unsure of the market for this product would be choice 2.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE C H A P T E R 1 Functions and Their Graphs Section 1.1

Functions...............................................................................................90

Section 1.2

Analyzing Graphs of Functions .........................................................100

Section 1.3

Transformations of Functions ............................................................116

Section 1.4

Combinations of Functions ................................................................126

Section 1.5

Inverse Functions................................................................................135

Section 1.6

Mathematical Modeling and Variation..............................................145

Review Exercises ........................................................................................................154 Chapter Test .............................................................................................................164 Problem Solving .........................................................................................................167

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE C H A P T E R 1 Functions and Their Graphs Section 1.1 Functions 1. domain; range; function

18. (a) The element c in A is matched with two elements, 2 and 3 of B, so it is not a function.

2. verbally; numerically; graphically; algebraically 3. independent; dependent

(b) Each element of A is matched with exactly one element of B, so it does represent a function.

4. piecewise-defined

(c) This is not a function from A to B (it represents a function from B to A instead). (d) Each element of A is matched with exactly one element of B, so it does represent a function.

5. implied domain 6. difference quotient 7. Yes, the relationship is a function. Each domain value is matched with exactly one range value.

19. Each is a function. For each year there corresponds one and only one circulation. 20. Reading from the graph, f 2002 is approximately 9

8. No, the relationship is not a function. The domain value of –1 is matched with two output values. 9. No, the relationship is not a function. The domain values are each matched with two range values. 10. Yes, it is a function. Each domain value is matched with only one range value.

million. 21. x 2 y 2

13. Yes, it does represent a function. Each input value is matched with exactly one output value. 14. No, the table does not represent a function. The input values of 0 and 1 are each matched with two different output values.

22. x 2 y 2

y

23. x 2 y

r 16 x 2

4 y

4 x2

Yes, y is a function of x. 24. y 4 x 2

y

36 4 x 2 36

Yes, y is a function of x. 25. 2 x 3 y

4 y

1 3

4

2 x

Yes, y is a function of x. 26. 2 x 5 y

16. Yes, the table does represent a function. Each input value is matched with exactly one output value.

y

17. (a) Each element of A is matched with exactly one element of B, so it does represent a function.

(c) Each element of A is matched with exactly one element of B, so it does represent a function.

16

No, y is not a function of x.

15. No, it does not represent a function. The input values of 10 and 7 are each matched with two output values.

(b) The element 1 in A is matched with two elements, –2 and 1 of B, so it does not represent a function.

r 4 x2

No, y is not a function of x.

11. No, the relationship is not a function. The domain values are each matched with three range values. 12. Yes, the relationship is a function. Each domain value is matched with exactly one range value.

4 y

10 2 x 2 5

Yes, y is a function of x. 27.

x

2 y 1 2

2

y

25 r

25 x 2 1 2

No, y is not a function of x.

(d) The element 2 in A is not matched with an element of B, so the relation does not represent a function.

INSTRUCTOR USE ONLY 90

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE

Section 1.1 1.

28.

x

2 y 2

4

y

r

2

41. V r 4 x 2

No, y is not a function of x.

x2 1 y

29. y 2

r

(b) V

x2 1

4S 3

32

(c) V 2r 42. S r

4

4S 2

(b) S

12

4S

(c) S 3r

x 5

43. g t

4 x

or

y

4 x

No, y is not a function of x. 35. x

12

4 2 3 2 5 2

15 (b) g t 2

4t 2 3t 2 5 2

(c) g t g 2 or

4t 2 3t 5 15 4t 2 3t 10

75 0 x

y

44. ht

Yes, y is a function of x. 37. y 5

5 or y

0

x

1

45. f y

21 3

(b) f 3 (c) f x 1 40. g y

9

2 x 1 3

2x 5

7 3y

(a) g 0 (b) g

1

2 3 3

7 3

7

0

(c) g s 2

7 3

2

x

0

21.5

2 2 x 2 2

3

4

x2 2 x

1

(b) f 0.25

3

0.25

(c) f 4 x 2

3

4 x2

46. f x

0.75

y

2.5

3 2 x

x 8 2

(a) f 8 (b) f 1

7 30

7 3

3

(a) f 4

2x 3

(a) f 1

1.5

(c) h x 2

No, this is not a function of x. 39. f x

22 2 2

(b) h1.5

0x 5

Yes, y is a function of x. 38. x 1

t 2 2t

(a) h 2

0

y

2

4t 2 19t 27

14

75

32 S r3 3

2

No, this is not a function of x. 36. y

8r 3

4t 2 3t 5

(a) g 2

Yes, y is a function of x. 4 x y

4 S 3

3

36S r 2

4 x

y

9 S 2

2

4S 3r

Yes, y is a function of x.

34.

S

Yes, y is a function of x.

33. y

4 S 27 3 8

2r

36S

16S

16 x 2

32. y

3

27

4S r 2

No, y is not a function of x. 31. y

4S 3

4 S 3

(a) S 2

r 4 x

y

3 3

4S 3 3 2

No, y is not a function of x. 30. x y 2

91

4S r 3 3

(a) V 3

2

Functions

8 1

(c) f x 8

8 2

8 2

x

2 5

8 8 2

x 2

7 3 s 2

INSTRUCTOR USE ONLY 7 3s 6

1 3s

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92

NOT FOR SALE

Chapter 1

Functions ctions and Their Graphs

1 x 9

47. q x

(a) q0

1 02 9

(b) q3

1 is undefined. 32 9

1 9

(a) f 2

1

(c) q y 3

48. qt

y

1 y2 6 y

3 9 2

2 2 3 2

2

83 4

2

11 4

20 3 2

(b) q0

(c) f 2

2 2 2

2 x 3

x

2x 3 x2 2

2

49. f x

x 2

(a) f 2

1

2 2

(b) f 2

(a) f 2

2

50. f x

1, if x 1 ® if x ! 1 ¯1,

x 1

(c) f x 1

x 1

x 4

(a) f 2

2 4

6

(b) f 2

2 4

6

(c) f x 2

x2 4

x2 4

51. f x

2 x 1, x 0 ® ¯2 x 2, x t 0

(a) f 1

2 1 1

1

(b) f 0

20 2

2

(c) f 2

2 2 2

6

3 10

3 2 1

(b) f 12

54. f x

32

9

4 5 x, x d 2 ° 2 x d 2 ®0, °x 2 1, x ! 2 ¯

4 5 3

4 2

(c) f 1

0

2

f 1

1

f 0

0

f 1

1

f 2

2

f x

1

19 17

x2 3

f 2

x

7

4

(b) f 4

55. f x

1

2

6

2

(a) f 3

x

2

3 x 1, x 1 ° 1 d x d 1 ®4, °x 2 , x !1 ¯

Division by zero is undefined. 2

2

1

(c) f 3

0 2

(c) q x

2 2

(b) f 1

53. f x

2t 3 t2 2

(a) q 2

2 °x 2, x d 1 ® 2 °¯2 x 2, x ! 1

52. f x

2

2

2

2

3

1

2

3

2

3 3

2

3 2

3

1

–2

1

0

1

2

1

–2

–3

–2

1

56. g x

x 3

g 3

33

0

g 4

43

1

g 5

53

2

g 6

63

3

g 7

7 3

x

3

4

g x

0

1

2 5

6 2

7 3

2

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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NOT FOR SALE

Section 1.1 1.

57. ht

58.

1 2

60. f x

t 3 1 2

5 3

1

h 4

1 2

4 3

1 2

f 1

9 1

h 3

1 2

3 3

0

f 2

9 2

h 2

1 2

2 3

1 2

f 3

h 1

1 2

1 3

1

3 3 4 3 5 3

–5

ht

1

f s f 0 f 1 § 3· f¨ ¸ © 2¹ §5· f¨ ¸ © 2¹ f 4

s

–4 1 2

59. f x

f 5

–2 1 2

0

0 2 1 2 1 2

2 2

1

1 1

1

3 2 2 3 2 2 5 2 2 5 2 2 4 2

2 2

4 2

–1

–1

1

5 2

–1

12 1 4

4 12

f 0

12 0 4

f 1

1 2

f 2

2

2

2

–2 5

5

8

5

0

1

2

5 5x 1 0 15

63.

3x 4 5 3x 4

0 0 4 3

1

12 x 2 5

f x

64.

12 x 2 5

0

x2

12 r 12

x 9 2

65. x 2 9

0

2

9

x

r3

x

4 1

2

2

4

x

x

° 1 x 4, x d 0 2 ® 2 °¯ x 2 , x ! 0

f 1

1

3

4

1

5

0

0

x 3 2

5

2

15

5x 1

1

1

8 2

1

3x

f x

1

12 12

12 2 4

f x

f x

62.

12 1 2

f 2

x

x

1

61. 15 3 x

s 2 0 2

2

–1

s 2

0 f s

f 4

–3

93

2 °9 x , x 3 ® °¯x 3, x t 3

h 5

t

Functions

f x

66.

0

x 2 8 x 15

–1

0

1

2

9 2

4

1

0

x

5 x 3

r2 3

x 2 8 x 15 0 0

x 5

0 x

5

x 3

0 x

3

x3 x

0

x x 1

0

x x 1 x 1

0

1,

or

67.

2

INSTRUCTOR USE ONLY x

0, x

x

1

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94

NOT FOR SALE

Chapter 1

Functions ctions and Their Graphs f x

68.

x3 x 2 4 x 4

x x 4x 4 3

2

x x 1 4 x 1 2

x

1 x 4 2

x 1

f x x2 x 2 2 x 1

x 2 x

x 3

1

x

r2

0 x 2

0

3

2

x f x

g x

x4 4x2

0

0

x x 4

0

x 2 x 2 x 2

0

2

1

2

2

2x2

2

0 x

0

x 2

0 x

2

x 2

0 x

2

x

g x

x 4

2 x

x 6

0

0

x 2

4

x 3

0

x

3, which is a contradiction, since

x 2

0

x

2 x

x represents the principal square root.

4 78. f t

5x2 2 x 1

t 4

3

Because f x is a polynomial, the domain is all real

Because f t is a cube root, the domain is all real

numbers x.

numbers t.

74. f x

1 2x2

79. g x

Because f x is a polynomial, the domain is all real

4 t

75. ht

s y

0.

3y y 5

The domain is all real numbers y except y

5.

2.

0, x

2.

10 x 2x

x2 2 x z 0 x x 2 z 0

81. f s

y z 5

0, x

2

The domain is all real numbers x except x

y 5 z 0

77. g y

h x

80.

The domain is all real numbers t except t

1 3 x x 2

The domain is all real numbers x except x

numbers x.

76.

0

0

f x

73. f x

0

3 x 2

x 2x

x

x 5x 6

x 2

2

x 3

7x 5

2

71.

0

x

x

0 x

0 x 1

72.

0

g x

x2

x

x 2x 1

0 x

x 4

69.

0

g x

2

0

2

f x

70.

s 1 s 4

Domain: s 1 t 0 s t 1 and s z 4 The domain consists of all real numbers s, such that s t 1 and s z 4.

y 10

Domain: y 10 t 0 y t 10

INSTRUCTOR USE ONLY The domain is all real numbers y such that y t 10.

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NOT FOR SALE

Section 1.1 1.

x 6 6 x

82. f x

Domain: x 6 t 0 x t 6 and x z 6 The domain is all real numbers x such that x ! 6 or 6, f . x 4 x

83. f x

The domain is all real numbers such that x ! 0 or 0, f . x 2 x 10

84. f x

x ! 10

The domain is all real numbers x such that x ! 10. f x

f 2

2 1

1

f 1

1 1

0

f 0

01

1

f 1

11

2

f 2

21

3

^2, 1 , 1, 0 , 0, 1 , 1, 2 , 2, 3 ` 89. No. The element 3 in the domain corresponds to two elements in the range.

91. A

x2

f 2

2

f 1

1

f 0

0

f 1

2

f 2

x 1

2

1

2

2

A

4

2

1

92. A

0

4

86.

f x

x

3

2 3

f 1

1 3 2 0

3

f 1

1 3

f 2

2 3

2

A

S¨

C2 4S

2

s2

b

2 s 2 64

b

1 bh 2 1 2 s 2 64 8 2

Thus, A

1

s 8

4 s 2 64

4

2

s

s 2 64

b2

^2, 25 , 1, 16 , 0, 9 , 1, 4 , 2, 1 ` 87.

2

9

2

s

2S r

b2 4

16

P 4

P2 16

§C · ¸ © 2S ¹

25

2

2

S r2, C

§b· 93. 82 ¨ ¸ © 2¹

2

f 2 f 0

§P· ¨ ¸ ©4¹

C 2S

^2, 4 , 1, 1 , 0, 0 , 1, 1 , 2, 4 `

4s

s 2 and P

r

1

2

95

90. An advantage to function notation is that it gives a name to the relationship so it can be easily referenced. When evaluating a function, you see both the input and output values.

x 10 ! 0

85.

f x

88.

Functions

8 s 2 64 square inches.

f x

x 2

f 2

2 2

4

f 1

1 2

3

f 0

0 2

2

f 1

1 2

3

f 2

2 2

4

94. A

s2

1 bh, and in an equilateral triangle b 2 §s· h2 ¨ ¸ © 2¹

2

h

§s· s2 ¨ ¸ © 2¹

h

4s 2 s2 4 4

^2, 4 , 1, 3 , 0, 2 , 1, 3 , 2, 4 `

1 s 2

s

3s 2

2

3s 2 3s 2 4

INSTRUCTOR USE ONLY A

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96

NOT FOR SALE

Chapter 1

95. (a)

Functions ctions and Their Graphs

Height, x

Volume, V

1

484

V

(b)

(c) V

1200

800

3

972

4

1024

2

Domain: 0 x 12

1000

Volume

2

x 24 2 x

800 600 400 200 x 1

5

980

6

864

4

5

6

V is a function of x.

96. (a) The maximum profit is $3375.

4 Revenue Cost

(c) Profit

price per unit number of units cost number of units ª¬90 x 100 0.15 º¼ x 60 x, x ! 100 90 0.15 x 15 x 60 x 105 0.15 x x 60 x

P 3400 3350

Profit

3

Height

The volume is maximum when x and V 1024 cubic centimeters. (b)

2

3300 3250 3200 3150 3100

105 x 0.15 x 2 60 x

x 110

130

150

170

Order size

45 x 0.15 x 2 , x ! 100

Yes, P is a function of x. 1 bh 2

97. A

1 xy 2

Because 0, y , 2, 1 , and x, 0 all lie on the same line, the slopes between any pair are equal. 1 y 2 0 1 y 2 y y

So, A

0 1 x 2 1 x 2 2 1 x 2 x x 2 1 § x · x¨ ¸ 2 © x 2¹

y 4

(0, y)

3 2

(2, 1) (x, 0)

1

x 1

2

3

4

x2 . 2 x 2

The domain of A includes x-values such that x 2 ª¬2 x 2 º¼ ! 0. By solving this inequality, the domain is x ! 2. 98. A

Aw

2 x y

2 xy

But y 36 x 2 , so A domain is 0 x 6.

99.

2 x 36 x 2 . The

y y30

1 x 2 3x 6 10 1 30 10 330 6 2

6 feet

If the child holds a glove at a height of 5 feet, then the ball will be over the child's head because it will be at a height of 6 feet

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NOT FOR SALE

Section 1.1 1.

Functions

97

10.6t 699, 0 d t d 4 ® ¯15.5t 637, 5 d t d 7

100. d t

2000: Use t

0 and find d 0 .

d 0

10.60 699

2001: Use t

1 and find d 1 .

d 1

10.61 699

2002: Use t

2 and find d 2 .

d 2

10.6 2 699

720.2 million

2003: Use t

3 and find d 3 .

d 3

10.63 699

730.8 million

2004: Use t

4 and find d 4 .

d 4

10.6 4 699

741.4 million

2005: Use t

5 and find d 5 .

d 5

15.55 637

714.5 million

2006: Use t

6 and find d 6 .

d 6

15.56 637

730.0 million

2007: Use t

7 and find d 7 .

d 7

15.57 637

745.5 million

699.0 million

709.6 million

°1.011t 2 12.38t 170.5, 8 d t d 13 ® 2 °¯6.950t 222.55t 1557.6, 14 d t d 17

101. pt

1998: Use t

8 and find p8 .

p8

1.0118 12.38t 170.5

136.164 thousand

$136,164

1999: Use t

9 and find p9 .

p9

1.0119 12.389 170.5

140.971 thousand

$140,971

2000: Use t

10 and find p10 .

p10

1.01110 12.3810 170.5

147.800 thousand

2001: Use t

11 and find p11 .

p11

1.01111 12.3811 170.5

156.651 thousand

2002: Use t

12 and find p12 .

p12

1.01112 12.3812 170.5

167.524 thousand

$167,524

2003: Use t

13 and find p13 .

p13

1.01113 12.3813 170.5

180.419 thousand

$180,419

2004: Use t

14 and find p14 .

p14

6.95014 222.5514 1557.6

195.900 thousand

$195,900

2005: Use t

15 and find p15 .

p15

6.95015 222.5515 1557.6

216.900 thousand

$216,900

2006: Use t

16 and find p16 .

p16

6.95016 222.5516 1557.6

224.000 thousand

$224,000

2007: Use t

17 and find p17 .

p17

6.95017 222.5517 1557.6

217.200 thousand

$217,200

2

2

2

2

2

2

2

2

2

2

102. (a) V Awh x yx x 2 y where 4 x y 108. So, y 108 4 x and x 2 108 4 x

V

(b)

12.30 x 98,000

C

R

Revenue Cost

(c) The dimensions that will maximize the volume of the package are 18 u 18 u 36. From the graph, the maximum volume occurs when x 18. To find the dimension for y, use the equation y 108 4 x. 108 4 x

P

17.98 x 12.30 x 98,000

P

5.68 x 98,000

104. (a) Model: Total cost

30 0

y

17.98 x

(c) Profit

12,000

0

price per unit u number of units

(b) Revenue

Domain: 0 x 27

108 418

108 72

36

Fixed costs

Labels:

C x

Variable costs

Total cost

C

Fixed cost

6000

Variable costs Equation: (b) C

$156,651

variable costs fixed costs

103. (a) Cost

108 x 2 4 x3.

$147,800

C

0.95 x

6000 0.95 x

6000 0.95 x x

6000 0.95 x

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98

NOT FOR SALE

Chapter 1

n rate

105. (a) R

n ª¬8.00 0.05 n 80 º¼ , n t 80

12.00n 0.05n 2

R

(b)

Functions ctions and Their Graphs

12n

240n n 2 , n t 80 20

n2 20

n

90

100

110

120

130

140

150

R n

$675

$700

$715

$720

$715

$700

$675

The revenue is maximum when 120 people take the trip. 106. F y

(a)

149.76 10 y 5 2

y

5

10

20

30

40

F y

26,474.08

149,760.00

847,170.49

2,334,527.36

4,792,320

The force, in tons, of the water against the dam increases with the depth of the water. (b) It appears that approximately 21 feet of water would produce 1,000,000 tons of force. (c)

149.76 10 y 5 2

1,000,000 1,000,000 149.76 10

y5 2

2111.56 | y 5 2 21.37 feet | y 107. (a)

f 2007 f 2000

80.0 35.4 | 6.37 2007 2000

2007 2000

Approximately 6.37 million more tax returns were made through e-file each year from 2000 to 2007. (b) Number of tax returns (in millions)

N 80 70 60 50 40 30 t 1

2

3

4

5

6

7

Year (0 ↔ 2000)

(c) Use the points 0, 35.4 and 7, 80.0 .

m N (d)

80.0 35.4 7 0 6.37t 35.4

6.37

t

0

1

2

3

4

5

6

7

N

35.4

41.8

48.1

54.5

60.9

67.3

73.6

80.0

(e ) Using a graphing utility yields the model N 6.56t 34.4 . Compared to the model in part (c), the model generated by the graphing utility produces values that reflect the data more accurately. 108. (a)

(b)

3000

2

h2 h

d2 d 2 3000

2

Domain: d t 3000 (because both d t 0 and d 2 3000 t 0 ) 2

d h

INSTRUCTOR STR STR R RU USE ONLY 3000 ft

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE

Section 1.1 1.

f x

109.

x2 x 1

f 2 h

2 h 2

f x

110.

2 h 1

f 5 h

55 h 5 h

h 3h 3

25 5h 25 10h h 2

2

21

f 2 h f 2

h 2 3h

f 2 h f 2

h 2 3h h

h

h 2 5h

3 f 5

55 5

f 5 h f 5

h 5h h h h 5

0

2

h f x

2

25 25

h 3, h z 0

h

111.

2

25 5h 25 10h h 2

2

99

5x x2

4 4h h 2 2 h 1 2

f 2

Functions

h 5 , h z 0

x3 2 x 1 ª x c 3 2 x c 1º x3 2 x 1 ¬ ¼ c

f x c f x c

x 3 3 x 2 c 3xc 2 c3 2 x 2c 1 x3 2 x 1 c 3 x 2 c 3 xc 2 c 3 2c c 3 x 3xc c 2, 2

f x

112.

2

c 3 x 2 3 xc c 2 2 c

c z 0

x3 x 1 ª x c 3 x c 1º x3 x 1 ¬ ¼ c

f x c f x c

x3 3x 2 c 3 xc 2 c3 x c 1 x3 x 1 c 3 x 2 c 3xc 2 c3 c c 3 x 3 xc c 1, 2

113. g x

c c z 0

3x 1

g x g 3

3 x

x 3

f 1 f t f 1 t 1 115. f x

1 t 1 1

3x 9 x 3

3 x 3 x 3

3, x z 3

1

1 1 t t 1

1 t t t t 1

5x

f x f 5 x 5

1 8 x 3

f t

114.

2

c 3x 2 3xc c 2 1

1t t t 1

§ 1· ¨ ¸ t 1 © t¹ t 1

1 , t 116.

5x 5 x 5

t z 1 f x

x2 3 1

f 8

82 3 1

f x f 8 x 8

5

x2 3 1 5 x 8

x2 3 4 x 8

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100

NOT FOR SALE

Chapter 1

Functions ctions and Their Graphs

117. By plotting the points, we have a parabola, so g x cx 2 . Because 4, 32 is on the graph, you

2. So, g x

c 4 c 2

have 32

127. f x

x 1

Domain: x t 1

g x

1 x 1

Domain: x ! 1

2 x 2 .

The value 1 may be included in the domain of f x

118. By plotting the data, you can see that they represent a

cx. Because 0, 0 and 1,

the line, the slope is 14 . So, f x

1 x. 4

a zero to occur in the denominator which results in the function being undefined. 128. Because f x is a function of an even root, the radicand

119. Because the function is undefined at 0, we have r x c x. Because 4, 8 is on the graph, you

have 8

as it is possible to find the square root of 0. However, 1 cannot be included in the domain of g x as it causes

14 are on

line, or f x

32. So, r x

c 4 c

cannot be negative. g x is an odd root, therefore the

32 x.

radicand can be any real number. So, the domain of g is all real numbers x and the domain of f is all real numbers x such that x t 2.

120. By plotting the data, you can see that they represent

h x

c

4

x . Because

2 and

and the corresponding y-values are 6 and 3, c h x

3

1

1, 3 and

x. x 2 4 is a relation between

121. False. The equation y 2

x and y. However, y a function.

r

x 2 4 does not represent

122. True. A function is a relation by definition. 123. True.

As long as all elements in the domain are matched with elements in the range, even if it is the same element, then the relation is a function. 124. False.

Each element in the domain is matched with exactly one element in the range in a function. 125. False. The range is >1, f . 126. True. The set represents a function. Each x-value is mapped to exactly one y-value.

129. No; x is the independent variable, f is the name of the function. 130. (a) Answers will vary. Sample answer: A relation is a rule of correspondence between two variables. A function is a particular relation that assigns exactly one output (y) to each input (x). (b) The domain of a function is the set of all inputs of the independent variable for which the function is defined. The range of a function is the set of all possible outputs. 131. (a) Yes. The amount that you pay in sales tax will increase as the price of the item purchased increases. (b) No. The length of time that you study the night before an exam does not necessarily determine your score on the exam. 132. (a) No. During the course of a year, for example, your salary may remain constant while your savings account balance may vary. That is, there may be two or more outputs (savings account balances) for one input (salary). (b) Yes. The greater the height from which the ball is dropped, the greater the speed with which the ball will strike the ground.

Section 1.2 Analyzing Graphs of Functions 1. ordered pairs

8. even

2. Vertical Line Test

9. Domain: f, 1@ >1, f

3. zeros

Range: >0, f

4. decreasing

10. Domain: f, f

5. maximum

Range: >0, f

6. step; greatest integer

11. Domain: >4, 4@

INSTRUCTOR USE ONLY 7. odd

Range: >0,, 4@

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© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.2

12. Domain: f, 1 1, f

A vertical line intersects the graph at most once, so y is a function of x.

13. Domain: f, f ; Range: >4, f

(a) f 2

0

(b) f 1

1

(c) f

12

(d) f 1

y 2

22. x

A vertical line intersects the graph more than once, so y is not a function of x. f x

23.

0

2 x 2 7 x 30

2 x 2 7 x 30

2

2 x

14. Domain: f, f ; Range: f, f

(a) f 1

0

5 x 6

2x 5 x

4

0

0

or

x 6

0

52

or

x

6

(b) f 2

4

(c) f 0

2

3x 2 22 x 16

0

0

3x 2 x 8

0

(d) f 1

f x

24.

15. Domain: f, f ; Range: 2, f

(a) f 2

0

(b) f 1

1

(c) f 3

2

(d) f 1

0

(c) f 0

1

(d) f 2

3

17. y

x 9x2 4 0 0

0

x 2

0 x

2

f x x

x 2x 3

x x 2

r

0

7

2

1 y

x 2 9 x 14 4x

0 x

1 x3 2

A vertical line intersects the graph at most once, so y is a function of x.

0 or

x

x

1 x3 2

0 20 0 x2 2

0

2

2

x

r 2

x

x 1

y is not a function of x. Some vertical lines intersect the graph twice. 20. x 2 y 2

8

x 7

27.

1 x3 4

19. x y

0 x

x 2 9 x 14 4x x 7 x 2

1 x2 2

2

x 8

f x

A vertical line intersects the graph at most once, so y is a function of x. 18. y

2 3

26.

3

(b) f 1

0 x

x 9x2 4 x

16. Domain: f, f ; Range: – f, 1@

3 x 2 22 x 16

3x 2

f x

25.

3

(a) f 2

101

2 xy 1

21. x 2

Range: 1, 1

Analyzing Graphs of Functions

25

f x

28.

x3 4 x 2 9 x 36

x3 4 x 2 9 x 36

0

x 2 x 4 9 x 4

0

4 x 2 9

0

A vertical line intersects the graph more than once, so y is not a function of x.

x x 4

0 x

4

INSTRUCTOR USE ONLY x 9 2

0 x

r3

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102

Chapter 1

29. f x

Functions ctions and Their Graphs

4 x3 24 x 2 x 6

0

4 x 2 x 6 1 x 6

0

x 6 4 x x 6 2 x 1 2 x x 6 x

6,

12 ,

9 x 25 x

2x 1

0

x

1 2

0

x 9 x 25

0

2

0 x

0

9 x 2 25

0 x

r 53

x x 7

0

x

0

x 7

0 x

−6

3

−1

11 2

Zero: x

2x 1

f x

(b)

2 x 11

0

2x

1

2 x 11

0

2x

1

2 x 11

0

x

1 2

36. (a)

3x 2

4 −4

3x 2

0

3x 2

0

23

x

28

−12

Zero: x 33. (a)

6

9

f x 5 3 x 3x 5 x

5 3

3

0

3 x 14

8

3 x 14

64

x

26

37. (a)

5 x

3x 14 8

3 x 14 8

−6

Zero: x

26 f x

(b)

−9

(b)

11 2

x

f x

7

5

2x 1

32.

7

x x 7

35. (a)

x2

f x

0, x

f x

(b)

9 x 4 25 x 2

2

31.

Zeros: x

0

0,

4

2

1

13

− 14

0

0, 2 x 1

f x

30.

1

x

3 −2

4 x3 24 x 2 x 6

2

34. (a)

2

−3

3

0 −2

0

Zero: x

5 3 (b)

f x 3x 1 x 6 3x 1

1 3

3x 1 x 6 0 0 1 3

INSTRUCTOR USE ONLY x

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.2

38. (a)

44. g x

10 −15

25

Analyzing Graphs of Functions

103

x

(a)

2

−3

3

− 30

Zeros: x

r2.1213

f x

2x2 9 3 x

(b)

2 x2 9 3 x

0

2 x2 9

0 x

39. f x

−2

Increasing on f, f (b)

r

3 2 2

–2

1

0

1

2

g x

–2

–1

0

1

2

r2.1213 s2 4

45. g s

3 x 2

The function is increasing on f, f . 40. f x

x

(a)

7

x2 4 x

The function is decreasing on f, 2 and increasing on

−6

2, f . 41. f x

Decreasing on f, 0 ; Increasing on 0, f x3 3x 2 2

(b)

s

–4

2

0

2

4

g s

4

1

0

1

4

The function is increasing on f, 0 and 2, f and decreasing on 0, 2 . 42. f x

46. h x

x2 1

The function is decreasing on f, 1 and increasing

x2 4

(a)

1 −4

on 1, f . 43. f x

6 −1

5

3

(a)

−5

Decreasing on f, 0 ; Increasing on 0, f

4

(b) −3

x

–2

1

0

1

2

h x

0

–3

–4

–3

0

3 0

Constant on f, f (b)

x f x

47. f t

–2

1

0

1

2

3

3

3

3

3

t 4

(a)

1 −3

3

−3

Increasing on f, 0 ; Decreasing on 0, f (b)

t f t

–2

1

0

1

2

–16

–1

0

–1

–16

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

104

Chapter 1

48. f x

NOT FOR SALE

Functions ctions and Their Graphs

52. f x

3x 4 6 x 2

(a)

x2 3

(a)

4

−6

6

6 −6

6

−4

−2

Increasing on 1, 0 , 1, f ; Decreasing on

f, 1 , 0, 1 (b)

x f x

Decreasing on – f, 0 ; Increasing on 0, f (b)

–2

1

0

1

2

24

–3

0

–3

24

x f x

53. g t 49. f x

1 x

3

–1

0

1

2

1.59

1

0

1

1.59

t 1

(a)

(a)

–2

2

3 −2

−4

−2

2

Increasing on f, f

−1

Decreasing on f, 1 (b)

4

x

2

–3

f x

(b)

2

–1 3

2

0

1

1

0

t

–2

1

0

1

2

g t

–1.44

–1.26

–1

0

1

54. f x 50. f x

x 3

x

(a)

3

x 5

(a)

5

9

−10

−9

−5

9

Increasing on f, f

−3

Increasing on 2, f ; Decreasing on 3, 2 (b)

x f x

–3

2

–1

0

1

0

–2

–1.414

0

2

(b)

x f x

55. f x 51. f x

(a)

5

x

32

–13

6

–5

–4

3

–2

–1

0

1

2

x 2 x 2

(a)

6

4

−9

9

−6 0

Increasing on 2, 2

6 0

Increasing on 0, f (b)

x

0

1

2

Constant on f, 2 and 2, f 3

4

(b)

x

–2

1

0

1

4

INSTRUCTOR NST TRUC U USE SE ONLY ON f x

0

1

2.8

5.2

8

f x

–4 4

–2 2

0

2

4

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.2

56. f x

59. f x

x 1 x 1

(a)

Analyzing Graphs of Functions

x

4 x 2

2

6

−8

−6

10

6

−10

−2

Decreasing on f, 1

Relative minimum: 1, 9

Constant on 1, 1

60. f x

Increasing on 1, f (b)

105

x f x

3x 2 2 x 5 3

–3

2

–1

0

1

2

3

6

4

2

2

2

4

6

−7

8

−7

x 3, x d 0 ° 0 x d 2 ® 3, °2 x 1, x ! 2 ¯

57. f x

(a)

Relative minimum: 61. f x

13 , 163

or 0.33, 5.33

x 2 3x 2

7

2

−3 −5

6

7 −1

−4

Increasing on f, 0 and 2, f

Relative maximum: 1.5, 0.25

Constant on 0, 2 (b)

x f x

62. f x

–2

1

0

1

2

3

4

1

2

3

3

3

5

7

12

− 12

°2 x 1, x d 1 ® 2 °¯x 2, x ! 1

58. f x

(a)

2 x 2 9 x

12 −4

Relative maximum: 2.25, 10.125

1 −3

3

63. f x

x x 2 x 3 10

−3

Increasing on f, 1 and 0, f

−12

Decreasing on 1, 0 (b)

x f x

12

−6

–2

1

12

–3

–1

–1.75

0

1

2

Relative minimum: 1.12, – 4.06

–2

–1

2

Relative maximum: 1.79, 8.21

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

106

NOT FOR SALE

Chapter 1

64. f x

Functions ctions and Their Graphs

x3 3x 2 x 1

70. f x

3 x 11 y

3

3 −7

8

−6

x

−3

6

9

−3 −6

−7

−9

Relative maximum: 0.15, 1.08 Relative minimum: 2.15, 5.08 65.

(0, −11)

−12

71. f x

x

22

3 4

y 3 2

− 10

10

−10

1 −3

−2

−1

Relative minimum: 1, 7

−2

72. f x

20

3

(0, − 43)

−3

Relative maximum: 2, 20 66.

x 2

1

−1

3x

5 2

y 4

− 10

10

3 2 1

−20

Relative minimum: 4, 17

x

−4 −3 −2 −1

2

−2

73. f x

10

4

(0, − 25)

Relative maximum: 0, 15 67.

3

16 x

5 2

y 1 −1

10

−3

−2

−1

−1

x 1

−1

3

(0, − 25)

−2

Relative minimum: 0.33, 0.38

2

−3 −4

68.

−5

5

−5

5

74. f x

23 x

5 6 y

− 10

4 3

Relative maximum: 2.67, 3.08 69. f x

−4 −3 −2 −1

1 2x

x 1

3

4

−3 −4

3

−3 −2 −1

(0, 65 )

−2

y

1

2

(0, 1) x

−1

1

2

3

INSTRUCTOR ST USE ONLY −2

−3 −3

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.2

75. f x

Analyzing Graphs of Functions

79. (a) f 5

1.8 2.5 x

4, f 2

17 4

3

m

2

−1

x 1

−1

3

y 4

3 x 5

y 4

3 x 15

y

3 x 11

f x

3 x 11

(0, −1.8)

−2

76. f x

2

10.2 3.1x

(b)

y

21 7

2 5

1 −2

17

5, 4 , 2, 17

y

−3

107

3

y 12 10

12

(0, 10.2)

10

8 6 4

4

2

2 −10 −8 −6

x x

−2 −2

77. (a) f 1

2

2

4

80. (a) f 3

4, f 0

y

2 x 6

f x

2 x 6

(b)

m

6 4 2 0 1 2 x 0

y 6

8

10

12

9, f 1

11

3, 9 , 1, 11

6

1, 4 , 0, 6 m

6

11 9 1 3

f x 9

5

5 x 3

f x

(b)

20 4

5x 6 y

2 1

y

x

−5 −4 −3 −2 −1

2 3 4 5

−2 −3 −4 −5 −6 −7

6 5 4 3 2 1 x −1

1

2

78. (a) f 3

3

4

5

6

81. (a) f 5

7

8, f 1

m

m 10 4

1 3

f x

(b)

5 2

1 1

0

y

1

f x

1 y

3

5 4

2

3

1

2

x −3 −2 −1

1 −4 −3 − 2 − 1

0

0 x 5

y 1

(b)

y

0 10

5 5

y 1

5 x 1 2 5 1 x 2 2

f x 2

1

5, 1 , 5, 1

2

3, 8 , 1, 2 2 8

1, f 5

1

2

3

x 1

2

3

4

−2

INSTRUCTOR ST USE S ONLY −33

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© Cengage Learning. All Rights Reserved.

108

NOT FOR SALE

Chapter 1

Functions ctions and Their Graphs

§ 2· 82. (a) f ¨ ¸ © 3¹

15 , f 4 2

86. g x

11

axb 1 y

§ 2 15 · ¨ , ¸, 4, 11 2¹ ©3

4

11 15 2

m

2

4 2 3

7 2 14 3

§ 7· ¨ ¸ © 2¹

−4

§ 3· ¨ ¸ © 14 ¹

3 x 4 4 3 x 8 4

f x 11 f x

(b)

6

x

−2

2

4

6

−4

3 4

−6

87. g x

ax 1b y 4 3 2

y

1 2 x

−2 −2

2

4

6

x

−4 −3

1

12 14

2

3

4

−2

−4

−3

−6

−4

−8

88. g x

− 12 −14

83. g x

ax 3b

a xb

y 2

y

1 x

−4 −3 −2 −1

4 3

1

2

3

4

−2

2

−3

x −4 −3 −2 −1

3

4 −6

−2 −3

84. g x

2 x 3, x 0 ® ¯ 3 x, x t 0

89. f x

−4

4a xb

y

y

4 3

16 12 8

1

4 x

− 4 − 3 −2

1

2

3

4

x –1

1

2

3

4

–1 –2

− 12 − 16

85. g x

90. g x

axb 2

° x 6, x d 4 ®1 °¯ 2 x 4, x ! 4

y y

2

4 2

1 −4 −3 −2 −1 −2

−5 −6

x 1

2

3

4

−4 − 2

x 2

8 10

−6 −8 −10 −12 −14 −16

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.2

° 4 x , x 0 ® °¯ 4 x , x t 0

91. f x

Analyzing Graphs of Functions

109

2 x 1, x d 1 ° 2 ®2 x 1, 1 x d 1 °1 x 2 , x ! 1 ¯

96. k x

y 5

y

4

3

3

2 1

1 x

–4 –3 –2 –1

1

2

3

4

x

−3 −2 −1

1

2

3

–2

−2

–3

−3

°1 x 1 2 , x d 2 ® x 2, x ! 2 °¯

92. f x y

97. f x

4 x

f x t 0 on f, 4@. y

4 3

5

2

4 3

1

2

x –1

1

2

3

4

5

1 x –1

2 x d1 °x 5, ® 2 °¯ x 4 x 3, x ! 1

93. f x y 10

1

98. f x

3

4

5

4x 2

f x t 0 on ª¬ 12 , f . 4x 2 t 0

8

4 x t 2 x t 12

4 2 x –4

2

–2

2

4

6

8

–2

ª 1 ¬ 2 , f

y

94. h x

°3 x 2 , x 0 ® 2 °¯x 2, x t 0

4 3

y

2

6 5 4

x –2

–1

1

2

2 1 −4 −3

x

−1

1

2

3

4

99. f x

9 x2

−2

95. h x

y

4 x 2 , x 2 ° ®3 x, 2 d x 0 °x 2 1, x t 0 ¯

10

6 4 2

y

x −6 −4 −2 −2

5

2

4

6

4 3 2 1 x −4 −3

−1

f x t 0 on >3, 3@

INSTRUCTOR NS USE ONLY 1

2

3

4

−22 −3 −3

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

110

NOT FOR SALE

Chapter 1

100. f x

Functions ctions and Their Graphs

x2 4x

f x t 0 on f, 0@ and >4, f .

y

1 2 x 2 , x d 2 ® ¯ x 8, x ! 2

105. f x

8 6 4 2

f x t 0 on (2, 8]

x2 4 x t 0

−8 −6 −4 −2

x x 4 t 0

x 2 4 6 8

−4 −6

f, 0@, >4, f y

2

° x 5, x ! 5 ® 2 °¯x x 1, x d 5

106. f x

x 1

3

–1

y

–2 30

–3

25 –4

20

101. f x

x 1

y

x

5

f x t 0 on >1, f .

−15 −10 −5 −5

4 3

x 1 t 0

1

x t1

x 5 t 0

x −1 −1

>1, f

1

2

3

4

5

x 5 t 0 x t 5

102. f x

x x 1t 0 2

y

x 2 4

f x t 0 on >2, f .

x

3

x 2 t 0

2

x t 2

x –2

>2, f 1 x

1

§ 1 ¨¨ f, 2 ©

2

y x

f x is never greater than 0.

f x

–1

–2

–1

1

21

5

5 º ª 1 », « 2 ¼ ¬

5

· , f ¸¸ ¹

2 14 x cde 14 xfgh

107. s x

2

(a)

0 for all x.

12 41 1

1 r 1 r 2

x 2 t 0

103. f x

10 15 20

f x t 0

2

x 1t 0

5

(b) Domain: f, f

8

Range: >0, 2

–2 –3 −9

9

–4

104. f x

1 2

(c) Sawtooth pattern

−4

2 x

y

108. g x

4

f x is always greater

2 14 x cde 14 xfgh

2

3

than 0. f, f

(a)

2

Range: >0, 2 x

–2

–1

(b) Domain: f, f

8

1

−9

9

2

(c) Sawtooth pattern

−4

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.2

4 12 x cde 12 xfgh

109. h x

(a)

Analyzing Graphs of Functions

111

y

113. 3 2

8

1 −3 −9

x

−1 −1

1

3

9

−3

−4

(a) Domain: >2, 2@

(b) Domain: f, f

(b) Range: >2, 0@

Range: >0, 4

(c) Increasing on 0, 2

(c) Sawtooth pattern

4 12 x cde 12 xfgh

110. k x

Decreasing on 2, 0

2

y

114.

(a)

8

6 4 2

−9

9

−6 −4

6

−6

(a) Domain: (f, 4]

Range: >0, 4

(b) Range: (f, 0]

(c) Sawtooth pattern

(c) Increasing on f, 4

y 6

115.

4

y 3

2

(− 1, 1) 2

x

−6 −4

4

−4

(b) Domain: f, f

111.

x

−4 −2

4

−2

6

(− 2, 0)

−4

−3

−6

(1,

3)

1 x

− 2 −1 −1

1

2

3

(0, 0)

−2

(a) Domain: All real numbers or f, f

−3

(b) Range: (f, 4]

(a) f 1

(c) Increasing on f, 0

(b) f 1

3

(c) f is increasing on 2, 1.6 and 0, f

Decreasing on 0, f 112.

1

f is decreasing on 1.6, 0 .

y 6

y

116.

4

3

2 −4 −2 −2

(1,

2 x 4

6

8

1

(− 2, 0) −3

−4 −6

−1

(− 1, −

(a) Domain: All real numbers or f, f

(2, 0) x 1

2

3

3 ) −2 −3

(b) Range: [1, f)

(a) g 1

(c) Increasing on 1, f

(b) g 1

3

Decreasing on f, 1

3)

3

(c) g is increasing on 1, 1 .

INSTRUCTOR USE ONLY g is decreasing on 2, 1 and an 1, 2 .

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© Cengage Learning. All Rights Reserved.

112

117.

NOT FOR SALE

Chapter 1 f x f x

Functions ctions and Their Graphs

x6 2 x 2 3

x

124.

2 x 3

6

2

g s

4s 2 3

g s

4 s 4s 2 3

x6 2 x 2 3

g s

f x

The function is even. y-axis symmetry.

The function is even. y-axis symmetry. 118.

h x

125.

x3 5

h x

x 3

y 8

5

6

x 5 3

4

z h x

2

z h x

g x g x

f x

2

The function is even. y

2

2 t 3

h x

−10

The graph of f x f x

9 f x

z h x

The function is even.

z h x The function is neither odd nor even. No symmetry. f x f x

127. f x

3x 2 y

x 1 x2 x 1 x

9 is symmetric to the y-axis,

which implies f x is even.

x 5

x 5 x

122.

6

−8

x 5

x

4

−6

The function is neither even nor odd. No symmetry. x

2

−4

z f t , z f t

h x

x

−6 −4 −2 −2

t 2 2t 3

121.

5

126.

t 2 2t 3

t

5 is symmetric to the y-axis,

f x

The function is odd. Origin symmetry.

f t

6

which implies f x is even.

5 x

g x

f t

4

The graph of f x

x3 5 x

x 3

2

−4

x3 5 x

120.

x

− 6 − 4 −2 −2

The function is neither odd nor even. No symmetry. 119.

23

4 3

2

2 1

x 1 x2

−4 −3 −2 −1

f x

4s3 2 4 s

2

3

4

−2

The function is odd. Origin symmetry. 123. f s

x 1

The graph displays no symmetry, which implies f x is neither odd nor even. f x

32

z f s

3 x 2 3 x 2 z f x

z f s

z f x

INSTRUCTOR USE ONLY The function is neither odd nor even. No symmetry.

The function is neither even nor odd.

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.2

128. f x

Analyzing Graphs of Functions

131. f x

5 3x

113

1 x

y

y 4

5

3

4

2

3 2 1

x

x − 4 − 3 − 2 −1

1

3

−3 −2 −1 −1

4

−2

1

2

3

−2

The graph displays no symmetry, which implies f x is

The graph displays no symmetry, which implies f x is

neither odd nor even.

neither odd nor even.

f x

5 3 x

f x

1 x

5 3x

1 x

z f x

z f x

z f x

z f x

The function is neither even nor odd. 129. h x

The function is neither even nor odd.

x 4 2

132. g t

3

t 1

y y

8 6

3

4

2

2

1

x −8 −6 −4

4

6

8

t − 2 −1

−6

The graph displays y-axis symmetry, which implies h x is even.

x 2

4

x2 4

h x

The graph displays no symmetry, which implies g t is neither odd nor even. g t

3

t

3

t 1

1

z g t

x 8 2

z g t

y − 8 − 6 −4 −2 −2

4

−3

The function is even. 130. f x

3

−2

−8

h x

2

The function is neither even nor odd.

x 2

4

6

8

133. f x

−4 −6

x 2 y 6 5 4 3

The graph displays y-axis symmetry, which implies f x is even. f x

x 8 2

The function is even.

x 8 2

f x

2 1 −4 −3 −2 −1

x 1

2

3

4

−2

The graph displays no symmetry, which implies f x is neither odd nor even. f x

x 2 z f x z f x

The function is neither even nor odd.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

114

NOT FOR SALE

Chapter 1

134. f x

Functions ctions and Their Graphs

x 5

0.294 x 2 97.744 x 664.875, 20 d x d 90

143. L

(a)

y

6000

2 1 x −1

1

2

3

5

7

−2

20

−3

90 0

−4 −5

2000 when x | 29.9645 | 30 watts.

(b) L

The graph displays no symmetry, which implies f x is

144. (a)

70

neither odd nor even. f x

x 5 x 5

0

z f x

(b) The model is an excellent fit.

z f x

The function is neither even nor odd.

top bottom 3 4 x x 2 3 4x x

2

top bottom

4 x x 2 2 x

3

24 .

20 to x

(e) Answers may vary. Temperatures will depend upon the weather patterns, which usually change from day to day. 145. (a)

top bottom 2

20 . Then the temperature increases

(d) The maximum temperature according to the model is about 63.93°F. According to the data, it is 64°F. The minimum temperature according to the model is about 33.98°F. According to the data, it is 34°F.

2x x2 138. h

6 to x

until 6 A.M. x

4 x 1 2

x2 4x 3

137. h

x

top bottom

x 2 136. h

(c) The temperature is increasing from 6 A.M. until noon x 0 to x 6 . Then it decreases until 2 A.M.

C

Cost of overnight delivery (in dollars)

135. h

24 0

x

60 50 40 30 20 10 x

139. L

right left 1 y2 2

140. L

0

1 2 3 4 5 6 7 8 9

Weight (in pounds)

1 y2 2

(b) C 9.25

23.40 3.759

right left 2

3

23.40 3.75a9.25b 57.15

2y

It costs $57.15 to mail a 9.25 pound package. right left 4 y2

142. L

right left 2 0 y 2 y

Flat fee fee per pound

146. (a) Cost

C x (b)

22.65 3.70a xb C

Cost of overnight delivery (in dollars)

141. L

60 50 40 30 20 10 x 1 2 3 4 5 6 7

INSTRUCTOR USE ONLY Weight (in pounds)

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.2

147. (a) For the average salaries of college professors, a scale of $10,000 would be appropriate.

(b) For the population of the United States, use a scale of 10,000,000. (c) For the percent of the civilian workforce that is unemployed, use a scale of 1%. 148.

x

x

x

(a) If f is even, another point is 4, 9 .

x

x x

8 8

4

64 2 x

155. 4, 9

x

1 2

(b) If f is odd, another point is 4, 9 .

2

156. 5, 1

Domain: 0 d x d 4 (b)

32 , 4 . (b) If f is odd, another point is 32 , 4 . (a) If f is even, another point is

(a) If f is even, another point is

8m

(a) A

53 , 7 . (b) If f is odd, another point is 53 , 7 .

x

x

115

153. 32 , 4

154. 53 , 7

8m

x

Analyzing Graphs of Functions

(a) If f is even, another point is 5, 1 .

80

(b) If f is odd, another point is 5, 1 .

0

157. (a)

x, y

158. (a)

2a, 2c

4 0

Range: 32 d A d 64 (c) When x 4

4, the resulting figure is a square. 8m

159. f x

4 4

4

s 4

4

4

By the Pythagorean Theorem, 42 42 s2 s 32 149. False. The function f x

x, y

(b)

2a, 2c

x 2, x 2 ° 0, 2 d x d 2 ® °ax b, x ! 2 ¯

(a) If f x is odd, then f x 8m

4

(b)

4 2 meters.

x 2 1 has a domain of

151. (a) Even. The graph is a reflection in the x-axis.

(b) Even. The graph is a reflection in the y-axis. (c) Even. The graph is a vertical translation of f. (d) Neither. The graph is a horizontal translation of f. 152. Yes, the graph of x y 2 1 in Exercise 19 does represent x as a function of y. Each y-value corresponds to only one x-value.

f 3

f 3 1

3a b

f 4

f 4 2

4a b

Solving the system: 4a b

2

3a b

1

yields a

1 and b

2

(b) If f x is even, then f x

all real numbers. 150. False. A piecewise-defined function is a function that is defined by two or more equations over a specified domain. That domain may or may not include x- and yintercepts.

f x

f x

f 3

f 3 1

3a b

f 4

f 4 2

4a b

Solving this system yields a

1 and b

2.

160. (a) Domain: >4, 5 ; Range: >0, 9@

(b)

3, 0

(c) Increasing 4, 0 3, 5 Decreasing 0, 3 (d) Relative Minimum 3, 0 Relative Maximum 0, 9

INSTRUCTOR USE ONLY (e) Neither

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116

NOT FOR SALE

Chapter 1

Functions ctions and Their Graphs

Section 1.3 Transformations of Functions 1. rigid

5. vertical stretch; vertical shrink

2. f x ; f x

6. (a) iv

(b) ii

3. nonrigid

(c) iii

4. horizontal shrink; horizontal stretch 7. (a) f x

x c

(d) i y

Vertical shifts

c

1 : f x

c

1 : f x

x 1

1 unit up

c

3 : f x

x 3

3 units up

x 1

c=3 c=1

6

1 unit down c = −1 x −2

−4

2

4

−2

(b) f x

x c

y

Horizontal shifts

c

1 : f x

c

1 : f x

x 1

1 unit right

c

3 : f x

x 3

3 units right

x 1

8

1 unit left

c = −1

6

c=1

c=3 −4

x

−2

2

4

6

−2

(c) f x

x 4 c

y

Horizontal shift four units left and a vertical shift c=3

c

1 : f x

c

1 : f x

x 4 1

1 unit up

c

3 : f x

x 4 3

3 units up

x 4 1

6

1 unit down

c=1 c = −1 x

−8

−6

−2 −2

8. (a) f x

x c 3 : f x

c

1 : f x

c

c=3 6

x 3

3 units down

c=1

4

c = −1

2

x 1

c = −3

1 unit down

x

2

−2

c

1 : f x

x 1

1 unit up

c

3 : f x

x 3

3 units up

(b) f x

y

Vertical shifts

8

10

12

−4 −6

x c

Horizontal shifts

c

3 : f x

x 3

3 units left

c

1 : f x

x 1

1 unit left

c

1 : f x

x 1

1 unit right

c

3 : f x

x 3

3 units right

y 8 6 4

c = −1 c = −3

2

c=1 −4 −2

−2

2

4

6

c=3 x 8

−4

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

Section 1.3 (c) f x

x 3 c 3 : f x

c

x 3 3

8

3 units down

1 : f x

c

1 : f x

x 3 1

1 unit up

c

3 : f x

x 3 3

3 units up

x 3 1

axb c

c

0:

c

2:

(b) f x

1 unit down

axb 2 f x a xb f x a xb 2

y

c=0

c=2

3

c = −2 x 3

Horizontal shifts

ax 2b f x a xb f x a x 2b

4

y

c=2

4

c=0

3

c = −2

2

x

−4

2 units left

ax 1b c 2 : f x a x 1b 2 0 : f x a x 1b 2 : f x a x 1b 2

3

Horizontal shift 1 unit right and a vertical shift

4

y

c=2 c=0

4

2 units down

3 2

c = −2

1

x −4 −3

2 units up

2 x 0 °x c, ® 2 °¯ x c, x t 0

(b) f x

c=1 y

4

° x c 2 , x 0 ® 2 °¯ x c , x t 0

c = −3 y c = −1 c= 1 c= 3

4

c = −3

c = −3

−4

−4

Parent function

10. (a) f x

4

2 units up

0:

c

2

Parent function

c

c

c = −1

2

2 units right

c

2 −2

2 units down

2 : f x

(c) f x

c=1

4

ax cb

2:

c=3

4

Vertical shifts

c

c

6

x

2 : f x

c

117

y

Horizontal shift 3 units right and a vertical shift

c

9. (a) f x

Transformations of Functions

2 x

−4 −3 −2

3

c = −1

x

−10 − 6 −2 −4 −6 −8 −10 −12

4

c=3 −4

6 8 10 12

c = −3 c = −1 c= 3 c= 1

(c) f x

x

1 c

c

2 : f x

c

0 : f x

x

1

c

2 : f x

x

1 2

3

x

y

Horizontal shift 1 unit to the left and a vertical shift 1 2 3

2 units down

4 3

c=2 c=0

3

x − 4 −3

3

2 units up

−1

−3

1

2

3

4

c = −2

INSTRUCTOR USE ONLY −4

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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118

NOT FOR SALE

Chapter 1

Functions ctions and Their Graphs

11. Parent function: f x

14. Parent function: f x

x2

(a) Vertical shift 1 unit downward g x

g x

g x

x 1

x 1 1 2

(c) Reflection in the x-axis, horizontal shift 2 units to the right, and a vertical shift 6 units upward g x

(a) Shifted down 3 units

2

(b) Reflection in the x-axis, horizontal shift 1 unit to the left, and a vertical shift 1 unit upward

x 2 6 2

x

g x

x 17

(c) Reflected in the x-axis and shifted to the right 5 units and upward 5 units g x

x 5 5

(d) Reflected about the x- and y-axis and shifted to the right 3 units and downward 4 units

5 3

g x

x 3 4

15. Parent function: f x

2

12. Parent function: f x

x 3

(b) Shifted downward 7 units and to the left 1 unit

(d) Horizontal shift 5 units to the right and a vertical shift 3 units downward g x

x

x3

Horizontal shift 2 units to the right

x3

y

x

2

3

(a) Reflected in the x-axis and shifted upward 1 unit g x

x3 1

16. Parent function: y

1 x3

Vertical shrink

(b) Shifted to the right 1 unit and upward 1 unit g x

x

1 1

x 3 1

x

10 4 3

13. Parent function: f x

x

x 3

x 2 4

x 6 1

axb

18. Parent function: y

y

(d) Reflection in the x-axis, horizontal shift 6 units to the right, and a vertical shift 1 unit downward g x

x2

x

Reflection in the x-axis and a vertical shift 1 unit upward

x 5

(c) Horizontal shift 2 units to the right and a vertical shift 4 units downward g x

Reflection in the x-axis

19. Parent function: f x

(b) Reflection in the x-axis and a horizontal shift 3 units to the left g x

x2

Vertical shift y a xb 4

(a) Vertical shift 5 units upward g x

1x 2

17. Parent function: f x

y

3

(d) Shifted to the right 10 units and downward 4 units g x

y

3

(c) Reflected in the x-axis and shifted to the left 3 units and downward 1 unit g x

x

x 1

20. Parent function: y

x

Horizontal shift y x 2 21. g x

12 x 2

(a) Parent function: f x

x2

(b) Reflection in the x-axis and a vertical shift 12 units upward (c)

y 12

4 −12 −8

x −4

8

12

−8 −12

INSTRUCTOR USE ONLY ((d)) g x

12 f x

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.3

22. g x

x

8

25. g x

2

(a) Parent function: f x

y

x2

(b) Horizontal shift of 8 units to the right (c)

y

Transformations of Functions

2 x 5

2

(a) Parent function: f x

x2

(b) Reflection in the x-axis, horizontal shift 5 units to the left, and a vertical shift 2 units upward y

(c)

16

119

4 12

3 2

8

1 x

4

−5 − 4

−7

−2 −1 −2

x 8

4

(d) g x 23. g x

12

1

−3

16

−4

f x 8

(d) g x

x 7

2 f x 5

3

(a) Parent function: f x

26. g x

x3

2

(a) Parent function: f x

(b) Vertical shift 7 units upward (c)

x 10 5 x2

(b) Reflection in the x-axis, horizontal shift 10 units to the left; vertical shift 5 units upward

y 10

(c)

8

y 10

4

5

2 −6

(d) g x 24. g x

x

x

−4

2

4

−20

6

−15

−10

−5 −5

f x 7

−10

(d) g x

x3 1

(a) Parent function: f x

x3

27. g x

(b) Reflection in the x-axis; vertical shift of 1 unit downward (c)

y

f x 10 5

x

1 2 3

(a) Parent function: f x

x3

(b) Horizontal shift 1 unit to the right and a vertical shift 2 units upward

3 2

y

(c)

1

5

x −3

−2

1

2

4

3

3

−2

2

−3

(d) g x

f x 1

1 x −2

−1

(d) g x

1

2

3

4

f x 1 2

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© Cengage Learning. All Rights Reserved.

120

NOT FOR SALE

Chapter 1

28. g x

x

Functions ctions and Their Graphs

31. g x

3 10 3

(a) Parent function: f x

x 4 8

(a) Parent function: f x

x3

x

(b) Horizontal shift of 3 units to the left; vertical shift of 10 units downward

(b) Reflection in the x-axis; horizontal shift 4 units to the left; and a vertical shift 8 units upward

(c)

(c)

y

y 8

x

−14 −12 −10 −8 − 6 −4 −2

6

2

4 2 −6

− 12

29. g x

(d) g x

f x 3 10

32. g x

x 2

(a) Parent function: f x

2

4

f x 4 8

x 3 9

(a) Parent function: f x

x

(b) Reflection in the x-axis; vertical shift 2 units downward (c)

x

−2 −2

− 14

(d) g x

−4

y

x

(b) Reflection in the y-axis; horizontal shift of 3 units to the right; vertical shift of 9 units upward (c)

y 12

1 −3

−2

−1

9

x −1

1

2

3

6

−2

3

−3 −4

x

(d) g x 30. g x

33. g x

(a) Parent function: f x

x 9

(c)

y 15

y

12

6

9

4

6 x

3 x 3

−4 −6

x

(b) Horizontal shift 9 units to the right

8

−10 −8 −6 −4 −2 −2

12

(a) Parent function: f x

x

(b) Reflection in the x-axis; horizontal shift of 5 units to the left; vertical shift of 6 units upward (c)

9

f x 3 9

(d) g x

f x 2

6 x 5

(d) g x

6

3

−5

(d) g x

6

9

12

15

f x 9

6 f x 5

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.3

34. g x

40. f x

x 4 8

(a) Parent function: f x

y 12

8

4

g x

or

x 7 2

x

y

x moved 6 units to the left and reflected in

x 6 x moved 9 units downward and reflected in

both the x-axis and the y-axis g x

45. f x

x 9

2

(a) Reflection in the x-axis and a vertical stretch (each y-value is multiplied by 3) 3x 2

(b) Vertical shift 3 units upward and a vertical stretch (each y-value is multiplied by 4)

x 2

8

g x

−2 −4

46. f x

−6

g x

(a) Parent function: f x

1 x3 4

g x

47. f x

y 4

2 x3

x

(a) Reflection in the x-axis and a vertical shrink

3

each y-value is multiplied by 12

2 1 x 1

2

3

g x

4

−3

12 x

(b) Vertical stretch (each y-value is multiplied by 3) and a vertical shift 3 units downward

−4

(d) g x

1 4

(b) Reflection in the x-axis and a vertical stretch each y -value is multiplied by 2

x

(b) Horizontal shift 3 units to the left, vertical shrink, reflection in the x-axis, vertical shift one unit down.

−2 −1

x3

(a) Vertical shrink each y -value is multiplied by

x 3 1

(c)

4x2 3

f 7 x 2

12

x2

g x

4

−4

x 4 8

44. f x

6

(b) Reflection in the y-axis, horizontal shift 7 units to the right, and a vertical shift 2 units downward

(d) g x

3

42. f x

f x 4 8

4

(a) Parent function: f x

−2

6 6

x 12

g x

2

7 x 2

36. g x

x

41. f x

x

−6 − 4 −2

(c)

g x

both the x- and y-axes.

2

35. g x

x3 moved 6 units to the left, 6 units downward,

43. f x

6

(d) g x

121

and reflected in the y-axis (in that order)

x

(b) Horizontal shift of 4 units to the left; vertical shift of 8 units upward (c)

Transformations of Functions

g x

12 f x 3 1

48. f x

37. f x

x

3 7

38. f x

x 2 9

39. f x

x3 moved 13 units to the right.

2

3 x 3 x

(a) Vertical stretch (each y-value is multiplied by 8)

g x

2

x

13

3

g x

8

x

(b) Reflection in the x-axis and a vertical shrink

each y-value is multiplied by 14

INSTRUCTOR USE E ONLY g x

14

x

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122

Chapter 1

NOT FOR SALE

Functions ctions and Their Graphs

49. Parent function: f x

53. Parent function: f x

x3

Vertical stretch (each y-value is multiplied by 2) g x

2x

Reflection in the x-axis, horizontal shift 2 units to the right and a vertical shift 2 units upward

3

g x

50. Parent function: f x

x

51. Parent function: f x

g x

x2

Reflection in the x-axis; vertical shrink

g x

axb

52. Parent function: y

x 4 2 x

Reflection in the x-axis and a vertical shift 3 units downward

12 x 2

x 3

56. Parent function: f x

Horizontal stretch (each x-value is multiplied by 2)

c 1 xf ed 2 hg

x2

Horizontal shift of 2 units to the right and a vertical shift of 4 units upward. g x

f x 2

57. (a) y

x

55. Parent function: f x

each y-value is multiplied by 12

g x

3

Horizontal shift of 4 units to the left and a vertical shift of 2 units downward

6 x

g x

x 2 2

54. Parent function: f x

Vertical stretch (each y-value is multiplied by 6) g x

x3

(b)

y Vertical shift 2 units upward

x

2 4 2

f x 2

y

(c)

Horizontal shift 2 units to the right

2 f x

y

Vertical stretch by a factor of 2.

y

5

y

(4, 4) 4

4

3

3

(3, 3)

2 1

1

2

3

4

−1

5

1

−2

f x

(e)

Reflection in the x-axis

y

2

3

4

5

(1, 0) x

−1

6

(2, −1)

1

−2

2

3

4

5

6

(0, −2)

−3

f x 3

(f )

Horizontal shift 3 units to the left

y

f x

y

Reflection in the y-axis y

y

2 1

1

(3, 0) x

x

(3, 2)

2

(5, 1)

1

(0, 1)

(d) y

3

(6, 2)

2

(1, 2)

(4, 4)

4

3

3

(0, 1)

(−4, 2)

(1, 2)

2

2

(1, 0)

x

1 −1

3

4

5

−2

(−3, −1)

−1

1

(−1, 0)

(−3, 1) x

−3

(4, −2)

−3

(0, 1)

(−2, 0)

(3, −1)

2

x −5

−4

−3

−2

−1

−1

(0, −1)

−2

−2

f 2 x

(g) y

Horizontal shink Each x-value is divided by 2. y

(2, 2)

2

1

( 32 , 1) ( 12 , 0)

x 2

3

INSTRUCTOR USE ONLY −1

(0, −1)

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.3

f x

58. (a) y

(b)

f x 4

y

y 10

8

(− 4, 6)

6 4 2 4

(0, −2)

6

−4

x

x 4

2

−10 −8 −6 −4

6

−4

f x 4

(e)

f x 3

y

y

(f )

10

10

8

8

8

6

6

4

4

(4, 2)

2 2

4

6

−4

−10 −8 −6

12

(0, −2)

4

(− 4, −1) 2 x

−4 −2

(10, −2)

(6, −1)

−2

(−2, −5)

−6

(0, − 4)

−6

y

10

(2, 2)

6

f x 1

y

y

6

4

−2

(−2, − 4)

−6

−6

(6, 4)

4 2

(0, 2)

−10 −8 −6 − 4 −2

8 10

(2, −2)

6

(− 4, 4)

(−2, 2) x

−6 −4

8

(6, 6)

6 4

(4, 2)

123

2 f x

y

10

8

(d) y

(c)

y

y 10

(− 6, 2)

Transformations of Functions

2

(−2, 1) 2

x

4

(0, −5)

−6

(0, 1) x

−8 −6 −4

−2

(−4, −3)

−4

2

6

8

(6, −3)

−6

f 2 x

(g) y

Horizontal shrink. Each x-value is divided by 2. y 3

(− 2, 2)

(3, 2)

2 1 x

−3 −2

2

−1

(−1, − 2)

3

4

(0, −2) −3 −4

f x 1

59. (a) y

(b)

Vertical shift 1 unit downward

f x 1

y

Horizontal shift 1 unit to the right

y

(2, 4)

4

3

(0, 2)

4

(1, 3)

3

3

x

(1, −1)

−1

1

1

3

(2, 0) −1

−2

1

f x 1

(e)

−3

4

(f )

Reflection about the x-axis and a horizontal shift 2 units to the right

2

−1

Vertical shrink by a factor of 12 y

(5, 1)

(−3, 4)

1

4

3

(3, 0)

(−1, 3)

3

1

2

x 4

(−2, 2) 2

5

−1

2

x −1

2 −1

(2, −1)

(1, 0) x

(2, −3)

−3 −4

(0, 23 )

1

−2

(0, 0) −2

1

f x

y

y

y

x

−1

(−3, −1)

(4, −1)

f x 2

y

(−1, 0)

x

2

−1

(3, −2)

Horizontal shift 1 unit to the left

−3

(0, 3)

2

−1

(d) y

Reflection about the y-axis

(−1, 4)

1 −2

f x

y

y

y

(−2, 3)

(c)

(0, − 4)

−2

−1

1 −1

(3, − ) 1 2

−2

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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124

Chapter 1

NOT FOR SALE

Functions ctions and Their Graphs

f 12 x

(g) y

Horizontal stretch Each x-value is multiplied by 2. y 6

(− 4, 4) 4

(0, 3)

2

(2, 0) x −4

−2

2

(6, −1)

−2 −4

f x 5

60. (a) y

(b)

f x 3

y

y

f x y

12

8

6

(2, 0) 2

4

(−3, 3)

x

− 4 −2

4

6

4

(e)

6

(5, 4)

2

(2, 0)

−2

4

6 x

(−1, −5)

−6

(− 6, − 4)

−8

y

x 8

(6, − 43 )

f x 10 y

(0, 5)

x −8 −6 −4 −2

−4

4

6

(−3, −10) (− 6, −14)

4

6

8

(0, −5)

x 8

(6, − 4)

−6

2

−4

(3, 0) 2

−8

−10

6

2

2

(−3, 0)

6

−8 −6 −4 −2

−4

(f )

8

4

4

−2

y

y

−10 −8 −6

(−6, − 43)

f x

y

2

−1

6

(0, −2)

−4

(3, 0)

−8 −6 −4 −2 x

f x 1

(− 4, 0)

(3, 3)

2

−10 −8 −6 −4

(11, − 4)

−6

1

(− 3, 0)

8 10 12

(−1, − 4)

(−7, 4)

(6, 7)

8

4

(8, 0)

(0, 53)

2

10

(− 6, 7)

(5, 5)

6

(g) y

1 3

y

y

10

(d) y

(c)

−8

(3, −10)

−12

(6, −14)

−14

f 12 x

Horizontal stretch. Each x-value is multiplied by 2. y 12 10 8 6 4 2

(−6, 0) −10

−2 −4 (−12, − 4) −6 −8 −10 −12

61. (a) g x

(0, 5) (6, 0)

x

2 4 6 8 10 12

(12, −4)

f x 2

(b)

Vertical shift 2 units upward

g x

f x 1

(c)

Vertical shift 1 unit downward

g x

f x

Reflection in the y-axis

y

y 7 6 5 4

y

4 3 2 1 x

2 1

−4 −3

g x

−4 −3 −2 −1 −2 −3

1 2 3 4 5 6

g

5 6

−2 −3 −4 −5 −6

g

7 6 5 4 3 2 x

−6 −5 −4 −3 −2 −1

1 2 3 4

−2 −3

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.3

(d) g x

2 f x

(e)

Reflection in the x-axis and a vertical stretch by a factor of 2

g x

f 2 x

Horizontal stretch Each x-value is multiplied by 2. y 8

2

6 4

1

g

x −4 −3 −2

1 −2 −3 −4 −5 −6

2

4 5 6

g

−2

1

−6 −4 −2

2

(b)

g x

1 2

(c)

g x

x

−4 −5 −6 −7 −8 −9

y

2 1

g

7 6 5 4 3 2

g 1 2 3 4 5 6 7

−6 −5 − 4 −3 −2 −1

−2 −3

(e)

3

g x

f 4 x

(f )

3

6

f 14 x

g x

Horizontal stretch Each x-value is multiplied by 4. y

y

x

−3

x 1 2 3 4

−2 −3

Horizontal shrink Each x-value is divided by 4.

y

g

x

−3 −2 −1

4 f x

8 10

f x

y

1 2 3 4 5 6

6

−8

7 6 5 4

−2

(d) g x

4

−6

f x

y 1 −3 −2 −1

2

−4

−2

f x 5

g x

x

−1 −1

62. (a) g x

125

f 12 x

g x

y

4 3 2 1

−3

(f )

Horizontal shrink Each x-value is divided by 2.

y

−6

Transformations of Functions

9

24

4

g

18

( − 34 , 3(

−6

2

−12

1

−2

12

3

(− 12, 3) 6

(0, 2)

(0, 2) (4, 0)

(24, 0) x

( ( ( 32 , 0(

−1

−18 − 12 − 6 −6

1 ,0 4

1

6

12 18 24

− 12

x

2

− 18

63. (a) Vertical stretch of 128.0 and vertical shift 527 units up 1200

0

16 0

(b) M

527 128.0 t 10; The graph is shifted 10 units to the left.

64. (a) Horizontal shift 5.99 units to the right, vertical stretch, vertical shift 5617 units up, reflection in the x-axis. 6000

0 4500

(b) 2015: Use t

7

15

N 15

24.7015 5.99 5617 | 3611.85 2

In the year 2015, there will be about 3,612,000 couples with stay-at-home mothers. Answers will vary. Sample answer: No, because the number of stay-at-home mothers has been increasing on average.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

126

Chapter 1

NOT FOR SALE

Functions ctions and Their Graphs x , the graphs of f x

65. True, because x

and f x

x 6

x 6 are identical.

Decreasing: f, 2 and 1, f

66. False. The point 2, 61 lies on the transformation.

(b) Increasing: 1, 2 Decreasing: f, 1 and 2, f

67. True. Let

68. (a) Increasing: 2, 1

g x

(c) Increasing: f, 1 and 2, f

f x c

Then g x

Decreasing: 1, 2

f x c f x c

(d) Increasing: 0, 3

Since f is even

g x Thus, g x

Decreasing: f, 0 and 3, f

f x c is also even.

(e) Increasing: f, 1 and 4, f Decreasing: 1, 4

Section 1.4 Combinations of Functions 1. addition; subtraction; multiplication; division

6. f x

2 x 5, g x

2 x

2. composition

(a)

f

g x

2x 5 2 x

3. g x

(b)

f

g x

2 x 5 2 x

x 3

2x 5 2 x 4. inner; outer 5. f x

(a)

f

3x 7

x 2, g x g x

x 2

(c)

g x

§f· (d) ¨ ¸ x ©g¹

f x g x

x

fg x

7. f x

f x g x

x

2x 5 2 x

Domain: all real numbers x except x

2 x 2

4

(c)

(a)

2 x 2

f

x 2 , g x

g x

g x

4x 5

x 2 4 x 5

x 4 f x

2

f x g x

2

§f· (d) ¨ ¸ x ©g¹

5 2 x

2 x 2 9 x 10

2 x 2

2x

f

2 x

4 x 2 x 2 10 5 x

f x g x

x (b)

fg x

x2 4 x 5

x 2 x 2

(b)

Domain: all real numbers x except x

f

g x

f x g x x 2 4 x 5

2

x2 4x 5 (c)

fg x

f x g x x 2 4 x 5 4 x3 5 x 2

§f· (d) ¨ ¸ x ©g¹

f x

g x

x2 4x 5

INSTRUCTOR USE ONLY Domain: all real numbers x except p x

5 4

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.4

8. f x

(a)

f

3x 1, g x g x

5x 4

11. f x

f x g x 3x 1 5 x 4

f

g x

f x g x

(c)

fg x

f x g x

3 x

g x

f x g x

1 1 2 x x

x 1 x2

(b)

f

g x

f x g x

1 1 2 x x

x 1 x2

(c)

fg x

15 x 2 7 x 4 f x

§f· (d) ¨ ¸ x ©g¹

12. f x 4 . 5

Domain: all real numbers x except x 9. f x

x 2 6, g x

1 x

(a)

f

g x

f x g x

x2 6

(b)

f

g x

f x g x

x 6

(c)

fg x

f x g x f x

§f· (d) ¨ ¸ x ©g¹

2

x2

1 x

x2

6 1 x 1 x

Domain: x 1 10. f x

x , g x x 1

0

x3

f

g x

x x3 x 1

x x 4 x3 x 1

(b)

f

g x

x x3 x 1

x x 4 x3 x 1

(c)

fg x

§f· (d) ¨ ¸ x ©g¹

x x3 x 1

x y x3 x 1

x4 x 1

x 1 x 1 x3

Domain: all real numbers x except x x 1

,

1 x 2 x 1 0 and

2

x x2 1

(a)

f

g x

x2 4

x2 x 1

(b)

f

g x

x2 4

x2 2 x 1

(c)

fg x

2

§ x2 · x2 4 ¨ 2 ¸ © x 1¹

x2

13.

f

g 2

14.

f

g 1

x2 4 x 1

1 2 4

3

f 1 g 1 1 1 4

7

2

15.

f

g 0

f 0 g 0

0 2

1 0 4

5

x2 4

x2

16.

f

g 1

f 1 g 1

1

Domain: x 2 4 t 0

2

1 1 4

1

x 2 t 4 x t 2 or x d 2

Domain: x t 2

22

1 1 5

x2 x 4 y 2 x 1 1

f 2 g 2

1 2

2

x2

x

For Exercises 13 – 23, f x = x 2 + 1 and g x = x – 4.

x 2 4, g x

§f· (d) ¨ ¸ x ©g¹

g x

x2 x

1x 1 x2

(a)

1 x

6 1 x

x2 6 1 x

g x

f x

1 x3

Domain: all real numbers x except x

3x 1 5x 4

g x

1§ 1 · ¨ ¸ x © x2 ¹

f x g x

§f· (d) ¨ ¸ x ©g¹

1 5 x 4

1 x2

f

3 x 1 5 x 4 2 x 5

1 , g x x

127

(a)

8x 3 (b)

Combinations of Functions

17.

f

g 3t

f 3t g 3t ª3t 2 1º 3t 4 ¬ ¼ 9t 2 3t 5

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

128 18.

Chapter 1

f

NOT FOR SALE

Functions ctions and Their Graphs

g t 2

f t 2 g t 2

t

27. f x

f

2 1 t 2 4 2

x 2 , g x

g x

t 4t 4 1 t 2 4 2

t 3t 1

fg 6

5 4

f 6 g 6

62

x 2x

y

2

19.

f

3

f+g

1 6 4

x –3 –2 –1

74

20.

fg 6

3

28. f x

ª 6 2 1º ª 6 4º ¼ ¬ ¼¬

f

37 10

4 x 2 , g x

g x

g 5

x

4 x2 x

4 x x2

y

370 f 5

4

g

–2

f 6 g 6

§f· 21. ¨ ¸5 ©g¹

2 x

2

6

g

52 1 5 4

26 x –6

–4

4

6

–2

f 0

§f· 22. ¨ ¸0 ©g¹

g 0

§f· 23. ¨ ¸ 1 g 3 ©g¹

02 1 0 4 f 1

1

2

1

1 4 2 1 5 24.

fg 5

f 4

f

1 x, 2

g x

g x

29. f x

f

3 4

3 x, g x

g x

3 x 2

x3 10

x3 10

3x 10

3 5

f −15

15

f+g

f 5 g 5 f 4

52

25. f x

f +g

–6

g 3

g 1

f

–4

1 4

1 5 4 42 1

−10

g

26 1 17

For 0 d x d 2, f x contributes most to the

43

magnitude. For x ! 6, g x contributes most to the magnitude.

x 1

y 4

1

f+g 2

f

1

x –2

1

2

3

x , g x 2 x f g x 2

30. f x

g

3

x x

4 10

f+g

26. f x

f

1 x, 3

g x

g x

1x 3

g

x 4 x 4

y

f

23 x 4

−4

14 −2

g x contributes most to the magnitude of the sum for

8

0 d x d 2. f x contributes most to the magnitude of

6

the sum for x ! 6.

f+g f

2

INSTRUCTOR STR USE ONLY x

2

–2

4

6

g

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.4

31. f x

f

3x 2, g x

g x

3x

x 5

(b)

g

Combinations of Functions

D f x

x 5 2

f+g −9

3

(c)

g

D g x

−6

x 1

x 1

3

x

9

g

f

g f x

g

6

3

1

1 1

x

g g x g x3 1

For 0 d x d 2, f x contributes most to the magnitude.

x3 1

For x ! 6, f x contributes most to the magnitude.

x9 3 x 6 3x3 2

32. f x

f

x 2 12 , g x

g x

36. f x

3x 2 1

2 x 2

3 2

f g x

§1· f¨ ¸ © x¹

§1· ¨ ¸ © x¹

(b)

g

D f x

g f x

g x3

1 x3

(c)

g

D g x

g g x

§1· g¨ ¸ © x¹

x

g

−4

For 0 d x d 2, g x contributes most to the magnitude. For x ! 6, g x contributes most to the magnitude. 33. f x

x , g x

x 1

2

f

D g x

f g x

f x 1

(b)

g

D f x

g f x

gx

(c)

g

D g x

g g x

g x 1

34. f x

f

3 x 5, g x D g x

2

x

1

x 1 2

f

x2

Domain: all real numbers x

D g x

g

35. f x

(a)

f

D g x 3

35 x 5

(b)

g

D f x

g f x g

g x

(a)

g f x

x 1, g x

D g x

f

3

g 5 x

x3 1

D g x

x 4

f g x f x3 1 3

x3 1 5

3

x3 4

x

(b)

g

D f x

g f x

f x 1

3

x3

2

all real numbers x

g

3

Domain: all real numbers x

f g x

x

x 4

Domain: all real numbers x

x 1

3

x 4

x 5

5 3x 5

3

3

x2 4

Domain: all real numbers x

38. f x

g g x

f x2

f g x

2

3 x (c)

1 x3

Domain: x t 4

f g x

g 3 x 5

3

Domain: x t 4

x 4

x 2

20 3x D f x

g x

5 x

f 5 x

g

37. f x

(a)

(a)

(b)

1 x

x 3 , g x D g x

f+g

(a)

1

f

f 6

3

(a) 4

−6

129

1 1 x

3

3

x 5

x 5

3

x 51

1 x 4

Domain: all real numbers x

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130

Chapter 1

39. f x

x2 1

g x

(a)

f

NOT FOR SALE

Functions ctions and Their Graphs

g x

Domain: x t 0

x

D g x

40. f x

Domain: all real numbers x

f g x

(a)

x x 1 f

(b)

x 1

D f x

g f x

x6

Domain: all real numbers x

D g x

f x6

f g x

x6

23

x4

g

D f x

g f x

g x2 3

x2 3

6

x4

Domain: all real numbers x

Domain: x t 0

g

Domain: all real numbers x

Domain: all real numbers x

2

(b)

f

x2 3

g x 1

41. f x

x 1

2

2

g x

Domain: all real numbers x

(a)

f

x

Domain: all real numbers x

x 6

Domain: all real numbers x

D g x

f g x

f x 6

x 6

Domain: all real numbers x (b)

g

D f x

g f x

g x

x 6

Domain: all real numbers x 42. f x

x 4

Domain: all real numbers x

g x

3 x

Domain: all real numbers x

(a)

f

D g x

f g x

f 3 x

3 x

4

x 1

Domain: all real numbers x (b)

g

D f x

g f x

g x 4

3

x 4

3 x 4

Domain: all real numbers x 43. f x

g x

(a)

f

1 x

Domain: all real numbers x except x

x 3

Domain: all real numbers x

D g x

f g x

1 x 3

f x 3

Domain: all real numbers x except x (b)

g

D f x

g f x

§1· g¨ ¸ © x¹

g x

(a)

f

g

0

3 x2 1

Domain: all real numbers x except x

x 1

Domain: all real numbers x

D g x

f g x

f x 1

Domain: all real numbers x except x (b)

3

1 3 x

Domain: all real numbers x except x 44. f x

D f x

g f x

0

§ 3 · g¨ 2 ¸ © x 1¹

Domain: all real numbers x except x

3

x

3 x2 2 x 1 1

1 1 2

0 and x

r1

3 x2 2x

2

3 1 x2 1

3 x2 1 x2 1

x2 2 x2 1

r1

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.4

45. h x

2 x

1

2

53.

One possibility: Let f x

then f D g x 46. h x

x and g x

2 x 1,

2

h x .

Combinations of Functions

x

0

1

2

3

f

2

3

1

2

g

–1

0

1 2

0

3

3 2

2

f g

1 x 3

One possibility: Let g x then f D g x

1 x and f x

x3 ,

1

131

y

h x .

4 3

47. h x

3

h

x 4 2

2

One possibility: Let f x then f D g x 48. h x

x and g x

1

x 2 4,

x

h x .

1

54.

9 x

One possibility: Let g x then f D g x 49. h x

3

9 x and f x

x,

h x .

2

4

x

–2

–1

0

1

2

f x

–2

0

–1

–1

1

g x

1

1

0

2

2

–1

1

–1

1

3

h x

1 x 2

3

f x g x y

One possibility: Let f x then f D g x

1 x and g x

x 2,

3

h x .

2

h 1

50. h x

4

5 x

2

x

2

–2

–1

1

2

–1

One possibility: Let g x then f D g x 51. h x

5 x 2 and f x

h x .

4 , x2

x2 3 4 x2

One possibility: Let f x then f D g x

h x .

x 3 and g x 4 x

x2 ,

55.

x

–2

0

1

2

4

f

2

0

1

2

4

g

4

2

1

0

2

f g

6

2

2

2

6

y 7

52. h x

One possibility: Let g x f x

6

27 x 3 6 x 10 27 x 3

5

h

27 x 6 x , then f D g x 10 27 x 3

2

x3 and

1 x

h x .

−2 −1

1

2

3

4

5

6

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© Cengage Learning. All Rights Reserved.

132

Chapter 1

NOT FOR SALE

Functions ctions and Their Graphs

56. The domain common to both functions is >1, 1@, which

59. (a)

f

D g 2

f g 2

f 2

0

(b)

g

D f 2

g f 2

g 0

4

60. (a)

f

D g 1

f g 1

f 3

2

(b)

g

D f 3

g f 3

g 2

2

is the domain of the sum.

x

–1

0

1

f x

0

1.5

3

g x

–1

–2

1

–1

– 0.5

4

h x

f x g x

61. (a) T x

y

Distance traveled (in feet)

3

h

1 x

− 4 −3 −2

3 x 4

1

2

3

4

250

T

200

B

150 100

R

50

x

−3

10

g 3

§f· (b) ¨ ¸ 2 ©g¹ 58. (a)

(b)

f

20

30

40

50

60

Speed (in miles per hour)

−4

f

1 x2 15

300

−2

57. (a)

(b)

4 2

R x B x

f 3 g 3 f 2 g 2

g 1

fg 4

0 2

(c) B x ; As x increases, B x increases at a faster

3

0

rate. 62. (a) R3

R1 R2 480 8t 0.8t 2 254 0.78t

f 1 g 1 f 4 g 4

21

23 40

1

0

734 7.22t 0.8t 2

(b)

800

R3 R1 R2 3

8 0

63. B D t

B t D t

0.197t 3 10.17t 2 128t 2043

This represents the number of births (in millions) more than the number of deaths in the United States from 1990 to 2006, where t 0 corresponds to 1990. 64. 2010: Use t

B 20

20

4388 million

This represents the total number of births in the United States in 2010. D 20

2413 million

This represents the total number of deaths in the United States in 2010.

B

D 20

1975 million

This represents the number of births more than the number of deaths in the United States in 2010. 2012: Use t B 22

22

4438.98 million

This represents the total number of births in the United States in 2012. D 22

2387.36 million

This represents the total number of deaths in the United States in 2012.

B

D 22

2051.62 million

This represents the number of births more than deaths in the United States in 2012.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.4

65. (a) ht

T t

P t

Combinations of Functions

133

0.0233t 4 0.3408t 3 1.556t 2 1.86t 22.8 2.78t 282.5

ht represents the number of people (in millions) playing tennis in the United States compared to the number of

people (in millions) in the United States, or the percent of the population that plays tennis expressed as a decimal. (b) h0

0.0807 million

h3

0.0822 million

h6

0.0810 million

66. (a) T is a function of t since for each time t there corresponds one and only one temperature T.

(b) T 4 | 60q; T 15 | 72q (c) H t

T t 1 ; All the temperature changes would be one hour later.

(d) H t

T t 1; The temperature would be decreased by one degree.

(e) The points at the endpoints of the individual functions that form each "piece" appear to be 0, 60 , 6, 60 , 7, 72 , 20, 72 , 21, 60 , and 24, 60 . Note that the value t 24 is chosen for the last ordered pair because that is when the day ends and the cycle starts over. From t

0 to t

6: This is the constant function T t

From t

6 to t

7: Use the points 6, 60 and 7, 72 .

72 60 12 7 6 y 60 12 x 6 y

60.

m

From t

7 to t

From t

20 to t

12 x 12, or

T t

12t 12

20: This is the constant function T t

72.

21: Use the points 20, 72 and 21, 60 .

72 60 12 20 21 y 60 12 x 21 y

m

60, ° °12t 12, ° ®72, °12t 312, ° °¯60,

A piecewise-defined function is T t

67. (a) r x

(b) A r (c)

0 d t d 6 6 t 7 7 d t d 20 . 20 t 21 21 d t d 24

(a)

S r2

A D r x

C

60 x 750, xt D x t

50t

C xt C 50t

A r x

§ x· A¨ ¸ © 2¹

§ x· S¨ ¸ © 2¹

2

6050t 750 3000t 750

represents the area of the circular base of

the tank on the square foundation with side length x.

A D r t

60.

69. C x

x 2

A D r x 68.

12t 312

24: This is the constant function T t

21 to t

From t

12 x 312, or T t

A r t

A0.6t

S 0.6t

2

0.36S t 2

A D r represents the area of the circle at the time t.

C D x t represents the cost of production as a function of time. (b) Use t x t

4 50 4

200

INSTRUCTOR USE ONLY In 4 hours, 200 units are produced.

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134

NOT FOR SALE

Chapter 1

Functions ctions and Their Graphs

(c) Find t when C D x t 15,000 t

(c) f x

x2 2x 1

3000t 750

f x

g x h x

4.75 hours

g x

1 ª f x f x º¼ 2¬ 1ª 2 2 x 2 x 1 x 2 x 1º ¼ 2¬ 1 2 ª x 2 x 1 x 2 2 x 1º¼ 2¬

15,000.

The cost of production for 4 hours 45 minutes is $15,000. 70. (a) f g x

f 0.03 x

(b) g f x

0.03 x 500,000

g x 500,000

0.03 x 500,000

g f x represents your bonus of 3% of an amount

h x

over $500,000. 71. False. f D g x

6 x 1 and g D f x

6 x 6.

72. True. The range of g must be a subset of the domain of f for f D g x to be defined. 73. (a) g x

1 ª f x f x º¼ 2¬

To determine if g x is even, show g x

g x .

1 ª f x f x º¼ 2¬ 1 ª f x f x º¼ 2¬ 1 ª f x f x º¼ 2¬

g x

f x

x2

k x

1 x 1 g x h x

k x

g x

g x 9 1 ª f x f x º¼ 2¬

h x

To determine if h x is odd show h x

f x

h x

a function

1 ªk x k x º¼ 2¬ 1ª 1 1 º 2 «¬ x 1 1 x »¼

º 1ª 2 x « » 2 ¬« x 1 1 x »¼ x x 1 1 x

function and h x is an odd function. g x h x 1 1 ª f x f x º¼ ª¬ f x f x º¼ 2¬ 2 1 1 1 1 f x f x f x f x 2 2 2 2 f x 9

1 1 x 1

1 ª1 x x 1 º « » 2 ¬« x 1 1 x ¼»

even function odd function

Using the result from part (a) g x is an even f x

1 2 x

1 ªk x k x º¼ 2¬ 1ª 1 1 º 2 «¬ x 1 x 1»¼ 1 ª1 x x 1º 2 «¬ x 1 1 x »¼

x

h x 9

(b) Let f x

1ª 2 º 2 «¬ x 1 1 x »¼ 1 1 x 1 x

h x

1 ª f x f x º¼ 2¬ 1 ª f x f x º¼ 2¬ 1 ª¬ f x f x º¼ 2

h x

1 ª2 x 2 2º¼ x2 1 2¬ 1 ª f x f x º¼ 2¬ 1ª 2 2 x 2 x 1 x 2 x 1 º» ¼ 2 «¬ 1 2 ª x 2 x 1 x 2 2 x 1º¼ 2¬ 1 >4 x@ 2 x 2

x k x

x 1 x 1

§ · § · 1 x ¨¨ x 1 x 1 ¸¸ ¨¨ x 1 x 1 ¸¸ © ¹ © ¹

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.5

§f· 74. (a) ¨ ¸ x ©g¹

(b)

Inver Inverse Functions

135

Domain: x ! 0

x3 2 ;

f

D g x

f g x

x;

Domain: x t 0

g

D f x

g f x

x;

Domain: all real numbers x.

No, they are not the same because negative numbers are in the domain of g D f but not in the domain of f D g .

Section 1.5 Inverse Functions 11. f x

1. inverse

3x 1 x 1 3

2. f 1

f 1 x

3. range; domain

f f 1 x

§ x 1· f¨ ¸ © 3 ¹

f 1 f x

f 1 3 x 1

4. y

x

5. one-to-one

12. f x

6. Horizontal 7. f x

6x x 6

f 1 x f f 1 x f 1 f x 8. f x

1 x 6 § x· f¨ ¸ ©6¹

f x

1 1

9. f x

f

1

f x

x

f f 1 x

f

3 x

1 3

3 x

x

3

1x 3

f

1

f f 1 x

f x

ª 1 º f « x 9 » 2 ¬ ¼

x

9

9 9

x

x

9 9

f x 4 f

1

x

4

x 4 4 x 4 4

1 2 x 2

ª 1 º 2 « x 9 » 9 2 ¬ ¼ x 1 ª¬ 2 x 9 9º¼ 2

x

x 1 5 5x 1 5x 1 1 5

f f 1 x

f 5 x 1

f 1 f x

§ x 1· f 1 ¨ ¸ © 5 ¹ x 11 x

x

x 4

9 9

f 1 2 x 9

x

f 1 x

x

x

3

1 x 9 2

13. f x

f x 9

1 1

y

f 1 f x

x 4

f 1 x

f

6x 6

x 9

x

10. f x

1

f 1 x

x 9

f 1 x ff

x

3x f 3 x

1

1 x 9 2

x

f f 1 x f

2 x 9 2 y 9

1x 3

f 1 x

1

y

§ x· 6¨ ¸ ©6¹

f 1 6 x

3 x

x

2 x 9

x

§ x 1· 3¨ ¸ 1 © 3 ¹

5x 5

x

§ x 1· 5¨ ¸ 1 © 5 ¹

x x

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136

NOT FOR SALE

Chapter 1

14. f x

y x 2x

Functions ctions and Their Graphs

4x 7 2 4x 7 2 4y 7 2 4y 7

1 2 x 7 4

y 1 2 x 7 4

f 1 x f f 1 x

ª1 º f « 2 x 7 » 4 ¬ ¼

f 1 f x

§ 4x 7 · f 1 ¨ ¸ © 2 ¹

15. f x

f

1

3

x

4 ª¬1 4 2 x 7 º¼ 7 2 º 1 ª § 4x 7 · 2¨ ¸ 7» 4 «¬ © 2 ¹ ¼

7 7 2

x

1 ª 4 x 7 7º¼ 4¬

x

18. f x

x x

2 x

3

f f 1 x

f x3

f 1 f x

f 1

3

x3

x x 3

3

x

x 5

(a) f g x

f x 5

g f x

g x 5

x 3

x 5, g x

x 5 5 x 5 5

x x

y

(b) 8

16. f x

2

f 1 x

5

x

x

f f 1 x

–8

f

x x 5

5

5

x5

6

8

f

x 19. f x

§ x· f¨ ¸ © 2¹

g f x

2

–4

x

x 2

2 x, g x

(a) f g x

5

–4 –2

–8

f 1 x5

f 1 f x 17. f x

g

6

x5

§ x· 2¨ ¸ x © 2¹ 2x x 2

g 2 x

x 1 7

7 x 1, g x

(a) f g x

§ x 1· f¨ ¸ © 7 ¹

g f x

g 7 x 1

(b)

§ x 1· 7¨ ¸ 1 © 7 ¹

7 x

1 1 7

x x

y

y

(b)

5 4

3 2

3

f

2

g

1

1 x

x –3

–2

1 –2

2

3

1

g

2

3

4

5

f

–3

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.5

20. f x

3 x 4

3 4 x, g x

§3 x· f¨ ¸ © 4 ¹

(a) f g x

§3 x· 3 4¨ ¸ © 4 ¹

3 3 x g f x

3 3 4 x

(b)

4x 4

4

x

137

1 x

(a) f g x

§1· f¨ ¸ © x¹

1 1x

1y

1 x

1

x 1

x

g f x

§1· g¨ ¸ © x¹

1 1x

1y

1 x

1

x 1

x

x

g 3 4 x

1 , g x x

22. f x

Inver Inverse Functions

y

(b) 3

y

2

f =g 1

f

x

4

g

1

2

2

3

x –8 –6 –4 –2

2

4

–4 –6 –8

21. f x

23. f x

x3 , g x 8

(a) f g x

f

3

3

(b)

8x

x2

3

8

§ x3 · g¨ ¸ ©8¹

g f x

3

3

§ x3 · 8¨ ¸ ©8¹

3

8x 8

x

x3

x

g f x

10

x 4 x 4

x

2

4

x

g

8

3

g 6

2 1 −4 −3

4 4

y

(b)

f

4

g

y

x 2 4, x t 0

f x 2 4 , x t 0

(a) f g x

8x

8x

x 4, g x

4 x 1

−1

2

3

4

f

2

x −3

2

4

6

8

10

−4

24. f x

1 x 3 , g x

(a) f g x

f

3

3

1 x

1 x

g 1 x3 3

25. f x

x3

f

3

1 x

3

f

6

g x

1 1 x3

–6 –4 –2

6 –2 –4 –6

x

9 x, x d 9

9 x ,x d 9

g 9 x 2 , x t 0

g f x

3

y

x

9 x 2 , x t 0; g x

(a) f g x

1

1 1 x g f x

(b)

9

9 x

9 9 x 2

2

x x

y

(b) 12 9 6

f g x

– 12 – 9 – 6 – 3

6

9 12

–6

INSTRUCTOR T USE ONLY –9

– 12

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138

NOT FOR SALE

Chapter 1

Functions ctions and Their Graphs

1 , x t 0; g x 1 x

26. f x

(a) f g x

§1 x · f¨ ¸ © x ¹

g f x

§ 1 · g¨ ¸ ©1 x ¹

(b)

1 x ,0 x d 1 x

1 §1 x · 1¨ ¸ © x ¹

1 x 1 x x x

1 1 x

§ 1 · 1¨ ¸ ©1 x ¹ § 1 · ¨ ¸ ©1 x ¹

1 x 1 1 x 1 x 1 1 x

x

x 1 x 1 1 x

x x 1 1 x 1

x

y 5 4

g 3 2 1

f x 1

2

3

5

x 1 , g x x 5

27. f x

5x 1 x 1 § 5x 1 · 1¸ ¨ © x 1 ¹ x 1 § 5x 1 · x 1 5¸ ¨ © x 1 ¹

5 x 1 x 1

5 x 1 5 x 1

ª § x 1· º «5¨ x 5 ¸ 1» x 5 © ¹ ¼ ¬ ªx 1 º x 5 «¬ x 5 1»¼

§ x 1· g¨ ¸ © x 5¹

g f x

(b)

§ 5x 1· f ¨ ¸ © x 1¹

(a) f g x

5 x 1 x 5

x 1 x 5

6 x 6

x

6x 6

x

y 10 8 6 4 2

f

g

28. f x

f x

−8 −6

2 4 6 8 10 −4 −6 −8 −10

g

x 3 , g x x 2

(a) f g x

g f x

29. No,

4

2x 3 x 1

§ 2x 3 · f¨ ¸ © x 1¹

§ x 3· g¨ ¸ © x 2¹

2x 3 3 x 1 2x 3 2 x 1 § x 3· 2¨ ¸ 3 © x 2¹ x 3 1 x 2

2 x 3 3x 3 x 1 2x 3 2x 2 x 1

^2, 1 , 1, 0 , 2, 1 , 1, 2 , 2, 3 , 6, 4 `

2x 6 x x 3 x

does

not represent a function. 2 and 1 are paired with two different values.

3x 6 2 x 2 2

5x 5

x

y

(b) 6 4

5x 5

x

g f

2

g f

x −4

−2

4

6

8

−4 −6

30. Yes,

^10, 3 , 6, 2 , 4, 1 , 1, 0 , 3, 2 , 10, 2 `

does represent a function.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

Section 1.5 31. Yes, because no horizontal line crosses the graph of f at more than one point, f has an inverse.

40. f x

1 8

x

Inver Inverse Functions

2 1 2

f does not pass the horizontal line test, so f does not have an inverse.

24

32. No, because some horizontal lines intersect the graph of f twice, f does not have an inverse. 33. No, because some horizontal lines cross the graph of f twice, f does not have an inverse. 34. Yes, because no horizontal lines intersect the graph, of f at more than one point, f has an inverse. 35. g x

−8

41. (a)

y

2x 3

8

2y 3

6

(b)

y

f f −1

4

x 3 2 x 3 2

2 x –2

2

4

6

8

–2

(c) The graph of f 1 is the reflection of the graph of f x. in the line y

10 f does not pass the horizontal line test, so f does not have an inverse.

−12

(d) The domains and ranges of f and f 1 are all real numbers. 42. (a)

12

x 4 x 4 10

− 10

10

h does not pass the horizontal line test, so h does not have an inverse.

3x 1

y

3x 1

3

3y 1

2

x 1 3

y

f 1 x

x 1 3

(b)

y

f f −1

1

x −3

−2

2

1

3

−2 −3

(c) The graph of f 1 is the reflection of f in the line y x. (d) The domains and ranges of f and f 1 are all real numbers.

−10

5

f x x

−2

x

2x 3

f 1 x

8

14

38. g x

f x

y

g passes the horizontal line test, so g has an inverse.

−4

37. h x

24

x

−4

36. f x

−24

4 x 6 4

139

3

43. (a)

g passes the horizontal line test, so g has an inverse.

4

−10

f x

x5 2

y

x5 2

x

y5 2

y

5

2

f 1 x

−4

5

x 2

y

(b) 3

f

2

f −1 x −3

−1

−1

2

3

x 2 −3

39. f x

2 x 16 x

2

20

− 12

12

f does not pass the horizontal line test, so f does not have an inverse.

(c) The graph of f 1 is the reflection of the graph of f x. in the line y (d) The domains and ranges of f and f 1 are all real numbers.

− 20

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140

NOT FOR SALE

Chapter 1

Functions ctions and Their Graphs

f x

44. (a)

x3 1

y

x 1

x

y 1

x

3

f

4

x –6

–4

x 1

2

4

6

x

y

(d) The domains and ranges of f and f 1 are all real numbers.

f 1 x

f x

4 x2 , 0 d x d 2

y

4 x

x

4 y2

x2

4 y2

y2

4 x2

f 1 x

(b)

f = f −1

3 2 1

x –3 –2 –1

1

2

3

4

–2 –3

(c) The graph of f 1 is the same as the graph of f. (d) The domains and ranges of f and f 1 are all real numbers except for 0.

2

48. (a)

2 x 2 x 2 y

f x

2

x

4 x2 , 0 d x d 2

y

(b)

y 3 2 1 −3

−2

−1

y f 1 x f = f −1

x 1

−1

2

f = f −1

−2

2 x 2 x

3

2

4

4 x 4 x

y

4 x

y

y

(b)

4

xy

–6

(c) The graph of f 1 is the reflection of f in the line y x.

45. (a)

4 x 4 x 4 y

f x y

f −1

2

y3 y

47. (a)

f

6

3

x 1 1

y

3

x 1 3

(b)

−3

(c) The graphs are the same.

1

(d) The domains and ranges of f and f 1 are all real numbers except for 0.

x 1

2

3

(c) The graph of f 1 is the same as the graph of f. (d) The domains and ranges of f and f 1 are all real numbers x such that 0 d x d 2. f x

46. (a)

r

49. (a)

f x y

x 2 2, x d 0

x 1 x 2 x 1 x 2 y 1 y 2

y

x2 2

x

x

y2 2

x y 2

y 1

x 2

y

xy 2 x

y 1

x

xy y

2x 1

y x 1

2x 1

f

1

(b)

x 2

y 4

f

y

3 2

f 1 x

1 x −4 −3

1

−3

2

3

4

y

(b)

6

f −1

4

f −1 −6 −4

2 −2

f x 4

−2

6

−4 −6

f

2x 1 x 1 2x 1 x 1

(c) The graph of f 1 is the reflection of graph of f in the line y x.

f −1

−4

(c) The graph of f 1 is the reflection of f in the line y x. (d) >2, f is the range of f and domain of f 1.

(d) The domain of f and the range of f 1 is all real numbers except 2. The range of f and the domain of f 1 is all real numbers umbers except 1.

INSTRUCTOR USE ONLY f, 0@

1

is the domain off f and the range of f .

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.5

x x x x y y

f x

50. (a)

y x xy 2 x y 3

y f

(b)

3 2 3 2 3 2

x

y x

6y 4

4 xy 5 x

6y 4

2 x 3 x 1 2 x 3 x 1

4 xy 6 y

5 x 4

y 4 x 6

5 x 4

f 1 x

4

(b)

f 2

4

y

6

(d) The domain of f and the range of f 1 is all real numbers except x 2.

y

3

x 1

x

3

x3

y 1

y

x3 1

f 1 x

x3 1

f −1

x –6

–4

2

4

6

54. (a) –6

(c) The graph of f 1 is the reflection of the graph of f in the line y x. (d) The domains and ranges of f and f 1 are all real numbers. 52. (a)

f x

x3 5

y

x3 5

x

y

(b) 3

−1

f x y x

y3 5

x5 3

y

–2

x5 3

–3

x –2

1

2

3

(c) The graph of f 1 is the reflection of the graph of f x. in the line y (d) The domains and ranges of f and f 1 are all real numbers.

4 6 4 6 4 6

6 x 4

f

–3

y 2 x 8 y

53

8x 2x 8x 2x 8y 2y

8y 4

1

y3 5

f −1

2 xy 6 x

2

x5 3 f 1 x

f

−2

The range of f and the domain of f 1 is all real 3 numbers except . 2

f

2

3

−3

4

y 1

2

(d) The domain of f and the range of f 1 is all real 5 numbers except . 4

y 6

1

(c) The graph of f 1 is the graph of f reflected in the x. line y

The range of f and the domain of f 1 is all real numbers x except x 1. (b)

x

−2

f −1

(c) The graph of f 1 is the reflection of the graph of f x. in the line y

x 1

f

1 −3

3

2

f

f −1

f x

5x 4 6 4x

3

x

51. (a)

5 x 4 4x 6 5 x 4 4x 6

y

141

4 5 4 5 4 5

x 4 y 5

y

f −1

2 x 3

6

f

6x 4x 6x 4x 6y 4y

f x

53. (a)

0

y x 1

1

Inver Inverse Functions

y

(b) f

8

f

4

x

f −1

8 12 16

f

−1

6 x 4 2x 8 3x 2 x 4

(c) The graph of f 1 is the graph of f reflected in the line y x. (d) The domain of f and the range of f 1 is the set of all real numbers x except x 3. The domain of f 1 and the range of f is the set of all real al numbers x except x 4.

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142

Chapter 1

NOT FOR SALE

Functions ctions and Their Graphs

55. f x

x4

y

x4

x

y4

y

y

r4 x

x

This does not represent y as a function of x. f does not have an inverse. 1 x2 1 x2 1 y2 1 x

56. f x

y x y2

3x 5 3x 5 3y 5 3y

f x

60.

5x 5x 4 5x 4 3

x y

58.

y

x

3 , x t 3 y t 0

y

x

3 , x t 3, y t 0

x

y

3 , y t 3, x t 0

x

y 3, y t 3, x t 0

3x 5

y x

3x 5 3y 5

4

y

4

2

2

x 3, x t 0, y t 3

f 1 x

x 3, x t 0

q x

x

5

2

y

x

5

2

x

y

5

r

x

y 5

5r

x

y

62.

2

This does not represent y as a function of x, so q does not have an inverse.

x 5 3y x 5 y 3 This is a function of x, so f has an inverse. x 5 f 1 x 3

59. p x

2

This is a function of x, so f has an inverse.

8x

f x

4

61. f x

This is a function of x, so g has an inverse. g 1 x

4

3y

y

x 8 x 8 y 8 8x

y

4

This is a function of x, so f has an inverse. 5x 4 f 1 x 3

1 y r x This does not represent y as a function of x. f does not have an inverse. 57. g x

4

x 3, x 0 ® ¯6 x, x t 0

63. f x

y 9 6 3 x

−6

3

6

9

−3

4 for all x, the graph is a horizontal line Because y and fails the Horizontal Line Test. p does not have an inverse.

−6

This graph fails the Horizontal Line Test, so f does not have an inverse.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.5

In Exercises 69 – 74, f x =

x d 0 x, ® 2 ¯ x 3 x, x ! 0

64. f x

Inver Inverse Functions

g x = x 3 , g –1 x =

3

143

– 3, f –1 x = 8 x + 3 ,

1x 8

x.

y

69.

4

f 1 D g 1 1

f 1 g 1 1

2

f 1

1

13

x −2

−4

2

−4

70.

g 1 D

f 1 3

4 2 x

g 1 0 71.

f 1 D

f 1 6

x 2 , x d 2, y t 0

x

y 2 , y d 2, x t 0 y 2 or

x 2 x

x

or 2 x

y

f 1 x

2x

y

2x

x

2y

2

2y

y

x 2

x

g 1 D g 1 4

2

3 3

f

3

f

f x

68.

x2 3 ,x t 0 2

x 2 x t 2, y t 0 x 2, x t 2, y t 0

x

y 2, y t 2, x t 0

x

x 2 2

y

3

x

1 3 y 8

3

x 3

1 y3 8

8 x 3

y3

8 x 3

y

D g

23 x 3

x

74. g 1 D f 1

4

1 x3 8

1

4

9

4

f x3

1 x3 8

3

g 1 f 1 x g 1 8 x 3 3

8 x 3

23 x 3 In Exercises 75 –78, f x = x + 4, f –1 x = x – 4,

y

2

3

f g x

D g x

This is a function of x, so f has an inverse. f 1 x

y 73.

600

g 1 g 1 4

y 2

2 x, x t 0

67. f x

0

8ª¬86 3 3º¼

g 1

3 3 x t ,y t 0 2 3 3, x t , y t 0 2 3 3, y t , x t 0 2 3 3, x t 0, y t 2 3 3 , x t 0, y t 2

0

f 1 8>6 3@

72.

The portion that satisfies the conditions y d 2 and x t 0 is 2 x y. This is a function of x, so f has an inverse.

3

f 1 f 1 6

x 2,x d 2 y t 0

y

32

g 1 8 3 3

The graph fails the Horizontal Line Test so h does not have an inverse. 66. f x

g 1 f 1 3

The graph fails the Horizontal Line Test, so f does not have an inverse. 65. h x

3

8

4

−2

3

g x = 2 x – 5, g –1 x =

75.

g 1 D

f 1 x

x + 5 . 2

g 1 f 1 x g 1 x 4

y 2, x t 0, y t 2

x

y, x t 0, y t 2

2 x 1 2

This is a function of x, so f has an inverse.

4 5

INSTRUCTOR USE ONLY f 1 x

x 2 2, x t 0

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144

76.

Chapter 1

NOT FOR SALE

Functions ctions and Their Graphs

f 1 D g 1 x

f 1 g 1 x

79. The inverse is a line through 1, 0 . Matches graph (c).

5· ¸ 2 ¹

80. The inverse is a line through 0, 6 and 6, 0 . Matches

4

81. The inverse is half a parabola starting at 1, 0 . Matches

§x f 1 ¨ © x 5 2 x 5 2 x 3 2 77.

f

graph (b).

8

graph (a). 82. The inverse is a third-degree equation through 0, 0 .

Matches graph (d)

f g x

D g x

83.

f 2 x 5

f

2 x 5 4

84.

2x 1

f

D g

1

x

x 1 2

78.

D g

g

1

x

D f x

g 1 D

1

x

85. (a)

f 1 x .

g f x g x 4 2 x 4 5

y x x 3 x 3 2

g 86. (a)

D f

1

x

Total cost

0

2

4

6

8

–2

–1

0

1

2

3

–10

–7

–4

–1

2

5

–3

–2

–1

0

1

2

y

10 0.75 x

x

10 0.75 y

x 10

0.75 y

x 10 0.75

y

x

(b) y

x 10 0.75

hourly wage, y 24.25 10 0.75

number of units produced 19

So, 19 units are produced. y x 3 2 § Cost of · § Cost of · ¨ first commodity ¸ ¨ second commodity ¸ © ¹ © ¹

(b)

x 50 x

Amount of second commodity Cost of first commodity

1.25 x

Cost of second commodity

y

1.6050 x

x y

1.25 x 1.6050 x

0 d y d 50 80 x d 50 0 d 0.35 0 d 80 x d 17.5

x x x 80 x 80 0.35

y

Amount of first commodity

(c)

–2

So, f 1 x

2x 8 5 2x 3 2x 3 2y 3 2y

Labels: Total cost

Equation: y

x

f 1 x

Note: Comparing Exercises 75 and 77,

f

x

(d)

80 73 0.35

1.25 y 1.6050 y 1.25 y 80 1.60 y 0.35 y y 80 x 0.35 total cost number of pounds of less expensive commodity y

20 pounds

80 d x d 62.5

INSTRUCTOR USE ONLY 62.5 d x d 80

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.6

90. (a) f 1 116,011

0.03x 2 245.50, 0 x 100

87. (a) y

245.50 y 545.50 0.03 y 245.50

x 245.50

0.03 y 2

(c) f t | 1140.32t 107,512.86

x 245.50 0.03

y2

(d) f 1 t

x 245.50 0.03

y, 245.50 x 545.50

(e) f 1 123,477 | 14 which represents 2014.

x

x 245.50 0.03 temperature in degrees Fahrenheit

y

percent load for a diesel engine

f 1 x

100

t 107,512.86 1140.32

91. False. f x

x 2 is even and does not have an inverse.

92. True. If f x has an inverse and it has a y-intercept at

0, b ,

then the point b, 0 , must be a point on the

graph of f 1 x . 93. Let f D g x

0

f

600 0

D g x

y. Then x y f g x g x

(c) 0.03 x 2 245.50 d 500 0.03 x 2 d 254.50 x 2 d 8483.33 x d 92.10

functions, f D g

88. (a) Yes.

(b) Given the population (in millions of people) you can determine the year. (c) P

357.5

25

because P 25

357.5

So, 357.5 million people are projected to be living in the United States in 2025. (d) No. The function would no longer be one-to-one because the projected population would be 357.5 million people in 2025 and 2050. 89. (a) f 1 113.5

5

(b) f 1 yields the year for a given amount spent on wireless communications services in the United States. (c) f t (d) f 1 t

f

D g

1

y .

Also,

y f 1 y

x

g 1 f 1 y

x

g 1 D

f 1 y .

Because f and g are both one-to-one

Thus, 0 x d 92.10.

1

145

7 which represents 2007.

(b) f 1 yields the year for a given number of households.

2

x

(b)

Mathematical Modeling aand Variation

1

g 1 D f 1.

94. The reciprocal, not the inverse, of f x was found.

Given f x f 1 x

x 6, then

x 2 6, x t 0.

95. This situation could be represented by a one-to-one function if the runner does not stop to rest. The inverse function would represent the time in hours for a given number of miles completed. 96. This situation could be represented by a one-to-one function if the population continues to increase. The inverse function would represent the year for a given population. 97. No. The function oscillates.

11.4t 55.4 98. No. After a certain age, height remains constant

t 55.4 11.4

(e) f 1 170.938

15 which represents 2015.

Section 1.6 Mathematical Modeling and Variation 1. variation; regression

4. correlation coefficient

2. sum of square differences

5. directly proportional

INSTRUCTOR USE ONLY 3. least squares regression

6. constant onstant of variation

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146

NOT FOR SALE

Chapter 1

Functions ctions and Their Graphs

7. directly proportional

13. (a)

y 240

Length (in feet)

8. inverse 9. combined 10. jointly proportional

220 200 180 160 140 t

Year

Actual Number (in thousands)

Model (in thousands)

1992

128,105

127,712

1993

129,200

129,408

1994

131,056

131,104

1995

132,304

132,800

y 162.3

1996

133,943

134,495

y

1997

136,297

136,191

(c) y | 1.01t 130.82

1998

137,673

137,887

(d) The models are similar

1999

139,368

139,583

2000

142,583

141,279

2001

143,734

142,975

2002

144,863

144,671

2003

146,510

146,367

2004

147,401

148,063

2005

149,320

149,759

2006

151,428

151,454

2007

153,124

153,150

20 28 36 44 52 60 68 76 84 92 100 108

Year (20 ↔ 1920)

(b) Using the points 32, 162.3 and 96, 227.7 :

m

227.7 162.3 96 32 | 1.02 1.02t 32 1.02t 129.66

(e) 2012 o use t

112

Model from part (b): y 1.02112 129.66

243.9 feet

Model from part (c) y 1.01112 130.82

243.94 feet

(f ) Answers will vary. 14. (a)

y

Total sales (in billions of dollars)

11.

24 20 16 12 8 4 t 1 2 3 4 5 6 7

y

Year (0 ↔ 2000)

Number of people (in thousands)

155,000

(b) Using the points 1, 15.700 and 7, 20.936 :

150,000 145,000 140,000

m

135,000 130,000 125,000

y 15.700

t 2

4

6

8 10 12 14 16 18

Year (2 ↔ 1992)

12. The model is not a good fit for the actual data.

(c) y

0.835 x 14.868

(d) The models are similar. (e) 2008 o use t

y 5.4 5.2

0.873 x 1

y | 0.873x 14.827

The model is a good fit for the actual data.

Winning time (in minutes)

20.936 15.700 7 1 | 0.873

8

Model from part (b): y 0.8738 14.827 | $21.811 billion

4.8 4.6 4.4 4.2 4.0 3.8 t

Model from part (c): y 0.8358 14.868 | $21.548 billion

0 8 16 24 32 40 48 54

Year (0 ↔ 1950)

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.6

15. (a)

Mathematical Modeling and a Variation

20. k

900

2

x kx 2

y 5

147

2

4

6

8

10

8

32

72

128

200

16 0

y

(b) S | 38.3t 224

200

(c)

160

900

120 80 40 5

x

16

2

0

The model is a good fit to the actual data. (d) 2007 o use t S 12

38.317 224 | $875.1 million

2009 o use t S 14

x

14

y

38.319 224 | $951.7 million

(e) Each year the annual gross ticket sales for Broadway shows in New York City increase by $38.3 million.

6

8

10

1 2

21. k

12

4

kx

2

2

4

6

8

10

2

8

18

32

50

y 50 40

16. (a) N | 9.29t 238.29

(b)

30 20

350

10 x 2

0

6

(c) 2008 o t N 8

8

9.298 238.29 | 312.6 million

6

8

10

1 4

22. k

0

4

x kx 2

y

(d) Answers will vary.

2

4

6

8

10

1

4

9

16

25

y

17. The graph appears to represent y

4 x, so y varies 25

inversely as x.

20

18. The graph appears to represent y

3 x, 2

directly with x. 19. k

so y varies

15 10 5 x

1

2

x y

kx

2

2

4

6

8

10

4

16

36

64

100

23. k

4

6

8

10

2 x k x2

y

y

2

4

6

8

10

1 2

1 8

1 18

1 32

1 50

100 80

y

60

5 10

40

4 10

20 x 2

4

6

8

10

3 10 2 10 1 10

INSTRUCTOR USE ONLY x

2

4

6

8

10

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148

NOT FOR SALE

Chapter 1

24. k

Functions ctions and Their Graphs

5

x k x2

y

2

4

6

8

10

5 4

5 16

5 36

5 64

1 20

28. y

kx

2

k5

2 5

k

y

2x 5

This equation checks with the other points given in the table.

y 5 4

y

29.

1 3 4

kx k 10

7

2 4

7 10

1 4

x 2

25. k

4

6

8

7 10 x

y

10

This equation checks with the other points given in the table.

10 x k x2

y

k

2

4

6

8

10

5 2

5 8

5 18

5 32

1 10

30.

k x k 5 k

y 24

y

120 5 2

120 x

y

2

This equation checks with the other points given in the table.

3 2

1 1 2

31.

x 2

26. k

4

6

8

10

20

x k x2

y

2

4

6

8

10

5

5 4

5 9

5 16

1 5

32.

y

kx

12

k 5

12 5

k

y

12 x 5

y

kx k 2

14 y 5

7

k

y

7x

4

y

33.

3 2

kx k 10

2050

1

205

x 2

27. y

1 5 y

4

6

8

10

k 205 x

y

k x k 5 k

y

34.

580 290 3

y

5 x

This equation checks with the other points given in the table.

kx k 6 k 290 x 3

35. A

kr 2

36. V

ke3

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.6

37. y

Mathematical Modeling aand Variation y

51.

k x2

149

5 4

k

38. h

s

2 1

39. F 40. z

kg r2

x 1

3

4

5

Using the points 0, 3 and 4, 4 , we have

kx 2 y 3 1 bh 2

The area of a triangle is jointly proportional to its base and height.

52.

3.

1 x 4

y 41. A

2

y 5 4

42. S

4S r

2

3 2

The surface area of a sphere varies directly as the square of the radius r.

1 x

43. V

4 3

Sr

1

3

4

5

The line appears to pass through (2, 5.5) and (6, 0.5), so 54 x 8. its equation is y

The volume of a sphere varies directly as the cube of its radius. 44. V

2

3

y

53.

S r 2h

5

The volume of a right circular cylinder is jointly proportional to the height and the square of the radius. 45. r

4

d t

2 1 x

Average speed is directly proportional to the distance and inversely proportional to the time. 46. Z

1

3

4

5

Using the points 2, 2 and 4, 1 , we have

kg W

12 x 3.

y

Z varies directly as the square root of g and inversely as

2

y

54.

the square root of W.

5

Note: The constant of proportionality is

4

k.

3

47. The data shown could be represented by a linear model which would be a good approximation.

1 x 1

48. The points do not follow a linear pattern. A linear model would be a poor approximation. A quadratic model would be better. 49. The points do not follow a linear pattern. A linear model would not be a good approximation. 50. The data shown could be represented by a linear model which would be a good approximation.

2

3

4

5

The line appears to pass through (0, 2) and (3, 3), so its 1 x 2. equation is y 3 55.

I 113.75 0.035 I

kP k 3250 k 0.035 P

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© Cengage Learning. All Rights Reserved.

150

Chapter 1

56.

I 0.0325

I

y

0.15

k 265

k

3 5300

k

0.0325 P

33

k 13

33 13

k

530 3

When x

20 inches, y | 50.8 centimeters.

kx

53

k 14

53 14

k

53 14

25 gallons: y

5

53 14

| 18.9 liters

5

F

d

67.

1.9 d

kF k 25 k

0.076

0.076 F

k

When the distance compressed is 3 inches, we have

0.0368 x

3

The property tax is $8280.

0.076 F F | 39.47.

No child over 39.47 pounds should use the toy. 68. d

1

kx

k

k 189.99

d

0.06 | k

y

0.06 x

y

0.06639.99

8 2

F

The sales tax is $38.40.

69.

k T Te

63. F

km1m2 r2

64. R

kS S L

A 9S

k V

62. R

kF k 15 1 15 1F 15 1 F 15

60 lb per spring

Combined lifting force

| $38.40

61. P

3 F 5500 3 F 5500

| 94.6 liters

$8280

y

k

The required force is 293 13 newtons.

0.0368 225,000

y

11.40

k 220

3 5500

k 150,000

y

60.

0.12

880 3

kx

0.0368

kF

0.16

5 gallons: y

5520

176 23 newtons

d

53 x 14

y

d

| 0.05 meter

F

F 66.

90

3 F 5300

(b) 0.1

10 inches, y | 25.4 centimeters.

59.

3 5300

(a) d

When x

y

3 F 5300

d

33 x 13

y

kF

k 6500

kx

y

58.

d

65.

kP

211.25

57.

NOT FOR SALE

Functions ctions and Their Graphs

70.

k 3 k

A

S r2

3 75

120 lb

kr 2

S

y

2F

2

k x k 25 k 75 x

INSTRUCTOR USE ONLY y

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 1.6

71.

k x k 4 k 28 x

y 7 28 y

72.

z

77.

k 11 3

4158

24 36 2 3 z

76.

0.32v 2

v2

0.12 0.32

v

3 2 2

3

3 8 6 | 0.61 mi hr 4

kv 2

If the velocity is doubled: d

k 2v

F

14rs 3

d

k 4v 2

kx y2

4kv 2 kv 2

k 42

d increases by a factor of 4 when velocity is doubled.

2

4

92 79.

k

k 18 x y2

66.17

Sd2

kl , A Sr2 A 4kl Sd2 41000 k § 0.0126 · ¸ © 12 ¹

4

2

S¨

k | 5.73 u 108

kx 2 y k 6

r r

k

P

6

0.12

14

P

z

0.32v 2

k

28 3 28 81 3 42 2 27 3 18

75.

d

78. d

krs 3

F

2

k

k 2 xy

151

kv 2 §1· k¨ ¸ © 4¹ 0.32

0.02

k 4 8

2 z

74.

d

kxy

64

73.

Mathematical Modeling and a Variation

r

45.73 u 108 l

2

§ 0.0126 · ¸ © 12 ¹

2

§ 0.0126 · ¸ © 12 ¹

2

S¨

4 k

33.5

45.73 u 108 l

S¨

k 2 3x 2 y v 1.5

kpq s2 k 4.1 6.3

1.2

1.5 1.44 4.1 6.3

k

2.16 25.83

k

k

2

2x2 3y

§ 0.0126 · 33.5S ¨ ¸ © 12 ¹ 45.73 u 108

l l | 506 feet

2

24 287 24 pq 287 s 2

INSTRUCTOR USE ONLY v

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152

NOT FOR SALE

Chapter 1

Functions ctions and Their Graphs

80. From Exercise 79.

81.

8

k | 5.73 u 10 .

2116.8

45.73 u 108 l

r

Sd

k

2

45.73 u 108 l

d

W When m have W

Sr 45.73 u 108 14

d

kmh k 120 1.8 2116.8

9.8

120 1.8 9.8mh

100 kilograms and h 1.5 meters, we 9.8100 1.5 1470 joules.

S 0.05

d | 0.0045 feet 82.

W

f

k

T l

440

k

T l

f

k

1.25T 1.2l

440l T

k

0.054 inch

where f

1.2 fl 1.25T

and

440l T

1.2 fl 1.25T

440l 1.25T

1.2 fl T

440 1.25T 242,000T 242,000 168,055.56

1.2 f

frequency, T

tension, and l

length of string

k

l

! 0

T 2

1.44 f T

T

! 0

1.44 f 2 f2

f | 409.95 vibrations per second

83. (a) v

v

k A k 0.75 A

84. Load

4§ k · ¨ ¸ 3© A¹

(a) load

The velocity is increased by one-third. (b) v v

k A k 4 A 3

k 2 w d 2

kwd 2 l

2l

The safe load is unchanged. (b) load

3§ k · ¨ ¸ 4© A ¹

kwd 2 l

k 2 w 2d

2

l

8kwd 2 l

The safe load is eight times as great.

The velocity is decreased by one-fourth.

(c) load

k 2 w 2d 2l

2

4kwd 2 l

The safe load is four times as great. (d) load

kw d 2 l

2

1 4 kwd 2 l

The safe load is one-fourth as great.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

Section 1.6 85. (a)

Mathematical Modeling aand Variation

153

C

Temperature (in degrees Celsius)

5 4 3 2 1 d 1000 2000 3000 4000 5000

Depth (in meters)

(b) Yes, the data appears to be modeled (approximately) by the inverse proportion model. k1 1000 4200 k1

k2 2000 3800 k2

4.2

k3 3000 4200 k3

1.4

4200 3800 4200 4800 4500 5

(c) Mean: k (d)

1.9

k4 4000 4800 k4

1.2

4300, Model: C

k5 5000 4500 k5

0.9

4300 d

6

0

6000 0

4300 d 4300 3

(e) 3 d

86. (a)

1 1433 meters 3

k d2 When the distance is doubled: k k I . 2 2 d 4 2d

Length (in centimeters)

x

88. I

7 6 5 4 3 2 1

The illumination is one-fourth as great. The model given k d 2. in Exercise 85 is very close to I

F 2

6

4

8

10 12

Force (in pounds)

The difference is probably due to measurement error.

(b) It appears to fit Hooke's Law.

(c)

k |

6.9 12

y

kF

9

0.575 F

89. False. “y varies directly as x” means y constant k.

0.575

kx for some nonzero

“y is inversely proportional to x” means y

F | 15.7 pounds

k for some x

nonzero constant k. 87. y

262.76 x 2.12

(a)

90. False.

“a is jointly proportional to y and z with the constant of kyz proportionality k” means a

0.2

25

91. False. E is jointly proportional (not “directly proportional”) to the mass of an object and the square of its velocity.

55 0

(b) y

262.76

92. False. The closer the value of r is to 1, the better the fit.

25 2.12 | 0.2857 microwatts per sq. cm.

93. The accuracy of the model in predicting prize winnings is questionable because the model is based on limited data. ata

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154

NOT FOR SALE

Chapter 1

Functions ctions and Their Graphs

94. As one variable increases, the other variable will also increase. Answers will vary.

96.

P

§9· kS ¨ ¸ © 2¹

8.78 48.78

§d · kS ¨ ¸ ©2¹

k S r 2

kA

95. (a) y will change by a factor of one-fourth.

(b) y will change by a factor of four.

2

2

k

81S

k | 0.138

However, we do not obtain $11.78 when d

12 inches.

2

P

§ 12 · 0.138S ¨ ¸ | $15.61 ©2¹

11.78 | 0.104. 36S

Instead, k

414.18

For the 15-inch pizza, k

225S

| 0.080.

The price is not directly proportional to the surface area. The best buy is the 15-inch pizza.

Review Exercises for Chapter 1 1. 16 x y 4 y

0

6. h x

4

16 x

y

r2 4 x

No, y is not a function of x. Some x-values correspond to two y-values. 2. 2 x y 3

0

2x 3

y

Yes, the equation represents y as a function of x. 3. y

1 x

2 x 1, x d 1 ® 2 ¯x 2, x ! 1

(a) h 2

2 2 1

3

(b) h 1

2 1 1

1

(c) h0

02 2

2

(d) h 2

22 2

6

7. f x

25 x 2 25 x 2 t 0

Domain:

Yes, the equation represents y as a function of x. Each x-value, x d 1, corresponds to only one y-value.

5 x 5 x

t 0

Critical numbers: x 4.

y y

x 2 corresponds to y

x 2 or

Test intervals: f, 5 , 5, 5 , 5, f

x 2.

Test: Is 25 x 2 t 0?

No, y is not a function of x. Some x-value correspond to two y-values. 5. f x

Solution set: 5 d x d 5 Domain: all real numbers x such that 5 d x d 5, or >5, 5@.

x2 1

(a) f 2

r5

2 2

1

y

5 10

(b) f 4 (c) f t 2 (d) f t 1

4 2

t 2

1

8

17

6

2

t

1

t4 1

4 2

1 1 2

−6 − 4

−2 −2

x 2

4

6

INSTRUCTOR USE ONLY t 2 2t 2

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NOT FOR SALE

Review Exercises ffor Chapter 1

5s 5 3s 9 5s 5 3 s 3

8. g s

7 6 5 4 3 2 1

s 2 4 6 8 10 12 14

− 5 −4 − 3 − 2 − 1

t 1 2 3 4

−2

11. vt

x x2 x 6 x x 2 x 3

y

32t 48

(a) v1

6 4

(b) 0

2

t

x 4

−2

Domain: All real numbers x 2, 3 except x

y

t 1

Domain: all real numbers

−6 −4 −2 −4 −6 −8 −10

Domain: All real numbers s except s 3.

9. h x

10. ht

y 10 8 6 4 2

155

6

(c) v 2

−4

16 feet per second 32t 48 48 32

1.5 seconds

16 feet per second

−6

12. (a) Model: 40% of 50 x 100% of x f x

Amount of acid in final mixture f x

0.450 x 1.0 x

amount of acid in final mixture

20 0.6 x

(b) Domain: 0 d x d 50 Range: 20 d y d 50 (c) 20 0.6 x 20 0.6 x 0.6 x x

13. f x

50%50 25 5 8 13 liters

2 x 2 3x 1

f x h f x h

ª2 x h 2 3 x h 1º 2 x 2 3 x 1 ¬ ¼ h 2 x 2 4 xh 2h 2 3 x 3h 1 2 x 2 3 x 1 h h 4 x 2h 3 h 4 x 2h 3, h z 0

f x

14.

f x h

x3 5 x 2 x

x

h 5 x h x h 3

2

x3 3x 2 h 3 xh 2 h3 5 x 2 10 xh 5h 2 x h f x h f x h

x3 3x 2 h 3 xh 2 h3 5 x 2 10 xh 5h 2 x h x3 5 x 2 x h 3 x 2 h 3 xh 2 h3 10 xh 5h 2 h h h3 x 2 3 xh h 2 10 x 5h 1 h 3 x 3 xh h 10 x 5h 1, h z 0 2

2

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156

NOT FOR SALE

Chapter 1

x

15. y

3

Functions ctions and Their Graphs

x2

22. f x

2

A vertical line intersects the graph no more than once, so y is a function of x.

4

2

28

4 y

16. x

A vertical line intersects the graph more than once, so y is not a function of x. 17. f x

3x 2 16 x 21

3x 2 16 x 21

0

3 x

0

7 x 3

3x 7 x 18. f x

0 or 7 3

6 −4

f is increasing on 2, 0 and 2, f . f is decreasing on f, 2 and 0, 2 . 23. f x

x 3

0

x

3

or

x2 2 x 1

Relative maximum: 1, 2 3

(1, 2)

5x2 4 x 1

5x2 4 x 1

5 x

−6

1 x 1

−3

0 0

−1

5x 1

0 x

1 5

x 1

0 x

1

24. f x

8x 3 11 x 8x 3 x

x4 4 x2 2

Relative minimum: 1.41, 6 , 1.41, 6

8x 3 11 x

19. f x

3

Relative maximum: 0, 2 1

0

−6

6

0

(0, −2)

83 (−1.41, −6) − 7 (1.41, −6)

20. f x

x3 x 2 25 x 25

x3 x 2 25 x 25

0

x 2 x 1 25 x 1

0

1 x 2 25

0

x

x3 6 x 4

Relative maximum: 0.125, 0.000488 | 0.13, 0.00 (0.1250, 0.000488) 0.25

x 1

0 x

1

x 25

0 x

r5

2

21. f x

25. f x

−0.75

0.75

x x 1

−0.75

f is increasing on 0, f . f is decreasing on f, 1 . f is constant on 1, 0 .

26. f x

x3 4 x 2 1

Relative minimum: 2.67, 10.48 . Relative maximum: 0, 1

5

4 −8

−5

4 −1

16

(0, −1)

(2.67, − 10.48) −12

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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NOT FOR SALE

Review Exercises ffor Chapter 1

27. f 2

6, f 1

y

3 4

Points: 2, 6 , 1, 3 3 6 1 2

m

y 6

9 3

3

x −4 −3 −2 −1

3 x 2

y 6

3 x 6

y

3 x

f x

3 x

1

2

3

y

x 2 2, x 2 ° 2 d x d 0 ®5, °8 x 5, x ! 0 ¯

32. f x

3

8 6 4 2 −8 −6 −4 −2

4

−2

−4

−3

−6

−4

−8

33.

f x

157

x 2

4

6

8

x5 4 x 7

f x

x 5

4 x 7

x 4x 7 5

28. f 0

5, f 4

8

x

3 4

3 x 0 4 3 x 5 4 3 x 5 4

f x 29. f x

z f x

1

40

y

z f x

2

0, 5 , 4, 8 8 5 m y 5

y

−7

−5 −4 − 3 − 2 −1

1

Neither even nor odd

−2 −3 −4 −5

34.

−6

f x f x

35.

f x f x

4

20 x

2

x 4 20 x 2

f x

x2 3

2x

2 x 2 x

y

x

2

3

x 3 2

f x

7 6 5

The function is odd.

4 3 2

36. x 1 2 3 4 5 6

−2

30. g x

x

The function is even.

axb 2

−1

x 4 20 x 2

f x

5

6 x2

f x

5

6 x

2

5

6 x2

f x

The function is even.

ax 4b

37. Parent function: f x

x3

y

Horizontal shift 4 units to the left and a vertical shift 4 units upward

4

38. Parent function: f x

2 1 − 5 − 4 −3 −2 −1 −1

x 1

x t 1 5 x 3, ® 4 x 5, x 1 ¯

31. f x

Vertical shift 4 units upward. 39. (a) f x

−2

(b) h x

x2 x2 9

Vertical shift 9 units downward y

(c)

y

x

2 −6 −4 6

x −2 −4

2

4

6

h

x −12

−6

6

12

− 10

−12

INSTRUCTOR USE ONLY (d) d) h x

f x 9

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158

NOT FOR SALE

Chapter 1

Functions ctions and Their Graphs

40. (a) f x

x3

(b) h x

x

43. h x

2

(a) f x

2 2 3

Horizontal shift 2 units to the right; vertical shift 2 units upward (c)

x 2 3 x2

(b) Horizontal shift two units to the left, vertical shift 3 units upward, reflection in the x-axis. (c)

y

y 4

5

h

4 3

−8 −6

−2

2

− 2 −1 −1

1

2

3

4

−8

f x 2 2

(d) h x 44. h x

x 4

(a) f x

1 2

f x 2 3

x

(a) f x

x

(b) Vertical shift 4 units upward, reflection in the x-axis (c)

−6

h

x

41. h x

4

−4

1

(d) h x

x 2

−2

y

1 2 2

x2

(b) Horizontal shift one unit to the right, vertical shrink, vertical shift 2 units downward. (c)

10

y

8

10

6

8

h

6

4

4

h

2

2

x 2

(d) h x 42. (a) f x

(b) h x

4

6

8

10

−6 −4

f x 4 (d) h x

x x 3 5

Horizontal shift 3 units to the left; vertical shift 5 units downward

(c)

2 x 2

4

−4 −6 −8

(d) h x

f x 1 2

Reflection in the x-axis and a vertical shift 6 units upward

4

−4 − 2 −2

6

axb a xb 6

(b) h x

6

−10 −8

x 4

−2

1 2

45. (a) f x

y

(c) h

−2

f x 3 5

y 9 6 5 4 3 2 1

h

x −2−1 −2 −3

(d) h x

1 2 3 4 5 6

9

f x 6

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NOT FOR SALE

Review Exercises ffor Chapter 1

46. (a) f x

49. (a) f x

x

(b) h x

Horizontal shift 9 units to the right and a vertical stretch (each y-value is multiplied by 5)

Reflection in the x-axis, a horizontal shift 1 unit to the left, and a vertical shift 9 units upward (c)

axb 5a x 9b

(b) h x

x 19

(c)

y

159

10

y 25 20 15

6

5

4

x

2

4

(d) h x

(b) h x

x 4 6

Reflection in both the x- and y-axes; horizontal shift of 4 unit to the right; vertical shift of 6 units upward (c)

10 12 14

5 f x 9

50. (a) f x

x

(b) h x

6

− 15

6

f x 1 9

47. (a) f x

4

− 10

x

−2

(d) h x

2

−5

2 −4

h

10

h

x3

13 x3

Reflection in the x-axis; vertical shrink each y-value is multiplied by 13

y

(c)

y

10

3

8

h

2

6

1

h

4

x −4

(d) h x 48. (a) f x

(b) h x

2

−2

4

6

x

− 3 − 2 −1 −1

2

2

3

−2

8

−3

f x 4 6

f x 4 6

x2

(d) h x 51. (a) f x

x 1 3 2

(b) h x

Reflection in the x-axis; horizontal shift 1 unit to the left; vertical shift 3 units downward (c) −6 −4 −2 −2

x 2

4

x 2

x 4

Reflection in the x-axis, a vertical stretch (each yvalue is multiplied by 2), and a horizontal shift 4 units to the right

y 2

13 f x

(c)

y

6 2

−4 −6 −8

x −2

h

−10

2

6

8

−2 −4

h

−6

(d) h x

f x 1 3

−8

(d) h x

2 f x 4

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160

NOT FOR SALE

Chapter 1

52. (a) f x

Functions ctions and Their Graphs x

(b) h x

(b)

g

g f x

D f x

13 x 3 3 13 x 3 1

x 1

1 2

g

Vertical shrink each y -value is multiplied by

1 2

;

x 91

vertical shift 1 unit downward (c)

x 8

y

Domain: all real numbers

3 2

56. f x

1

h

x

−3 −2

2

(a)

−3

53. f x

(a) (b) (c)

1 2

g x

g

g f x

D f x

x 4, g x

57. h x

3 x

3

x3 4 7

3

x3 3

3 x

(b)

f

g x

f x g x

x 4

3 x

(c)

fg x

x

f x g x

g x

f x

2

2

4

3 x

x 4 , x 3 3 x

g x

3x 1

f g x

f g x 58. h x

x3 and g x

f 1 2 x 3

1

2 x

3

1 2 x.

h x .

x 2

Answer is not unique. One possibility: Let g x

59. (a)

1 3

x

1 3

x

8 3

3

r

f x 2

c t

x 2 and f x 3

x 2

r t c t

3

x.

h x

178.8t 856

This represents the average annual expenditures for both residential and cellular phone services from 2001 to 2006.

f 3 x 1

3 x

3

One possibility: Let f x

f g x

The domains of f x and g x are all real numbers.

1 3

2 x

2

3, g x

D g x

1

Answer is not unique. x2 4

f

(b)

Domain: all real numbers

f x g x

(a)

4

Domain: all real numbers

x 3 1 , Domain: x z 2x 1 2

g x

55. f x

3

2

2

1x 3

x 3

f

f

x 7

3

x 7 4 2x 1

(a)

(d)

x 7

f g x

D g x

x 2 3, g x

§f· (d) ¨ ¸ x ©g¹

54. f x

f

f x 1

x 2 3 2 x 1 x2 2 x 2 f g x x 2 3 2 x 1 x 2 2 x 4 fg x x 2 3 2 x 1 2 x3 x 2 6 x 3 f

3

The domains of f x and g x are all real numbers.

3

−2

(d) h x

x3 4, g x

(b)

2200

(r + c)(t) r(t) c(t)

Domain: all real numbers

7

0 0

(c)

r

c 13

178.813 856

$3180.40

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NOT FOR SALE

Review Exercises ffor Chapter 1

60. (a) N T t

161

25 2t 1 50 2t 1 300, 2 d t d 20 2

25 4t 2 4t 1 100t 50 300 100t 2 100t 25 100t 250 100t 2 275

The composition N T t represents the number of bacteria in the food as a function of time. (b) When N 750 100t 2

750, 100t 2 275 475

t2 t

4.75 2.18 hours.

After about 2.18 hours, the bacterial count will reach 750. 61.

f x

3x 8

3x 8 3y 8 3y x 8 y 3 1 y x 8 3 1 So f 1 x x 8 3 y x x 8

f f 1 x

§1 · f ¨ x 8 ¸ 3 © ¹

f 1 f x

f 1 3 x 8

62. f x

y x 5x y

§1 · 3¨ x 8 ¸ 8 x 88 3 © ¹ 1 1 3x 8 8 3 x x 3 3

x

x 4 5 x 4 5 y 4 5 y 4 5x 4

So, f 1 x f f 1 x f 1 f x

5x 4

5x 4 4 5x x 5 5 § x 4· § x 4· f 1 ¨ x 4 4 ¸ 5¨ ¸ 4 © 5 ¹ © 5 ¹

f 5 x 4

63. Yes, the function has an inverse because no horizontal lines intersect the graph at more than one point.

The function has an inverse. 64. No, the function does not have an inverse because some horizontal lines intersect the graph twice.

x 65. f x

4 13 x

Yes, the function has an inverse because no horizontal lines intersect the graph at more than one point. The function has an inverse. 6

INSTRUCTOR USE ONLY −4

8

−2 − 2

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162

Chapter 1

66. f x

x

NOT FOR SALE

Functions ctions and Their Graphs

1

70. f x

2

5x 7

(a)

No, the function does not have an inverse because some horizontal lines intersect the graph twice.

y

5x 7

x

5y 7 5y

f

−2

f

−1

x –8 –6 –4 –2

2

4

6

8

–4

x 7 5

x

–6 –8

(c) The graph of f 1 is the reflection of the graph of f in the line y x.

2 t 3

67. ht

4

y

7

1

f

2

x 7 5

−5

y 8 6

x 7

6

(b)

Yes, the function has an inverse because no horizontal lines intersect the graph at more than one point. The function has an inverse.

(d) The domains and ranges of f and f 1 are the set of all real numbers.

4

71. (a)

−4

f x

x 1

y

x 1

x

y 1

8

x

−4

68. g x

y 1

2

x 1 2

x 6

f

Yes, the function has an inverse because no horizontal lines intersect the graph at more than one point.

1

y

x

x 2 1, x t 0

Note: The inverse must have a restricted domain.

(b)

6

y

f −1

5 4 3

−8

4

f

2 −2

x

69. (a)

f x

–1

1x 2

3

y

1x 2

3

x

1y 2

3

x 3 2 x 3 f 1 x

1y 2

y 2x 6

2

3

4

5

–1

(b)

y 8 6 2 −10 − 8 −6

−2

(c) The graph of f 1 is the reflection of the graph of f in the line y x.

f −1

f x 8

−6 −8 −10

(c) The graph of f 1 is the reflection of the graph of f in the line y x. (d) The domains and ranges of f and f 1 are the set of all real numbers.

(d) The domain of f and the range of f 1 is [ 1, f). The range of f and the domain of f 1 is [0, f). 72. f x

(a)

3

x3 2 y

x3 2

x

y3 2

x 2

y3

x 2

y

f 1 x

3

y

(b) 4

f

3

f −1 1 x –4 –3 –2

x 2

1

3

4

–2 –3 –4

1

(c) The graph of f is the reflection of the graph of f in the line y

x.

(d) The domains and ranges of f and f 1 are the set of all real numbers.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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NOT FOR SALE

Review Exercises ffor Chapter 1

73. f x

2 x 4 is increasing on 4, f . 2

Let f x

77.

2 x 4 , x ! 4 and y ! 0.

D

km

4

2.5k

2

1.6

k

y

2 x 4

x

2 y 4 , x ! 0, y ! 4

In 2 miles:

x 2

y

D

x 2

y 4

4

8 m 5

D

2

2

or D

1.6 2

1.6m

3.2 kilometers

In 10 miles:

x 4 2

k 27

750

x 2

16 kilometers

kS 3

P

78.

x 4, x ! 0 2

3

k

0.03810395

P

0.03810395 40

3

2438.7 kilowatts

Increasing on 2, f Let f x

1.610

D

y

f 1 x 74. f x

2

ks 2

79. F

x 2, x ! 2, y ! 0.

If speed is doubled,

y

x 2

F

k 2s

x

4ks 2 .

2

y 2, x ! 0, y ! 2

F

x 2

y, x ! 0, y ! 2

f 1 x

So, the force will be changed by a factor of 4.

x 2, x ! 0

75. V

80.

0.742t 13.62

(a)

k p

x

k 5 4000

800

V

Value of shipments (in billions of dollars)

14

k

13 12 11

4000 | 667 boxes 6

x

10 9 8 7

81. T

t 1 2 3 4 5 6 7 8

Year (0 ↔ 2000)

3 (b) The model is a good fit to the actual data. k 76. (a)

1760

T

k r k 65 365

12

T 16.22t 1568.8

195

195 r

When r 0 1600

(b) H

163

80 mph,

195 2.4375 hours 80 | 2 hours, 26 minutes.

The model is a good fit to the actual data. (c) 2020 o use H 20

t

82.

20

16.22 20 1568.8

1893.2 hours

(d) The projected number of hours of television usage in the United States increases by about 16.22 hours per year.

C 28.80

khw2 k 16 6

2

k

0.05

C

0.05 14 8 2 $44.80

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164

NOT FOR SALE

Chapter 1

Functions ctions and Their Graphs k x

83. y is inversely proportional to x : y

y

y

9 when x

5.5

9

k k 5.5

9 5.5

cx

x

−9 −6 −3 −3

3

6

9

−6 −9 −12 −18

y

6

c 9 4

6

18c

1 3

c

86. True. If f x

x3 and g x

3

x , then the domain

of g is all real numbers, which is equal to the range of f and vice versa.

1 x 3

F

3

49.5

49.5 x

So, y 84. F

85. False. The graph is reflected in the x-axis, shifted 9 units to the left, then shifted 13 units down.

87. The Vertical Line Test is used to determine if a graph of y is a function of x. The Horizontal Line Test is used to determine if a function has an inverse function.

y

88. A function from a set A to a set B is a relation that assigns to each element x in the set A exactly one element y in the set B. 89. If y

kx, then the y-intercept is (0, 0).

Chapter Test for Chapter 1 4x 3 x

2

(a) f 7

(a) 0, 3

4 32

(b) f 5

2 56

1 8

(b) 1 28

x

(c) f x 9 2. f x

4. f x

x 9 x 81

1. f x

x

10

9 81 2

10

−2

x x 2 18 x

4

−10

(c) Increasing on f, 2 Decreasing on 2, 3

3 x

Domain: 3 x t 0 x t 3 x d 3

(d) The function is neither odd nor even. 5. f x

x 5

(a) –5 3. f x

2 x6 5 x 4 x 2

(b)

10

(a) 0, r 0.4314 (b)

0.1 − 12 −1

6 −2

1

(c) Increasing on 5, f Decreasing on f, 5

− 0.1

(c) Increasing on 0.31, 0 , 0.31, f

(d) The function is neither odd nor even.

Decreasing on f, 0.31 , 0, 0.31 (d) y -axis symmetry The function is even.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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NOT FOR SALE

Chapter Test ffor Chapter 1

9. h x

3x 7, x d 3 ® 2 ¯4 x 1, x ! 3

6. f x

a xb

Parent function: f x

y

165

axb

Transformation: Reflection in the x-axis

30

y

20

6

10

4

x

−2 − 10

2

4

6

−20

x −4

−30

7. (a) f 10

12, f 16

1 12 16 10

(b)

13 26

f x

6

−6

1 x 16 2 1 x 7 2

f x 1

4 −2 −4

1

10, 12 , 16, 1 m

−2

1 2

10. h x

x 5 8

Parent function: f x

x

Transformation: Reflection in the x-axis, a horizontal shift 5 units to the left, and a vertical shift 8 units upward y 10 8

y 16 14 12 10

4 2 x −6 − 4 −2 −2

6 4 2

11. h x

x −2

2 4 6 8 10 12 14

2

4

6

2 x 5 3 3

Parent function: f x §1· 8. (a) f ¨ ¸ © 2¹

6, f 4

3

Transformation: Reflection in the x-axis, vertical stretch, a horizontal shift 5 units to the right, and a vertical shift 3 units upward

§1 · ¨ , 6 ¸, 4, 3 ©2 ¹ 3 72

4 1 2

y 3 y 3 y f x

(b)

y

3 6

m

x3

6 x 4 7 6 24 x 7 7 6 45 x 7 7 6 45 x 7 7

6

6 7

4 2 −2

x −2

2

4

8

10

−4 −6

y 1 x −1 −2 −3 −4 −5 −6

1 2 3 4 5 6

8 9

−8 −9

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166

NOT FOR SALE

Chapter 1

12. f x

(a) (b) (c)

Functions ctions and Their Graphs

3 x 2 7, g x

x2 4 x 5

3x 2 7 x2 4 x 5 f g x 3x 2 7 x 2 4 x 5 fg x 3x 2 7 x 2 4 x 5 3x 4 f

g x

§f· (d) ¨ ¸ x ©g¹

12 x3 22 x 2 28 x 35

3x 2 7 , x z 5, 1 x 4x 5

f

D g x

f g x

f x 2 4 x 5

(f )

g

D f x

g f x

g 3 x 2 7

1 , g x x

2

1 2 x

x

1 2 x3 2 ,x ! 0 x

(b)

f

g x

1 2 x

x

1 2 x3 2 ,x ! 0 x

(c)

fg x

x

1 x

§f· (d) ¨ ¸ x ©g¹

2

x x

1

2

x

2x

1 ,x ! 0 2 x3 2

x

f

D g x

f g x

f 2

(f)

g

D f x

g f x

§1· g¨ ¸ © x¹

x

1 2 2

1 x

x ,x ! 0 2x

x 2 x

Since f is one-to-one, f has an inverse.

3

x

y3 8

3

x 8

y3

x 8

y

f 1 x

3

15. f x

2

x x

,x ! 0 16. f x

x3 8

x 8

9 x 4 30 x 2 16

,x ! 0

(e)

y

3 x 4 24 x3 18 x 2 120 x 68

x

g x

2

2

f

§1· ¨ ¸2 © x¹

3 x 2 4 x 5 7

3 x 2 7 43x 2 7 5

(a)

14. f x

4 x 2 4 x 12

2

(e)

13. f x

2x2 4x 2

3x

x

3x3 2

Domain: >0, f Range: >0, f The graph of f x passes the Horizontal Line Test, so f x is one-to-one and has an inverse.

x 8

f x

3 x3 2

y

3x3 2

3

x

3 y3 2

2

x 3

y3 2

1

x2 3 6

Since f is not one-to-one, f does not have an inverse. § x· ¨ ¸ © 3¹

x

23

f 1 x

y

1

y

2

3

23

§ x· ¨ ¸ ,x t 0 © 3¹

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE

Problem Solving ffor Chapter 1

17.

v

k

24

18

k

v

6 s

A

7

kxy k 15 8

500

k 120

48 b

(b) S (c)

0.0297t 3 1.175t 2 12.96t 79.0

55

k

A

32

17 30

500

b

55

k 16

6

25 6

19.

20. (a)

s

167

25 xy 6

7

17 30

The model is a good fit for the data.

k a k 1.5 k

25: S 25 | $25.3 billion

(d) For 2015, use t

No, this does not seem reasonable. The model decreases sharply after 2009.

48 a

Problem Solving for Chapter 1 1. Mapping numbers onto letters is not a function. Each number between 2 and 9 is mapped to more than one letter.

^2, A , 2, B , 2, C , 3, D , 3, E , 3, F , 4, G , 4, H , 4, I , 5, J , 5, K , 5, L , 6, M , 6, N , 6, O , 7, P , 7, Q , 7, R , 7, S , 8, T , 8, U , 8, V , 9, W , 9, X , 9, Y , 9, Z ` Mapping letters onto numbers is a function. Each letter is only mapped to one number.

^ A, 2 , B, 2 , C , 2 , D, 3 , E , 3 , F , 3 , G, 4 , H , 4 , I , 4 , J , 5 , K , 5 , L, 5 , M , 6 , N , 6 , O, 6 , P, 7 , Q, 7 , R, 7 , S , 7 , T , 8 , U , 8 , V , 8 , W , 9 , X , 9 , Y , 9 , Z , 9 ` 2. (a) Let f x and g x be two even functions.

Then define h x h x

f x r g x .

f x r g x

Then define h x h x

f x r g x .

f x r g x

f x r g x because f and g are even

f x r g x because f and g are odd

h x

h x So, h x is also odd. If f x z g x

So, h x is also even. (c) Let f x be odd and g x be even. Then define h x h x

(b) Let f x and g x be two odd functions.

f x r g x .

f x r g x

f x r g x because f is odd and g is even z h x z h x So, h x is neither odd nor even.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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168

Chapter 1 f x

3.

f x

NOT FOR SALE

Functions ctions and Their Graphs

a2 n 1 x 2 n 1 a2 n 1 x 2 n 1 " a3 x3 a1 x a2 n 1 x

2 n 1

a2 n 1 x

2 n 1

" a3 x a1 x 3

f x

a2 n 1 x 2 n 1 a2 n 1 x 2 n 1 " a3 x3 a1 x Therefore, f x is odd. f x

4.

f x

a2 n x 2 n a2 n 2 x 2 n 2 " a2 x 2 a0 a2 n x

2n

a2 n 2 x

2n 2

" a2 x a0 2

a2 n x 2 n a2 n 2 x 2 n 2 " a2 x 2 a0 f x Therefore, f x is even. 5. (a) y

(b) y

x

x2

4

−6

6

−6

4

−6

6

6

−4

x4

(e) y

−4

x5

4

−6

x3

4

−4

(d) y

(c) y

(f ) y

x6 4

4

6

−6

−4

−6

6

6

−4

−4

All the graphs pass through the origin. The graphs of the odd powers of x are symmetric with respect to the origin and the graphs of the even powers are symmetric with respect to the y-axis. As the powers increase, the graphs become flatter in the interval 1 x 1. (g) The graph of y The graph of y

x 7 will pass through the origin and will be symmetric with the origin. x8 will pass through the origin and will be symmetric with respect to the y-axis.

4

−6

4

6

−6

−4

−4

6. If you consider the x-axis to be a mirror, the graph of y f x is the mirror image of the graph of

y

f x .

6

7. y

f x 2 1

Horizontal shift 2 units to the left and a vertical shift 1 unit downward.

0, 1 o 0 2, 1 1 2, 0 1, 2 o 1 2, 2 1 1, 1 2, 3 o 2 2, 3 1 0, 2

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE

Problem Solving ffor Chapter 1

8. Let f x and g x be two odd functions and define

h x

f x g x . Then

h x

f x g x

12.

ª ¬ f x º¼ ª ¬ g x º¼ Since f and g are odd f x g x h x

x

f x

x

f 1 x

–2

–5

–5

–2

–1

–2

–2

–1

1

2

2

1

3

3

3

3

y

Thus, h x is even.

5

Let f x and g x be two even functions and define h x

f x g x . Then

h x

f x g x f x g x

4 3 2 1 1

−1

2

3

−2

Since f and g are even

−3

13.

Thus, h x is even. 9. Let f x be an odd function, g x be an even function h x

x

−5 −4 −3

h x

and define h x

169

f x g x . Then

x

–2

–1

3

4

f

6

0

–2

–3

x f 1 x

f x g x ª ¬ f x º¼ g x Since f is odd and g is even.

–3

–2

0

6

4

3

–1

–2

y

f x g x

6

h x

4 2

Thus, h is odd.

x –4

10. (a) The profits were only

g t

3 4

f t

3 4

2

4

–4

(c) There was a 2-year delay: g t

f t 2

14.

x

f x

–4

3

–2

4

x

1

3

4

6

0

0

f

1

2

6

7

3

–1

x f 1 x

1

2

6

7

1

3

4

6

The graph does not pass the Horizontal Line Test, so f 1 x does not exist. 15. If f x

y

f 1 3

8

k 2 x x3 has an inverse and 2, then f 2

f 2

6 4

6

–2

as large as expected:

(b) The profits were $10,000 greater than predicted: g t f t 10,000

11.

–2

3. So,

k 2 2 2

k 2 2 8

3

12k

3

3

3

2 x 2

4

6

8

k

3 12

1 4

INSTRUCTOR USE ONLY So, k

1. 4

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170

NOT FOR SALE

Chapter 1

Functions ctions and Their Graphs

22 x 2 , and the length of the trip over land is

16. (a) The length of the trip in the water is

1 3 x . 2

The total time is 1 3 x

4 x2 2

T x 1 2

2

4

1 4 x2 4

x 2 6 x 10

(b) Domain of T x : 0 d x d 3 (c)

3

0

3 0

(d) T x is a minimum when x

1.

(e) Answers will vary. Sample answer: To reach point Q in the shortest amount of time, you should row to a point one mile down the coast, and then walk the rest of the way. 1, x t 0 ® ¯0, x 0

17. H x

y 4 3 2 1 x

−3 −2 − 1

1

2

3

4

−2 −3 −4

(a) H x 2

(b)

H x 2

y

−3

−2

−1

(c)

H x y

y

3

3

3

2

2

2

1

1

1

2

−3 −2 −1 −1

3

−1

−3

(d) H x

1 x

x

(e)

1 H 2

1

2

−3 −2 −1

3

x

−2

−2

−3

−3

x

(f )

3

3

2

2

−2 −1

−1

1

2

3

3

y 3

1

x

2

H x 2 2

y

y

−3

1

−1

1 x

−3 −2 −1 −1

1

2

3

−3 −2 −1

x −1

−2

−2

−2

−3

−3

−3

1

2

3

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© Cengage Learning. All Rights Reserved.

NOT FOR SALE

Problem Solving ffor Chapter 1

18. f x

1 1 x

y

(a) Domain: all real numbers x except x Range: all real numbers y except y § 1 · f¨ ¸ ©1 x ¹ 1 § 1 · 1¨ ¸ ©1 x ¹

(b) f f x

1 x x

f D g D h x f D g D h x

1 1 x

x

1, 1 .

1 1 x 1 1 x

y 2 1 x

f g D h x

D g h x

20. (a) f x 1

1 § x 1· 1¨ ¸ © x ¹

The graph is not a line. It has holes at 0, 0 and

x 1 x

f

§ x 1· f¨ ¸ © x ¹

0

Domain: all real numbers x except x x 1

19.

(c) f f f x

1

171

−2

0 and

1

2

−2

f g h x

f g h x

(b)

f

D g D h x

f

D g D h x

f x 1

(c)

y

2 f x y

y

4

4

3

3

4

1 −4 −3

x

−1

1

2

3

−4 −3

4

x 1

−1

3

x

− 4 − 3 −2

4

−2

−2

−2

−3

−3

−3

−4

−4

−4

(d) f x

(e)

f x

(f )

y

2

3

4

2

3

4

f x

y

y

4

4

4

3

3

3

2

2

2

1 − 4 − 3 −2

(g) f x

x 2

−1

3

x

− 4 −3 −2

4

2

3

4

− 4 −3 −2 −1

−2

−2

−3

−3

−3

−4

−4

−4

x 1

−2

y 4 3 2

− 4 −3 −2

x −1

2

3

4

−2 −3 −4

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172 21.

Chapter 1

NOT FOR SALE

Functions ctions and Their Graphs

x

f f 1 x

2

–4

f f 1 4

f 2

4

1

–2

f f 1 2

f 0

–2

1

0

0

f f 1 0

–1

0

—

4

f f 1 4

0

–2

–1

1

–3

–2

x

f

2

–4

—

–3

f 3 f 1 3

4 1 1 0

0

–3

f

2 1

2

0

x

f x

f 1 x

–4

—

–3

(a)

(c)

x

f

4

–3

f 3 f 1 3

41

5

2

–2

f 2 f 1 2

1 0

1

f 1

0

0

f 0 f 1 0

f 3

4

1

f 1 f 1 1

(b)

f 1 x

3

—

—

–2

f 2 f

4

—

–3

0

f 0 f 1 0

1

f 1 f

1

1

2

1

(d)

3 2

4

6

f 1 x

2 1 3 2

x

f 1 x

–4

f 1 4

2

2

3

1

1

f 1 0

1

1

4

3

3

4

f

1

1

3 5

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© Cengage Learning. All Rights Reserved.

NOT FOR SALE C H A P T E R 2 Polynomial and Rational Functions Section 2.1

Quadratic Functions and Models .......................................................174

Section 2.2

Polynomial Functions of Higher Degree...........................................188

Section 2.3

Polynomial and Synthetic Division ...................................................201

Section 2.4

Complex Numbers..............................................................................215

Section 2.5

The Fundamental Theorem of Algebra .............................................220

Section 2.6

Rational Functions..............................................................................237

Review Exercises ........................................................................................................252 Chapter Test .............................................................................................................270 Problem Solving .........................................................................................................272

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE C H A P T E R 2 Polynomial and Rational Functions Section 2.1 Quadratic Functions and Models 8. f x

1. polynomial

x

4 opens upward and has vertex 4, 0 . 2

Matches graph (c).

2. nonnegative integer; real

9. f x

3. quadratic; parabola

x 2 2 opens upward and has vertex 0, 2 .

Matches graph (b).

4. axis

10. f x

5. positive; minimum

1, 2 .

6. negative; maximum 7. f x

11. f x

2 opens upward and has vertex 2, 0 .

x

x

2

18 x 2

y

(c)

2

Matches graph (d) 3 2 x 2

y

(d)

3 x 2

y

y

y

5

6

5

6

4

4

4

4

3

3 2

−6 −4

1 1

2

x 4

−2

2

6

3

Vertical shrink

Vertical shrink and reflection in the x-axis

x2 1

(b)

y

1

−1

2

(c)

(d)

x2 3

y

y

y

5

4

10

8

4

3

8

6

3

2

6

4

2

1 x

x − 3 −2 −1 −1

1

2

–2

2

3

3

Vertical shift one unit upward

6

Vertical stretch and reflection in the x-axis

x2 3

y

y

–3

4

3

Vertical stretch

x2 1

y

x

2

x −3 −2 −1

−6

–1

x

− 6 −4 − 2

1

−4

x –3 –2 –1

14. (a) y

x 2 4 opens

2

y

y

2

x 4 opens downward and has vertex

4, 0 . (b)

Matches graph (a).

4 x 2

12. f x

1 x2 2

2

downward and has vertex 2, 4 . Matches graph (f).

Matches graph (e).

13. (a) y

1 2 opens upward and has vertex

–2

Vertical shift one unit downward

− 6 −4 −2 −2

x 2

4

6

Vertical shift three units upward

− 6 –4

–2

4

6

−4

Vertical shift three units downward

INSTRUCTOR USE ONLY 174

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© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.1

x

15. (a) y

1

2

(b)

3 x 2

y

y 5 4 3

1

(c)

Quadratic Function Functions and Models

13 x

y

y

2

3

(d)

x

y

3

y

175

2

y

5

8

10

4

6

8

3

4

x x –2

–1

1

2

3

x −3 −2 −1 −1

4

–1

Horizontal shift one unit to the right

1

2

−2 −2

3

6

2

x −8 −6 −4 −2 −2

−4

Horizontal shrink and a vertical shift

12 x 2 1 2

16. (a) y

−6

Horizontal stretch and a vertical shift three one unit upward

2

4

Horizontal shift three units to the left units downward

12 x 2 1 2

(c) y

y

y 8

6

6

4

4

2

6

x

−8 −6 −4

x

−6 −4 −2

2

4

6

8 10

−4 −6 −8

Horizontal shift two units to the right, vertical shrink

Horizontal shift two units to the left, vertical shrink

each y-value is multiplied by , reflection in the

each y-value is multiplied by 12 , reflection in

x-axis, and vertical shift one unit upward

x-axis, and vertical shift one unit downward

1 2

ª 1 x 1 º 3 ¬2 ¼ 2

(b) y

¬ª2 x 1 º¼ 4 2

(d) y

y

y 7

10 8 6

4

4

3 2

x

−8 −6 −4

2

6

8

1 x

−4

−4 − 3 − 2 −1 −1

−6

1

2

3

4

Horizontal shift one unit to the left, horizontal shrink

Horizontal shift one unit to the right, horizontal stretch each x-value is multiplied by 2 , and vertical

each x-value is multiplied by 12 , and vertical shift

shift three units downward

four units upward

17. f x

18. f x

1 x2

Vertex: 0, 1 Axis of symmetry: x or the y-axis

Vertex: 0, 8

y 4

0

2 x −4 −3 −2

0

1

x

r1

x

2

x-intercepts: 1, 0 , 1, 0

−1

1

2

3

4

x2 8

0

−3

2

8

x

r2

x

6

0

4 2 x −8 −6 −4

Find x-intercepts:

−2 −4

y

Axis of symmetry: x or the y-axis

3

Find x-intercepts: 1 x2

x2 8

2

4

6

8

−10

2

x-intercepts: 2 2, 0 , 2 2, 0

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176

NOT FOR SALE

Chapter 2

19. f x

Polynomial ynomial and Rational Function Functions

23. f x

x2 7

Vertex: 0, 7

y

Axis of symmetry: x or the y-axis

4 3 2

Vertex: 4, 3

14

0

x

12

0

2

7

2

4

6

8

x

4 3

3

x

4

12 x

r 3 4 r

2

Vertex: 0, 12

1

0 x

−8 −6 −4 −2

x-intercepts: none 20. f x

3

Find x-intercepts:

6 4

x2

4

4

Axis of symmetry: x

2

Find x-intercepts: x2 7

y

−4 −3

−1 −1

x 1

−2 −3

2

−4

x 4

3

x

x-intercepts: 4

y

−7 −6

2

3, 0 , 4

3, 0

14

Axis of symmetry: x or the y-axis

0

4

x2

x

r2

−2

2

6

8

4

1 2

x

2

0

2

3

4

8

x

6

2

8

–5

x

r 8

4 2

2

x 2 8 x 16

Axis of symmetry: x

x

4

2

x −2 −2

x

4

2

4

6

y

4

16 12 8

0

x

4

4 x –4

4

8

26. g x

x2 2 x 1

x

1

y

Axis of symmetry: x

y 6

1

5

18

0

x2

64

x

r8

4

Find x-intercepts:

0

Find x-intercepts: 1 x2 4

16

2

Vertex: 1, 0

14 x 2 16

Vertex: 0, 16

16

12

x-intercept: 4, 0

Axis of symmetry: x or the y-axis

8 10 12 14

2

0

x 4

r2 2

1 x2 4

6

Find x-intercepts:

–2 –3

16

6 8

20 x 1

0

x

8

2

1 –1

x-intercepts:

22. f x

12

Vertex: 4, 0

2

–4 –3

Find x-intercepts:

2, 0 , 2 2, 0

0

25. h x

3

2

6

x-intercepts: none

0 4 y

Axis of symmetry: x or the y-axis

x

Axis of symmetry: x

10

3

Vertex: 0, 4

4

y

Find x-intercepts:

x-intercepts: 2 3, 0 , 2 3, 0

1 x2 2

2

14

x −8 −6

12

1 2 x 2

6 8

2

0

21. f x

x

Vertex: 6, 8

6

Find x-intercepts: 12 x 2

24. f x

8

x

12 9

1

2

x 1

6 3 −9 −6 −3 −3

x 3

6

9

x

3

0

2 1

0 1

x –4

–3

–2

–1

1

2

x-intercept: 1, 0

x-intercepts: 8, 0 , 8, 0

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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NOT FOR SALE Section 2.1

27. f x

29. f x

5 4 1· 1 5 § 2 ¨x x ¸ 4¹ 4 4 © x2 x

x2 2 x 5

x 1 6 2

Vertex: 1, 6

2

Axis of symmetry: x Find x-intercepts: 5

1 2

Axis of symmetry: x Find x-intercepts:

4

x2 2 x 5

0

3

x 2x 5

0

1

x

2

x

0

–2

1r

x

1

y

§1 · Vertex: ¨ , 1¸ ©2 ¹

5 4

177

x 2 2 x 1 1 5

1· § ¨x ¸ 1 2¹ ©

x2 x

Quadratic Function Functions and Models

–1

1

2

3

2r

4 20 2

1r

6

x-intercepts: 1

15 2

6, 0 , 1

6, 0

y

Not a real number

6

No x-intercepts 28. f x

1 4 9· 9 1 § 2 ¨ x 3x ¸ 4 4 4 © ¹ x 2 3x

x –4

2

6

–2 –4

2

3· § ¨x ¸ 2 2¹ ©

30. f x

x2 4x 1

x 2 4 x 4 4 1

§ 3 · Vertex: ¨ , 2 ¸ © 2 ¹

x 2 5 2

Axis of symmetry: x

Vertex: 2, 5

3 2

1 4

Find x-intercepts: x 4 x 1

0

x 4x 1

0

2

0

2

3 r

x

2

Axis of symmetry: x

Find x-intercepts: x 2 3x

x 2 4 x 1

9 1

x

2

3 r 2

§ 3 x-intercepts: ¨ © 2

4 r 2 r

2

· § 3 2, 0 ¸, ¨ ¹ © 2

· 2, 0 ¸ ¹

x-intercepts: 2

5, 0 , 2

5, 0

16 4 2 5

y 5

y

4

4 2

3

1

2

x

1

–6 –5 x

–5 –4 –3 –2 –1

1

2

–3 –2 –1

1

2

–2 –3

–2 –3

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178

NOT FOR SALE

Chapter 2

31. h x

Polynomial ynomial and Rational Function Functions

34. f x

4 x 2 4 x 21

13 x 2 9 x 6

1· § §1· 4¨ x 2 x ¸ 4¨ ¸ 21 4¹ © © 4¹

13 x 92 Vertex: 92 , 34

13 x 2 9 x

2

1· § 4¨ x ¸ 20 2¹ ©

y

§1 · Vertex: ¨ , 20 ¸ ©2 ¹ Axis of symmetry: x

1 2

10

4 x 2 4 x 21

–4

4

8

0 4r

x

16 336 2 4

No x-intercepts 2 x2 x 1

0

4 3

1 4

1 x –3

–2

–1

1

2

18 2 2

1

2

1 4

14 30

2 x 12

x 8 x 16

1 4

x

1 4

16

Vertex: 4, 16

x-intercepts: 4 r 38. f x

y 4

4

x –8

4

Find x-intercepts: 0

4 x 12

0

x

4 or x

–12 –16 –20

12

10

−80

x

4 5 2

14

4

Axis of symmetry: x

12

2

x

−10

12

x 2 8 x 11

4 16

0

35

121 4

Vertex: 4, 5

2

7

x 2 x 30

x 12 Vertex: 12 , 121 4

37. g x

No x-intercepts

x 8 x 48

−8

x-intercepts: 6, 0 , 5, 0

Not a real number

2 x 12

2

5

Axis of symmetry: x

2

x 1 4

3

0

Axis of symmetry: x

−4

x 2 2 x 3

x2 x

5

1 4

10

x 2 x 30

6

Find x-intercepts:

1 x2 4

0

x 3 x 6

y

§1 7· Vertex: ¨ , ¸ ©4 8¹

1 2 x 4

x 2 9 x 18

36. f x

2

33. f x

8

−5

1· 7 § 2¨ x ¸ 4¹ 8 ©

x

6

x-intercepts: 3, 0 , 1, 0

2

1r

4

−6

0

Axis of symmetry: x

1· § §1· 2¨ x ¸ 2¨ ¸ 1 4¹ © © 16 ¹

2x x 1

x −2 −2

9 2

Vertex: 1, 4

1 · § 2¨ x 2 x ¸ 1 2 ¹ ©

2

y 2

13 x 2 3 x 6

35. f x

Axis of symmetry: x

3 4

x-intercepts: 3, 0 , 6, 0

Not a real number

32. f x

13 814 6

Find x-intercepts: x

–8

2

81 4

Axis of symmetry: x

20

Find x-intercepts:

13 x 2 3 x 6

8

16

5, 0

−18

12

−6

x 2 10 x 14

x2 x

10 x 25 25 14 5 11 2

5 −20

Vertex: 5, 11 Axis of symmetry: x

x-intercepts: 5 r

10

5 11, 0

−15

INSTRUCTOR USE ONLY x-intercepts: x-intercepts: 4, 0 , 12, 0

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.1

39. f x

2 x 2 16 x 31

f x

2

Vertex: 4, 1

−6

Axis of symmetry: x

40. f x

1 2

12

2, 0

4 x 2 24 x 41 4 x 2 6 x 41

2

4 x 2

1 2

x

x

3 5

x

2 6, 0

2

−8

3 2

27 5

42 5

10

−10

2

Because the graph passes through 1, 0 , a1 1 4 2

4a

1

a.

So, y

a 1 2 2 2

a.

2

2

a x 2 0 2

2

a3 2

2

a.

a x 2

2

2

2 x 2 . 2

47. 2, 5 is the vertex.

a x 2 5 2

Because the graph passes through 0, 9 , a 0 2 5 2

4

4a

1

a.

So, f x

1 x 2 5 2

x

2 5. 2

48. 4, 1 is the vertex.

f x

1 x 1 4

2 x 2 2.

46. 2, 0 is the vertex.

9

a x 1 4

4

2

f x

3

43. 1, 4 is the vertex.

0

2

a x 2 2

So, y

6

14, 0

2 1.

Because the graph passes through 3, 2 ,

3

−14

x-intercepts: 3 r

x

f x −4

Axis of symmetry: x

y

a.

So, y 4

6 x 9

Vertex: 3, 42 5

1

2

2 3

x 2 6 x 5 2

4a

0

4

x-intercepts: 2 r

3 5

4a 1

−20

Axis of symmetry: x

3 5

3 4

Because the graph passes through 1, 0 ,

Vertex: 2, 3

42. f x

2

6

3

No x-intercepts

x2

a 0 2 1

y

Axis of symmetry: x

1 2

3

45. 2, 2 is the vertex.

0 0

Vertex: 3, 5

41. g x

2

So, y

4 x 2 6 x 9 36 41 4 x 3 5

a x 2 1

Because the graph passes through 0, 3 ,

−12

4

179

44. 2, 1 is the vertex.

48

2 x 4 1

x-intercepts: 4 r

Quadratic Function Functions and Models

x 1 4. 2

a x 4 1 2

Because the graph passes through 2, 3 , 3

a 2 4 1

3

4a 1

4

4a

1

a.

2

So, f x

x

4 1. 2

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180

NOT FOR SALE

Chapter 2

Polynomial ynomial and Rational Function Functions

49. 1, 2 is the vertex.

f x

54.

a x 1 2 2

Because the graph passes through 1, 14 , 14

a 1 1 2

14

4a 2

16

4a

4

a.

f x

4 x 1 2. 2

a x 2 3 2

2

a0 2 3

2

4a 3

2

1 14

a 2

4

81 a 34 4 81 a 4

19 81

14 x 2 3.

f x

a x

5 2

2

34 .

2

5 2

2

16 x 3

a7 5 12

f x

5 2

. 2

a x 6 6 2

Because the graph passes through

2

3 . 4

5 12. 2

52. 2, 2 is the vertex.

Because the graph passes through 1, 0 , a 1 2 2

0

a 2

2

a.

a

3 2

1 a 100

92

1 a 100

2

2

1061, 23 ,

6

6

a.

So, f x

2

0

1061 6

3 2

450

a x 2 2

So, f x

3 4

56. 6, 6 is the vertex.

Because the graph passes through 7, 15 ,

f x

So, f x

a x 5 12

x

2

a.

2

3 4

3 4

55. 52 , 0 is the vertex.

a.

So, f x

x 52

16 3

4a a

2

a.

a 72

2

3 2

5 2

16 3

f x

Because the graph passes through 72 , 16 , 3

51. 5, 12 is the vertex.

f x

4

4a

So, f x

5 2

Because the graph passes through 2, 4 ,

So, f x

Because the graph passes through 0, 2 ,

53. 14 ,

a x

19 4 19 81

50. 2, 3 is the vertex.

3

f x

2

So, f x

15

52 , 34 is the vertex.

57. f x

450 x 6 6. 2

x2 4 x

4

x-intercepts: 0, 0 , 4, 0

2 x 2 2. 2

is the vertex. a x 14 32

−4

0

x2 4 x

0

x x 4

x

0

or

x

8

−4

4

The x-intercepts and the solutions of f x

0 are the

same.

2

Because the graph passes through 2, 0 , 0 32

a 2 49 a 16

So, f x

1 4

a

2

3 2

24 . 49

2

INSTRUCTOR USE ONLY 24 x 49

1 4

23 .

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.1

58. f x

63. f x

14

2 x 2 10 x

2 x 2 10 x

0

2 x x 5

2 x

−1

x 5

g x

x2 2x 3

0 are the

Note: f x x 2 9 x 18

0

x 2 9 x 18

0

x

x

3 or x

64. f x −8

3 x 6

x 25, opens upward

−4

6

0 are the

g x

f x , opens downward

g x

x 2 25

x 2 8 x 20

x-intercepts: 2, 0 , 10, 0 0

x 2 8 x 20

0

x

−4

12

65. f x g x

2

opens upward

0 x

x 0 x 10

opens downward

x 10 x 2

10

Note: f x

0 are the

a x 0 x 10

ax x 10 has

x-intercepts 0, 0 and 10, 0 for all real

same.

numbers a z 0.

61. f x

2 x 7 x 30 2

x-intercepts: 52 , 0 , 6, 0 0

2 x 2 7 x 30

0

2 x

x

52

10 −5

x

2

−40

g x

f x , opens downward

g x

x 2 12 x 32

10 −18

ª¬ x 3 º¼ ª x 12 º 2 ¬ ¼

x

4

x

12 x 45

3 x

1 2

opens upward

2

3 2 x 1

2x 7x 3 2

15 x 3 0 x

a x 4 x 8 has x-intercepts 4, 0

and 8,0 for all real numbers a z 0.

0 are the 67. f x

12 x 45

0 x

4 x 8

Note: f x

6

x-intercepts: 15, 0 , 3, 0

x2

x

x 12 x 32, opens upward

same. 7 10

66. f x

2

5 x 6 or x

10

The x-intercepts and the solutions of f x

x 3

0 x 10

x 10 x −40

The x-intercepts and the solutions of f x

x 15

x 2

0 x

x 10

and 5, 0 for all real numbers a z 0.

10

2 x 10

x 2

x

a x 2 25 has x-intercepts 5, 0

Note: f x

60. f x

0

5 x 5

2

same.

7 10

ª¬ x 5 º¼ x 5

x

16

The x-intercepts and the solutions of f x

a x 1 x 3 has x-intercepts 1, 0

and 3, 0 for all real numbers a z 0.

12

x-intercepts: 3, 0 , 6, 0

0

ª¬ x 1 º¼ x 3 opens downward x 2 2 x 3

5

same.

62. f x

1 x 3

x 1 x 3

The x-intercepts and the solutions of f x

59. f x

opens upward

x 2x 3

0

0 x

181

2

6

−6

0 x

ª¬ x 1 º¼ x 3

x

x-intercepts: 0, 0 , 5, 0 0

Quadratic Function Functions and Models

−60

15

g x

2 x 2 7 x 3

opens downward

= 2x2 7 x 3

3

The x-intercepts and the solutions of f x

Note: f x

a x 3 2 x 1 has x-intercepts

INSTRUCTOR USE E ONLY same. same

0 are the

3,, 0

and 12 , 0 for all real numbers a z 0.

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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182

NOT FOR SALE

Chapter 2

Polynomial ynomial and Rational Function Functions

2 x 52 x 2 2 x 12 x 5

68. f x

2 ª x 52 º x 2 ¬ ¼

(a) From the graph it appears that there are no xintercepts.

2

(b) There are no x-intercepts and there are no real solutions to the equation x 2 3 x 3 0.

2 x 2 x 10, opens upward g x

f x , opens downward

g x

2 x x 10

(c) x 2 3 x 3

0

x 3x 3

0

2

2

Note: f x

a x

5 2

x 2 has x-intercepts

52 , 0 and 2, 0 for all real numbers a z 0. 69. y

x 2 3x 3

72. y

32 41 3

3 r

2

3 2

Not a real number No x-intercepts 73. f x

x2 4 x 5

3 r

x

ax 2 bx c

(a) x-intercepts: 5, 0 , 1, 0

b · § a¨ x 2 x ¸ c a ¹ ©

(b) The x-intercepts and the solutions of the equation are the same.

§ b b2 b2 · a¨ x 2 x ¸ c 2 a 4a 4a 2 ¹ ©

x2 4x 5

(c) 0 0

x

x

5 or x

§ § b2 · b b2 · a¨ 2 ¸ c a¨ x 2 x 2¸ a 4a ¹ © © 4a ¹

5 x 1 1

2

b · b2 4ac § a¨ x ¸ 2a ¹ 4a 4a ©

The x-intercepts are 5, 0 and 1, 0 .

2

70. y

b · 4ac b 2 § a¨ x ¸ 2a ¹ 4a ©

2x2 5x 3

(a) From the graph it appears that the x-intercepts are 12 , 0 and 3, 0 . (b) The x-intercepts and solutions of 2 x 2 5 x 3 0 are the same. 2x2 5x 3

0

2 x 1 x 3

0

(c)

1 2

x

§ b · f ¨ ¸ © 2a ¹

12 , 0

and 3, 0 . 71. y

(a) From the graph it appears that the x-intercept is 1, 0 . (b) The x-intercept and the solution to x 2 2 x 1 0 are the same. (c) x 2 2 x 1

0

x 2x 1

0

x

1

2

x 1 x

b2 2b 2 4ac 4a 4a 4a b 2 4ac 4a

x2 2 x 1

2

2

§ b · § b · a ¨ ¸ b¨ ¸ c © 2a ¹ © 2a ¹ § b2 · b2 a¨ 2 ¸ c © 4a ¹ 2a

3 The x-intercepts are

or x

§ b 4ac b 2 · The vertex is ¨ , ¸. 4a © 2a ¹

4ac b 2 4a § b Thus, the vertex occurs at ¨ , © 2a

§ b ·· f ¨ ¸ ¸. © 2a ¹ ¹

74. (a) Yes, it is possible for a quadratic equation to have only one x-intercept. That happens when the vertex is the x-intercept.

0 0 1 The x-intercept is at 1, 0 .

(b) Yes. If the vertex is above the x-axis and the parabola opens upward, or if the vertex is below the x-axis and the parabola opens downward, then the graph of the quadratic equation will have no x-intercepts. Examples: f x

x 2 4; g x

x2 1

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.1

75. Let x

the first number and y

the second number.

78. Let x

110 y

The product is P x

P x

110 x. xy

x110 x

110 x x 2 .

x 2 110 x x 2 110 x 3025 3025 2 ª x 55 3025º ¬ ¼

x 55 3025 2

x y

S y

The product is P x P x

42 21 3

and y

the second number.

S x.

the first number and y

The maximum value of the product occurs at the vertex of P x and is 147. This happens when x 21

The maximum value of the product occurs at the vertex y 55. of P x and is 3025. This happens when x 76. Let x the first number and y Then the sum is

79. x S x

xy

Sx x 2 .

7. So, the numbers are 21 and 7.

x

y

y

Sx x 2 x 2 Sx

x

2x 2 y

§ S S · ¨ x 2 Sx ¸ 4 4 ¹ © 2

2

(a) A x

2

50 x x50 x

xy

Domain: 0 x 50

The maximum value of the product occurs at the vertex of P x and is S 2 4. This happens when y

100

y

S· S2 § ¨ x ¸ 2¹ 4 ©

x

183

the second number. 42 x Then the sum is x 3 y 42 y . 3 § 42 x · xy x¨ The product is P x ¸. © 3 ¹ 1 P x x2 42 x 3 1 x 2 42 x 441 441 3 1ª 1 2 2 x 21 441º x 21 147 ¼ 3¬ 3

Then the sum is

x y

Quadratic Function Functions and Models

(b)

A 700

S 2.

560 420

77. Let x

the first number and y

the second number.

280 140

Then the sum is

x 2y

24 y

The product is P x

P x

x

24 x . 2

xy

10

§ 24 x · x¨ ¸. © 2 ¹

1 x2 24 x 2 1 x 2 24 x 144 144 2 1ª 1 2 2 x 12 144º x 12 72 ¼ 2¬ 2

The maximum value of the product occurs at the vertex of P x and is 72. This happens when x 12 and

y

24 12

20

30

40

50

(c) The area is maximum (625 square feet) when x y 25. The rectangle has dimensions 25 ft u 25 ft. 80. Let x

length of rectangle and y

2x 2 y y (a) A x

36 18 x x18 x

xy

Domain: 0 x 18 (b)

(c) The area is maximum (81 square meters) when x y 9 meters. The rectangle has dimensions 9 meters u 9 meters.

A 100

2

6. So, the numbers are 12 and 6.

width of rectangle.

80 60 40 20

INSTRUCTOR USE ONLY x

4

8

12

16

20

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184

Chapter 2

NOT FOR SALE

Polynomial ynomial and Rational Function Functions

4 24 x2 x 12 9 9

81. y

The vertex occurs at

82. y

24 9 2 4 9

b 2a

3. The maximum height is y3

4 2 24 3 3 12 9 9

16 feet.

16 2 9 x x 1.5 2025 5

(a) The ball height when it is punted is the y-intercept.

y

16 9 2 0 0 1.5 2025 5

(b) The vertex occurs at x

1.5 feet

b 2a

§ 3645 · The maximum height is f ¨ ¸ © 32 ¹

95 2 16 2025

3645 . 32 2

16 § 3645 · 9 § 3645 · ¨ ¸ ¨ ¸ 1.5 2025 © 32 ¹ 5 © 32 ¹

6561 6561 1.5 64 32

6561 13,122 96 64 64 64

6657 feet | 104.02 feet. 64

(c) The length of the punt is the positive x-intercept.

0

16 2 9 x x 1.5 2025 5

9 5 r

x

9 5 2

4 1.5 16 2025

32 2025

|

1.8 r 1.81312 0.01580247

x | 0.83031 or x | 228.64

The punt is about 228.64 feet. 800 10 x 0.25 x 2

83. C

0.25 x 2 10 x 800

The vertex occurs at x

b 2a

The cost is minimum when x

10 20.25

20.

20 fixtures.

230 20 x 0.5 x 2

84. P

20 2 0.5

20.

Because x is in hundreds of dollars, 20 u 100 2000 dollars is the amount spent on advertising that gives maximum profit. 85. R p

12 p 2 150 p

(a) R$4

12$4 150$4

$408

R$6

12$6 150$6

$468

R$8

12$8 150$8

$432

2 2

2

(b) The vertex occurs at b 2a

The vertex occurs at x

86. R p

p

b 2a

25 p 1200 p

(a) R 20

$14,000 thousand

$14,000,000

R 25

$14,375 thousand

$14,375,000

R30

$13,500 thousand

$13,500,000

150 2 12

$6.25.

Revenue is maximum when price

$6.25 per pet.

The maximum revenue is R$6.25

2

12$6.25 150$6.25 2

$468.75.

(b) The revenue is a maximum at the vertex.

b 2a

R 24

1200 2 25

24

14,400

The unit price that will yield a maximum revenue of $14,400 thousand is $24.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE 87. (a)

(b)

x

(c)

600 1066

2 3

1400

20

1600

30

4 50 x 3 8 x50 x

A

ª4 º 2 x « 50 x » ¬3 ¼

2 xy

8 x50 x 3

This area is maximum when x 25 feet and 100 1 y 33 feet. 3 3

A

15

25

1 200 4 x 3

200 y

185

3

x

5 10

Quadratic Function Functions and Models

4x 3y y

x

Section 2.1

1666

8 x50 x 3 8 x 2 50 x 3 8 2 x 50 x 625 625 3 8ª 2 x 25 625º ¼ 3¬ 8 5000 2 x 25 3 3

(d) A

2 3

The maximum area occurs at the vertex and is 5000 3 square feet. This happens when x 25 feet

1600

200 425

and y

2000

3

100 3 feet. 1 50 feet by 33 feet. 3

The dimensions are 2 x (e) They are all identical. 0

60 0

This area is maximum when x 100 1 y 33 feet. 3 3 x

88. (a)

(d) Area of rectangular region: A

y

1 y 2 Distance around two semicircular parts of track: §1 · d 2S r 2S ¨ y ¸ S y ©2 ¹

(b) Radius of semicircular ends of track: r

(c) Distance traveled around track in one lap: S y 2 x 200 d

Sy

200 2 x

m

S

2

S 2

S 2

S

x2

100 x

x2

100 x 2500 2500

x

50

S

S

2

5000

S 50 and

S

attendance, or the number of tickets sold. 100, 20, 1500

y 1500

100 x 20

y 1500 y

100 x 2000 100 x 3500

R x

§ 200 2 x · x¨ ¸ S © ¹ 1 200 x 2 x 2

number of tickets sold price per ticket

89. (a) Revenue

Let y

xy

The area is maximum when x 200 250 100 y .

200 2 x

y

1 33 feet 3

25 feet and y

x

25 feet and

R x

y x 100 x

R x

100 x 3500 x

3500 x

INSTRUCTOR USE ONLY 2

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186

NOT FOR SALE

Chapter 2

Polynomial ynomial and Rational Function Functions

(b) The revenue is at a maximum at the vertex.

3500 2 100

b 2a

R17.5

17.5

10017.5 350017.5 2

$30,625

A ticket price of $17.50 will yield a maximum revenue of $30,625. Area of rectangle Area of semicircle

90. (a) Area of window

xy

1 2 S radius 2

xy

1 § x· S¨ ¸ 2 © 2¹

xy

2

Sx2 8

To eliminate the y in the equation for area, introduce a secondary equation. perimeter of rectangle perimeter of semicircle

Perimeter

1 circumference 2 1 2 y x 2S radius 2 § x· 2y x S ¨ ¸ © 2¹

2y x

16 16 16 y

8

Sx 1 x 2 4

Substitute the secondary equation into the area equation. xy

Area

S x2 8

S x · S x2 1 § x¨ 8 x ¸ 2 4 ¹ 8 © 8x

S x2 1 2 S x2 x 2 4 8

8x

1 2 S x2 x 2 8

1 64 x 4 x 2 S x 2 8 (b) The area is maximum at the vertex. 8x

Area

1 2 S x2 x 2 8

§ 1 S· 2 ¨ ¸ x 8x 8¹ © 2 b 2a

x

y

8

8 | 4.48 § 1 S· 2¨ ¸ 8¹ © 2

S 4.48 1 | 2.24 4.48 2 4

The area will be at a maximum when the width is about 4.48 feet and the length is about 2.24 feet.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.1

91. (a)

Quadratic Function Functions and Models

187

4200

0

55 0

(b) The maximum annual consumption occurs at the point 16.9, 4074.813 . 4075 cigarettes 1966 o t

16

The maximum consumption occurred in 1966. After that year, the consumption decreases. It is likely that the warning was responsible for the decrease in consumption. (c) Annual Consumption per smoker

Annual consumption in 2005 total population total number of smokers in 2005 1487.9 296,329,000 59,858,458 7365.8

About 7366 cigarettes per smoker annually Number of cigarettes per year Number of days per year

Daily Consumption per smoker

7366 365 | 20.2 About 20 cigarettes per day 92. (a), (b)

95. True. The negative leading coefficient causes the parabola to open downward, making the vertex the maximum point on the graph.

7

0

7 0

0.0408 x 2 0.715 x 2.82

y

(c) The model is a good fit to the actual data. (d) The greatest sales occurred in the year 2007. (e) Sales will be at a maximum at the vertex.

x

b 2a

0.715 | 8.76 2 0.0408

96. True. The positive leading coefficient causes the parabola to open upward, making the vertex the minimum point on the graph. 97. f x

x 2 bx 75, maximum value: 25

The maximum value, 25, is the y-coordinate of the vertex. Find the x-coordinate of the vertex: x

Sometime during 2008. (f ) 2011 o Use x y11

f x

11.

0.040811 0.71511 2.82 | 5.75 2

Sales in the year 2011 will be about $5.75 billion. 93. True. The equation 12 x 1 0 has no real solution, so the graph has no x-intercepts. 2

94. True. The vertex of f x is 54 ,

g x is 54 , 71 . 4

53 4

and the vertex of

§b· f¨ ¸ © 2¹ 25

b 2a

b 2

x 2 bx 75 2

§b· §b· ¨ ¸ b¨ ¸ 75 2 © ¹ © 2¹

b2 b2 75 4 2

400

b2 4 b2

r20

b

100

b 2 1

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188

NOT FOR SALE

Chapter 2

98. f x

Polynomial ynomial and Rational Function Functions

x 2 bx 16, maximum value: 48

100. f x

x 2 bx 25 , minimum value: –50

The maximum value, 48, is the y-coordinate of the vertex.

The minimum value, –50, is the y-coordinate of the vertex.

Find the x-coordinate of the vertex:

Find the x-coordinate:

b 2a

x

f x

b 2 1

b 2

b 2a

f x

x 2 bx 16 2

§b· f¨ ¸ © 2¹

x

§b· §b· ¨ ¸ b¨ ¸ 16 2 © ¹ © 2¹

b 21

b 2

x 2 bx 25 2

§ b· f ¨ ¸ © 2¹

§ b· § b· ¨ ¸ b¨ ¸ 25 2 © ¹ © 2¹

25

256

b2 4 b2

100

b2 b2 25 4 2 b 2 4 b2

r16

b

r10

b

48 64

99. f x

b2 b2 16 4 2

50

x 2 bx 26 , minimum value: 10

The minimum value, 10, is the y-coordinate of the vertex. Find the x-coordinate of the vertex: x

b 2a

f x § b· f ¨ ¸ © 2¹

b 21

b 2

§ b· § b· ¨ ¸ b¨ ¸ 26 © 2¹ © 2¹

r8

b

ax 2 bx c has two real zeros, then by the

Quadratic Formula they are x

2

64

16

103. If f x

2

b b 26 4 2 b2 4 b2

10

y value is adjusted by a factor of a, and the parabola becomes narrower or wider. Every point on the parabola is shifted up k units. 102. Conditions (a) and (d) are preferable because profits would be increasing.

x 2 bx 26

2

101. The graph of f x is moved h units to the right. Every

b r

b 2 4ac . 2a

The average of the zeros of f is b

b b 2 4ac 2a 2

b 2 4ac 2a

2b 2a 2 b . 2a

This is the x-coordinate of the vertex of the graph.

Section 2.2 Polynomial Functions of Higher Degree 9. f x

1. continuous

Matches graph (c).

2. Leading Coefficient Test

10. f x

3. x n

4, 0

4. n; n 1 5. (a) solution; (b) x a ; (c) x-intercept 6. repeated; multiplicity 7. touches; crosses 8. standard

2 x 3 is a line with y-intercept 0, 3 .

11. f x

x 2 4 x is a parabola with intercepts 0, 0 and

and opens upward. Matches graph (g). 2 x 2 5 x is a parabola with x-intercepts 0, 0

and 52 , 0 and opens downward. Matches graph (h). 12. f x

2 x3 3 x 1 has intercepts

0, 1 , 1, 0 , 12

1 2

3, 0 and 12

1 2

3, 0 .

INSTRUCTOR USE ONLY Matches graph (f).

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.2

13. f x

r2

14 x 4 3 x 2 has intercepts 0, 0 and

13 x3 x 2

4 3

(a) f x

x

1

5

y

has y-intercept 0, 43 .

4 3

Matches graph (e).

2 1

x 4 2 x3 has intercepts 0, 0 and 2, 0 .

15. f x

189

x5

18. y

3, 0 . Matches graph (a).

14. f x

Polynomial Functions of Hi Higher Degree H

x – 4 –3

1

2

3

4

Matches graph (d). –3

16. f x

1 x5 5

2x 3

9 x 5

has intercepts

0, 0 , 1, 0 , 1, 0 , 3, 0 , 3, 0 . x3

17. y

(a) f x

–4

Matches graph (b).

Horizontal shift one unit to the left (b) f x

x5 1

y

x

4

3

y

4

4

3

3 2

2 1 x

−2

1

2

4

5

x

6

–4 –3 –2

1

2

3

4

−2 −3

–3

−4

–4

Vertical shift one unit upward

Horizontal shift four units to the right (b) f x

(c) f x

x3 4

1

1 x5 2

y

y 4

2

3

1

2

x

−3 −2

1

2

3

4 x

−2

–4 –3 –2

−3

2

3

4

–3 −6

–4

Reflection in the x-axis, vertical shrink each y -value is multiplied by 12 , and vertical shift one unit upward

Vertical shift four units downward (c) f x

1 x3 4

(d) f x

y

12 x 1

4

5

y

3 2

4

1

3 x

−4 −3 −2

2

3

2

4

1

−2

x

−3

1

–5 –4 –3 –2

2

3

−4 –3

Reflection in the x-axis and a vertical shrink each y -value is multiplied by 14

(d) f x

–4

x

4 4

Refection in the x -axis, vertical shrink each y -value is multiplied by 12 , and

y

3

horizontal shift one unit to the left

2 1 x

−2

1

2

3

4

5

6

−2 −3 −4 −5 −6

INSTRUCTOR USE ONLY Horizontal shift four units to the right and vertical shift four units downward

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190

NOT FOR SALE

Chapter 2

Polynomial ynomial and Rational Function Functions

x4

19. y

(a) f x

x

3

4

(b)

f x

x4 3

y

f x

4 x4 y

6

4

6

5

3

5

4

2

3

f

(c)

y

1

3 x

2 –4 –3 –2

1

2

3

2

4

1

x –5 –4 –3 –2 –1

1

2

x

3

–4 –3 –2

–2

–4

Horizontal shift three units to the left (d) f x

1 2

x

1

(e)

f x

2

3

4

–2

Vertical shift three units downward

4

1

–1

2 x 4

y

1

Reflection in the x-axis and then a vertical shift four units upward (f )

f x

12 x

y

4

2

y

6

6 5 4 3 2 1

x –4 –3 –2 –1

1

2

3

−4 −3 −2 −1 −1

4

Horizontal shift one unit to the right and a vertical shrink each y -value is multiplied by

(a)

1

2

3

−4 −3

4

x

−1 −1

1

3

4

−2

–2

20. y

x

1 2

Vertical shift two units downward and a horizontal stretch each y -value

Vertical shift one unit upward and a horizontal shrink each y -value is multiplied by 16

1 is multipied by 16

x6 f x

18 x 6

(b)

f x

x

y

2 4 6

(c)

x6 5

y

y

4

3

3

2

2

1 x

1 x –4 –3 –2

f x

2

–1

3

−4 −3 −2

x

4

–5 –4

–2

1

2

1

−1

2

3

4

3

–2 –3 –4

–4

Vertical shrink each y -value is multiplied by

1 8

Horizontal shift two units to the left and a vertical shift four units downward

and

reflection in the x-axis

(d) f x

14 x 6 1

(e)

14 x

f x

y

6

2

Vertical shift five units downward

(f )

f x

2 x 6

y

1

y

4 3 2 x –4 –3 –2

–1

2

3

4 x

–2

−8 − 6

–3

2

6

8

− 4 − 3 −2 −1

Reflection in the x-axis, vertical shrink each y -value 1 4

, and

Horizontal stretch each x-value is multiplied by 4 , and vertical shift two units downward

x 1

2

3

4

−2

−4

–4

is multiplied by

−2

Horizontal shrink each x -value

is multiplied by 12 , and vertical shift one unit downward

INSTRUCTOR USE ONLY vertical shift one unit upward

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.2

21. f x

1 x3 5

4x

29. f x

Degree: 3 Leading coefficient:

2 x 3x 1 2

Degree: 2 Leading coefficient: 2 The degree is even and the leading coefficient is positive. The graph rises to the left and rises to the right. 23. g x

5

7 x 2

3 2 t 3t 6 4

Leading coefficient:

3 4

The degree is even and the leading coefficient is negative. The graph falls to the left and falls to the right. 78 s 3 5s 2 7 s 1

30. f s

Degree: 3 Leading coefficient: 78 The degree is odd and the leading coefficient is negative.

3x 2

The graph rises to the left and falls to the right.

Degree: 2 Leading coefficient: 3

31. f x

3x 3 9 x 1; g x

The degree is even and the leading coefficient is negative. The graph falls to the left and falls to the right. 24. h x

191

Degree: 2

1 5

The degree is odd and the leading coefficient is positive. The graph falls to the left and rises to the right. 22. f x

Polynomial Functions of Hi Higher Degree H

8

g

f

−4

1 x6

Degree: 6 Leading coefficient: 1 The degree is even and the leading coefficient is negative. The graph falls to the left and falls to the right.

3x3

4

−8

13 x3 3 x 2 , g x

32. f x

13 x3

6

25. f x

2.1x 4 x 2 5

g

3

f −9

Degree: 5 Leading coefficient: 2.1

−6

The degree is odd and the leading coefficient is negative. The graph rises to the left and falls to the right. 26. f x

33. f x

4 x5 7 x 6.5

x 4 4 x3 16 x ; g x

x4

12

Degree: 5 Leading coefficient: 4 The degree is odd and the leading coefficient is positive. The graph falls to the left and rises to the right. 27. f x

9

−8

f − 20

34. f x

6 2 x 4 x 2 5 x3

8

g

3x 4 6 x 2 , g x

3x4

5

Degree: 3 Leading coefficient: 5

f

The degree is odd and the leading coefficient is negative.

g −6

6

The graph rises to the left and falls to the right. 3x 4 2 x 5 4 Degree: 4 3 Leading coefficient: 4 The degree is even and the leading coefficient is positive.

−3

28. f x

The graph rises to the left and rises to the right.

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192

NOT FOR SALE

Chapter 2

35. f x

Polynomial ynomial and Rational Function Functions

x 2 36 x 2 36

(a) 0

x

0

38. f x

x 2 10 x 25

(a) 0

x 2 10 x 25

6 x 6

x 6

0

0

x

6

6

x

(b) x

5

2

5

Zero: x

x 6

x

5 has a multiplicity of 2 (even multiplicity).

Turning points: 1 (the vertex of the parabola)

Zeros: r6

(c)

25

(b) Each zero has a multiplicity of one (odd multiplicity). Turning points: 1 (the vertex of the parabola) (c)

− 25

15 −5

6 − 12

12

39. f x

(a) 0

− 42

36. f x

81 x 2

(a) 0

81 x 2

0

9

1 x2 3 1 2 x 3

13 x

1 3

x2

1 3

x

13 x

2 3

x 2

2 x 1 2, x

Zeros: x

x 9 x

2 3

1

(b) Each zero has a multiplicity of 1 (odd multiplicity).

9 x

0

9 x

9

x

x

0

Turning points: 1 (the vertex of the parabola)

9

(c)

4

Zeros: r 9 −6

(b) Each zero has a multiplicity of one (odd multiplicity).

6

−4

Turning points: 1 (the vertex of the parabola) (c)

90

40. f x − 15

(a) For

15 −9

37. ht

t 6t 9 2

(a) 0

t 6t 9 2

Zero: t (b) t

1 2 5 3 x x 2 2 2

3

3

5 r 2

37 4

Zeros: x

5 ,c 2

3 . 2

2

§5· § 1 ·§ 3 · ¨ ¸ 4¨ ¸¨ ¸ © 2¹ © 2 ¹© 2 ¹ 1

3 has a multiplicity of 2 (even multiplicity).

10

1 ,b 2

5 r 2

x

t

0, a

2

Turning points: 1 (the vertex of the parabola) (c)

1 2 5 3 x x 2 2 2

5 r 37 2

(b) Each zero has a multiplicity of 1 (odd multiplicity). Turning points: 1 (the vertex of the parabola) (c)

−6

3

12 −2

−8

4

−5

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© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.2

41. f x

(a) 0

44. f x

3 x 3 12 x 2 3 x

3x x 2 4 x 1

3 x3 12 x 2 3 x

Zeros: x

2r

0, x

Polynomial Functions of Hi Higher Degree H

Formula) (b) Each zero has a multiplicity of 1 (odd multiplicity).

x 4 x3 30 x 2

0

x 2 x 2 x 30

0

x 2 x 6 x 5

x2

0

x 6

0

x 5

x

0

x

6

x

Turning points: 2 (c)

x 4 x3 30 x 2

(a) 0

3 (by the Quadratic

8

Zeros: x −6

0, x

− 24

5 x x 2 2 x 1

(a) 0 0

5

5

6, x

0 is 2 (even multiplicity).

The multiplicity of x

6 is 1 (odd multiplicity).

The multiplicity of x

5 is 1 (odd multiplicity).

Turning points: 3 (c)

5 x x 2 2 x 1

60 −9

9

x x 2 2 x 1

For x 2 2 x 1 x

0

6

(b) The multiplicity of x

42. g x

193

2 r

0, a

2 21

2

2, c

1, b

1.

41 1

45. g t

tt

0, x

1r

Zeros: t

2

(b) Each zero has a multiplicity of 1 (odd multiplicity).

(b) t t

Turning points: 2

t t 4 6t 2 9

t 5 6t 3 9t

2

Zeros: x

(c)

t 5 6t 3 9t

(a) 0

2r 8 2 1r

−300

3

t 2

0, t

3

r

t t 2 3

2

2

3

0 has a multiplicity of 1 (odd multiplicity). r

3 each have a multiplicity of 2 (even

multiplicity).

16

Turning points: 4 −6

(c)

6

−9

− 16

43. f t

6

9

t 3 8t 2 16t

(a) 0

−6

t 8t 16t 3

2

46. (a) f x

0

t t 2 8t 16

0

t t 4 t 4

t

0

t 4

0

t 4

t

0

t

4

t

0, t

4

Zeros: t

x5 x3 6 x

0

x x 4 x 2 6

0

0

x x 2 3 x 2 2

4

Zeros: x

0, r

2

(b) Each zero has a multiplicity of 1 (odd multiplicity). 0 is 1 (odd multiplicity).

(b) The multiplicity of t

4 is 2 (even multiplicity).

The multiplicity of t

Turning points: 2 (c)

6

Turning points: 2 10

(c)

−9

9

INSTRUCTOR USE ONLY −6

−9

9

−2 − 2

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194

NOT FOR SALE

Chapter 2

47. f x

(a) 0

Polynomial ynomial and Rational Function Functions

3x 4 9 x 2 6 3x 4 9 x 2 6

0

3 x 3 x 2

0

3 x 1 x 2

4

50. f x

x 3 4 x 2 25 x 100

(a) 0

x 2 x 4 25 x 4

2

2

2

No real zeros

x2

0

x

25 x 4 5 x 5 x 4 r 5, 4

Zeros: x

(b) Turning points: 1 (c)

0

(b) Each zero has a multiplicity of 1 (odd multiplicity).

21

Turning points: 2 (c)

−6

140

6 −3

−9

48. f x

2 x 4 2 x 2 40

(a) 0

2 x 4 2 x 2 40

0

2 x 4 x 2 20

0

2 x 2 4 x 2 5

Zeros: x

r

51. y

4 x3 20 x 2 25 x

(a)

12

−2

5

(b) x-intercepts: 0, 0 ,

Turning points: 3 20 −6

(a) 0

x3 3 x 2 4 x 12

x3 3 x 2 4 x 12

x2 Zeros: x

4 x 3 r 2, x

x

x 2 x 3 4 x 3 2 x 2 x 3

4 x3 20 x 2 25 x

0

x 4 x 2 20 x 25

0

x 2 x 5

x

0,

2

5 2

(d) The solutions are the same as the x-coordinates of the x-intercepts. 52. y

4 x3 4 x 2 8 x 8

(a)

3

52 , 0

(c) 0 6

− 60

49. g x

6 −4

(b) Each zero has a multiplicity of 1 (odd multiplicity). (c)

9 − 20

2 −3

3

(b) Each zero has a multiplicity of 1 (odd multiplicity). Turning points: 2 (c)

4 −8

− 11 7

(b)

1, 0 ,

(c) 0 −16

0

2, 0

4 x3 4 x 2 8 x 8 4 x 2 x 1 8 x 1

0

4 x 2 4 x 2

x

r

0

2, 0 ,

8 x 1 2 x 1

2, 1

(d) The solutions are the same as the x-coordinates of the x-intercepts.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.2

59. f x

x5 5 x3 4 x

53. y

(a)

Polynomial Functions of Higher Hi H Degree

x

195

0 x 4 x 5

x x 2 9 x 20

4

x3 9 x 2 20 x −6

Note: f x

6

ax x 4 x 5 has zeros 0, 4, and

5 for all real numbers a z 0. −4

(b) x-intercepts: 0, 0 , r1, 0 , r2, 0 x x 2 1 x 2 4

0

x x 1 x 1 x 2 x 2

x

0, r1, r 2

(d) The solutions are the same as the x-coordinates of the x-intercepts.

x2

9

(a)

Note: f x 61. f x

62. f x

x

x x 4 5 x 2 4

0 x 8

Note: f x

63. f x

x 8x

x

x

x

ax x 7 has zeros 0 and 7 for all real

Note: f x 1

64. f x

2 x 6

x 2 4 x 12

Note: f x

a x 2 x 6 has zeros 2 and 6 for

x

3º ¼

3 ºª ¼¬ x 1

3

3 º¼

2

a x 2 2 x 2 has zeros 3 for all real numbers a z 0.

x

2 ª x 4 ¬ x 2 ª¬ x 4

5 ºª x 4 ¼¬ 5 ºª ¼¬ x 4

5º ¼ 5 º¼

2 2 ª x 4 5º ¬ ¼

x x 4 5 x 2 x 4 10 2

4 x 5

2

x3 8 x 2 16 x 5 x 2 x 2 16 x 32 10 x3 10 x 2 27 x 22

x 2 x 20

Note: f x

2

3 and 1

x

all real numbers a z 0. 58. f x

1

x2 2x 2

numbers a z 0. 57. f x

3 ºª x 1 ¼¬

x2 2x 1 3

0 x 7

x2 7 x

Note: f x

ªx 1 ¬ ª x 1 ¬

ax x 8 has zeros 0 and 8 for all real

numbers a z 0. 56. f x

ax x 2 x 1 x 1 x 2 has

zeros 2, 1, 0, 1, and 2 for all real numbers a z 0.

2

Note: f x

x 2 x 1 x 0 x 1 x 2

x 5 5 x3 4 x

0, r 3

x

a x 4 4 x3 9 x 2 36 x has zeros

x x 2 4 x 2 1

9

(d) The solutions are the same as the x-coordinates of the x-intercepts. 55. f x

4 x 2 9 x

x x 2 x 1 x 1 x 2

(b) x-intercepts: 0, 0 , 3, 0 , 3, 0

x2

4 x 3 x 3 x 0

4, 3, 3, and 0 for all real numbers a z 0. 18

1 x3 4

x x

Note: f x

−12

(c) 0

ax x 1 x 10 has zeros 0, 1, and

10 for all real numbers a z 0.

x 4 4 x3 9 x 2 36 x

12

−18

0 x 1 x 10

x3 11x 2 10 x

0

1 x3 4

x

x x 2 11x 10

x5 5 x3 4 x

(c) 0

54. y

60. f x

a x 4 x 5 has zeros 4 and 5 for

Note: f x

a x3 10 x 2 27 x 22 has zeros

INSTRUCTOR USE ONLY all real numbers a z 0.

2, 4

5, and 4

5 forr all real numbers a z 00.

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

196

NOT FOR SALE

Chapter 2

65. f x

x

Polynomial ynomial and Rational Function Functions

68. f x

3 x 3

x2 6 x 9

a x 2 6 x 9 , a z 0, has degree 2 and

Note: f x

zero x 66. f x

12 x 6

69. f x

x 18 x 72

Note: f x

a x 2 18 x 72 , a z 0, has degree 2

and zeros x

12 and 6.

67. f x

x

x 4x 5x

and zeros x

0, 5, and 1.

3 x 3

3 x

3

x3 3x

Note: f x

a x3 3x , a z 0, has degree 3 and

zeros x

3, and

3.

70. f x

0,

x

2

ax x 2 4 x 5 , a z 0, has degree 3

0 x

x x 2 4 x 5 Note: f x

x

x x

0 x 5 x 1

3

2, 4, and 7.

degree 3 and zeros x

2

x3 9 x 2 6 x 56

a x3 9 x 2 6 x 56 , a z 0, has

Note: f x

3.

x

x 2 x 4 x 7 x 2 x 2 11x 28

0 x 2 2 ª x 2 2 º ¬ ¼

x x 2 2 x 2 2

x x 2 8 x3 8 x

Note: f x

a x3 8 x has these zeros for all real

numbers a z 0. 71. f x

x

1 ª¬ x 2 º¼ ª x 1 ¬

x

1 x 2 ª¬ x 1

x2 x2

3 ºª x 1 ¼¬ 3 ºª ¼¬ x 1

3º ¼ 3 º¼

2 x 2 ª x 1 3º ¬ ¼

x 2 x 2 2 x 2

x 4 x3 6 x 2 2 x 4

a x 4 x3 6 x 2 2 x 4 has these zeros for all real numbers a z 0.

Note: f x 72. f x

x

3 ª¬ x 2 º¼ ª x 2 ¬

x

3 x 2 ª¬ x 2

x2 x2

5 ºª x 2 ¼¬ 5 ºª ¼¬ x 2

5º ¼ 5 º¼

2 x 6 ª x 2 5º ¬ ¼

x 6 x 2 4 x 1

x 4 5 x3 3 x 2 25 x 6

Note: f x 73. f x

a x 4 5 x3 3 x 2 25 x 6 has these zeros for all real numbers a z 0.

x 4 x 4

x5 4 x 4

or f x

x3 x 4

or f x

x 2 x 4

or f x

x x 4

4

2 3

x5 8 x 4 16 x3 x5 12 x 4 48 x3 64 x 2 x5 16 x 4 96 x3 256 x 2 256 x

Note: Any nonzero scalar multiple of these functions would also have degree 5 and zeros x

0 and 4.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.2

74. f x

x

or f x

1 x 4 x 7 x 8 2

x x x

or f x or f x

Polynomial Functions of Hi Higher Degree H

x5 17 x 4 79 x3 11x 2 332 x 224

1 x 4 x 7 x 8

x5 22 x 4 169 x3 496 x 2 208 x 896

1 x 4 x 7 x 8

x5 25 x 4 223 x3 787 x 2 532 x 1568

1 x 4 x 7 x 8

x5 26 x 4 241x3 884 x 2 640 x 1792

2

2

2

1, 4, 7, and 8.

Note: Any nonzero scalar multiple of these functions would also have degree 5 and zeros x 75. f x

78. g x

x x 5 x 5

x3 25 x

x 2 x 8

x 2 10 x 16

(a) Falls to the left; rises to the right

(a) Falls to the left; falls to the right

(b) Zeros: 0, 5, 5

(b) Zeros: 2, 8

(c)

x f x

(d)

197

(c)

2

1

0

1

2

42

24

0

24

42 (d)

y

x

1

3

5

7

9

g x

7

5

9

5

7

y 10

48

8

(−5, 0) −2

2

6

(5, 0)

(0, 0)

− 8 −6

4

6

x 4

8

− 24

2

− 36

(2, 0)

− 48

76. f x

x 2 x 3 x 3

x4 9 x2

79. f x

(a) Rises to the left; rises to the right

x f x

(d)

6

x

10

x 2 x 2

x3 2 x 2

(a) Falls to the left; rises to the right

(b) Zeros: 3, 0, 3 (c)

(8, 0)

4

(b) Zeros: 0, 2

2

1

0

1

2

24

8

0

8

24

(c)

x f x

y

(d)

1

0

1 2

1

2

3

3

0

83

1

0

9

y

15 4

10

(− 3, 0) 5 (0, 0) −4

−2 − 1

1

(3, 0) 2

3 2

x

4

1

(0, 0) (2, 0)

−4 −3 −2 −1

3

x

4

−20 −25

77. f t

1 4

t 2

2t 15

1 4

t

1 2

7 2

80. f x

(a) Rises to the left; rises to the right

t f t

x 4 2 x x 2

(a) Rises to the left; falls to the right

(b) No real zeros (no x-intercepts) (c)

2

8 x3

(b) Zero: 2

1

0

1

2

3

4.5

3.75

3.5

3.75

4.5

72 .

(d) The graph is a parabola with vertex 1, y

(c)

x f x

(d)

2

1

0

1

2

16

9

8

7

0

y 14

8

12 10

6 6 4 2

INSTRUCTOR NSTR STR TR USE SE ONLY 2

t

–44 –

–2

2

−44 −33 −2 2 −1 1 −1

(2, 0)

1

3

x

4

4

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© Cengage Learning. All Rights Reserved.

198

NOT FOR SALE

Chapter 2

81. f x

Polynomial ynomial and Rational Function Functions 3 x x 2 x 3

3x3 15 x 2 18 x

82. f x

4 x3 4 x 2 15 x

(a) Falls to the left; rises to the right

x 4 x 2 4 x 15

(b) Zeros: 0, 2, 3

x 2 x 5 2 x 3

(c)

x f x

(d)

0

1

2

2.5

3

3.5

(a) Rises to the left; falls to the right

0

6

0

1.875

0

7.875

(b) Zeros: 32 , 0, (c)

y 7 6 5 4 3 2

f x (d)

2

1

0

1

2

3

99

18

7

0

15

14

27

(0, 0)

(4, 0)

20

(3, 0) 1

3

y

(2, 0)

(0, 0) −3 −2 −1

x

16

x

4 5 6

12

−2

8

(− 23, 0)

4

(0, 0)

−4 −3 −2

83. f x

(d)

f x

2

3

x

4

y

(b) Zeros: 0, 5 x

1

( 25, 0)

x 2 5 x

5 x 2 x3

(a) Rises to the left; falls to the right

(c)

5 2

5

(−5, 0) −15

5

4

3

2

1

0

1

0

16

18

12

4

0

6

(0, 0) x

−10

5

10

−20

84. f x

48 x 2 3 x 4

3x 2 x 2 16

(a) Rises to the left; rises to the right

(d)

(b) Zeros; 0, r 4 (c)

x f x

y

(− 4, 0) 100

4

3

2

1

0

1

2

3

4

5

675

0

189

144

45

0

45

–144

–189

0

675

−6

−2

2

x

6

−200 −300

85. f x

x 2 x 4

(a) Falls to the left; rises to the right

(d)

y

(b) Zeros: 0, 4 (c)

x f x

2

(0, 0) –4

1

0

1

2

3

4

5

5

0

3

8

9

0

25

–2

2

(4, 0) 6

x 8

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.2

86. h x

1 x3 3

x

4

Polynomial Functions of Hi Higher Degree H

89. f x

2

x3 16 x

(a) Falls to the left; rises to the right

199

x x 4 x 4

32

(b) Zeros: 0, 4 −6

(c)

x

1

0

1

h x

25 3

0

3

(d)

2 32 3

3

4

5

9

0

125 3

6

−32

Zeros: 0 of multiplicity 1; 4 of multiplicity 1; and 4 of

y

multiplicity 1.

14 12

90. f x

10 8

2 x2

1 x4 4

6 6

4

(0, 0)

(4, 0)

–4 –2

87. g t

2

4

6

x

8 10 12

−9

14 t 2 t 2 2

2 −6

Zeros: 2.828 and 2.828 of multiplicity 1; 0 of multiplicity 2

(a) Falls to the left; falls to the right (b) Zeros: 2, 2 (c)

9

91. g x

t

3

2

1

0

1

2

3

g t

25 4

0

94

4

94

0

25 4

1 5

x

1 x 3 2 x 9 2

14

y

(d)

(− 2, 0)

(2, 0)

−12

t –3

–1 –1

1

2

18

3 −6

–2

Zeros: 1 of multiplicity 2; 3 of multiplicity 1;

1 10

x

of

multiplicity 1

–5 –6

88. g x

9 2

92. h x

1 x 3 2

1 5

x

2 3x 5 2

2

3 21

(a) Falls to the left; rises to the right (b) Zeros: 1, 3 (c)

−12

x

2

1

0

1

2

4

g x

12.5

0

2.7

3.2

0.9

2.5

(d)

12 −3

Zeros: 2, 53 , both with multiplicity 2

y 6 4 2

(−1, 0) –6 –4 –2

(3, 0) 4

6

x 8

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© Cengage Learning. All Rights Reserved.

200

NOT FOR SALE

Chapter 2

93. f x

Polynomial ynomial and Rational Function Functions

x 4 ; f x is even.

l wh

95. (a) Volume

height length

y 5

x width

So, V x

4 3

36 2 x

36 2 x 36 2 x x

x36 2 x . 2

(b) Domain: 0 x 18

2 1

The length and width must be positive.

x –3

–2

–1

1

2

3

(c)

–1

(a) g x

f x 2

Vertical shift two units upward g x

f x 2

f x 2

g x

Even (b) g x

f x 2

Horizontal shift two units to the left

Box Height

Box Width

Box Volume, V

1

36 21

1ª¬36 21 º¼

2

36 2 2

2 ª¬36 2 2 º¼

3

36 23

3ª¬36 23 º¼

4

36 2 4

4 ª¬36 2 4 º¼

5

36 25

5ª¬36 25 º¼

2

6

36 26

6 ª¬36 26 º¼

2

7

36 27

7 ª¬36 27 º¼

2

Neither odd nor even (c) g x

f x

x 4

x4

Reflection in the y-axis. The graph looks the same. Even (d) g x

f x

x

Reflection in the x-axis (e) g x

f

1156 2

2 2

2048 2700 3136 3380 3456 3388

The volume is a maximum of 3456 cubic inches when the height is 6 inches and the length and width are each 24 inches. So the dimensions are 6 u 24 u 24 inches.

4

Even 1 x 2

2

(d)

3600

1 4 x 16

Horizontal stretch Even (f ) g x

1 2

f x

0

1 4 x 2

The maximum point on the graph occurs at x

Vertical shrink f x3 4

x3 4

4

x 4

3

4

x3 , x t 0

24 2 x 24 212 x 46 x x 8 x12 x 6 x l wh

96. (a) Volume

Neither (h) g x

f

D f x

(b) x ! 0,

f f x f x4

x 4

6.

This agrees with the maximum found in part (c).

Even (g) g x

18 0

4 x x

12 x ! 0,

6 x ! 0

x 12

x 6

Domain: 0 x 6

4

(c)

V

720

16

x

600

Even

480 360

94. R

1 x3 600 x 2 100,000

240 120 x

The point of diminishing returns (where the graph changes from curving upward to curving downward) occurs when x 200. The point is 200, 160 which

1

2

3

4

5

6

x | 2.5 corresponds to a maximum of 665 cubic inches.

corresponds to spending $2,000,000 on advertising to obtain a revenue of $160 million.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.3

97. False. A fifth-degree polynomial can have at most four turning points. 98. True. f x

x

1 has one repeated solution. 6

Polynomial and Synthetic Synth Division

201

100. (a) Degree: 3

Leading coefficient: Positive (b) Degree: 2 Leading coefficient: Positive

99. True. A polynomial of degree 7 with a negative leading coefficient rises to the left and falls to the right.

(c) Degree: 4 Leading coefficient: Positive (d) Degree: 5 Leading coefficient: Positive

Section 2.3 Polynomial and Synthetic Division 1. f x is the dividend; d x is the divisor: q x is the

quotient: r x is the remainder

9. y1

x2 2 x 1 , y2 x 3

(a) and (b)

2. improper; proper

x 1

2 x 3

3 −9

9

3. improper 4. synthetic division −9

5. Factor

x 1 (c) x 3 x 2 2 x 1 x 2 3x x 1 x 3 2

6. Remainder 7. y1

x2 and y2 x 2

x 2

4 x 2

x 2 x 2 x2 0x 0 x2 2x 2 x 0 2 x 4 4

So,

8. y1

x2 x 2

10. y1 4 and y1 x 2

x 2

x 4 3x 2 1 and y2 x2 5

x2 2 4 x 5 x 3x 2 x4 5x2 8 x 2 8 x 2 x 3x 1 So, x2 5 4

2

x2 2x 1 x 3

x 1

x4 x2 1 , y2 x2 1

x2

So,

x2 8

(a) and (b)

y2 . 39 x2 5

2 and y1 x 3

y2 .

1 x2 1

6

−6

6 −2

8 1

x2

(c) x 0 x 1 x 0 x x 2 0 x 1 x 4 0 x3 x 2 1 2

1 40 39 39 and y1 x 8 2 x 5 2

So, y 2.

4

3

x4 x2 1 x2 1

x2

1 and y1 x2 1

y2 .

2x 4 11. x 3 2 x 2 10 x 12

2 x2 6x 4 x 12 4 x 12 0 2 x 2 10 x 12 x 3

2 x 4, x z 3

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202

Chapter 2

NOT FOR SALE

Polynomial ynomial and Rational Function Functions 5x 3

x 2 3x 9

12. x 4 5 x 17 x 12 5 x 2 20 x 3 x 12 3 x 12 0

17. x 3 x 0 x 2 0 x 27 x3 3x 2 3x 2 0 x 3x 2 9 x 9 x 27 9 x 27 0

2

5 x 2 17 x 12 x 4

3

5 x 3, x z 4

x

x3 27 x 3

3x 1

2

13. 4 x 5 4 x3 7 x 2 11x 5 4 x3 5 x 2 12 x 2 11x 12 x 2 15 x 4x 5 4x 5 0 4 x3 7 x 2 11x 5 4x 5

x 2 5 x 25

x 2 3x 1, x z

18. x 5 x3 0 x 2 0 x 125 x3 5 x 2 5 x 2 0 x 5 x 2 25 x 25 x 125 25 x 125 0

5 4

x3 125 x 5

2x 4x 3 2

14. 3x 2 6 x3 16 x 2 17 x 6

12 x 2 17 x 12 x 2 8 x

7x 3 x 2

9x 6 9x 6 0 2 x 2 4 x 3, x z

2 3

x3 3x 2 1 15. x 2 x 4 5 x3 6 x 2 x 2 x 4 2 x3 3x3 6 x 2 3x3 6 x 2

16. x 3 x 4 x 2

x3 3x 2 1, x z 2

3 x 12

x 3x 7 x 2 3x 7 x 2 21x 18 x 12 18 x 54 42 3

2

x3 4 x 2 3 x 12 x 3

11 x 2

4 20. 2 x 1 8 x 5 8x 4 9 8x 5 2x 1

x3 9 x2 1

x 2 7 x 18 3

7

4

9 2x 1

x 21. x 2 0 x 1 x3 0 x 2 0 x 9 x3 0 x 2 x x 9

x 2 x 2 0

x 4 5 x3 6 x 2 x 2 x 2

x 2 5 x 25, x z 5

7 19. x 2 7 x 3 7 x 14 11

6 x3 4 x 2

6 x3 16 x 2 17 x 6 3x 2

x 2 3 x 9, x z 3

x 9 x2 1

x2 22. x3 0 x 2 0 x 1 x5 0 x 4 0 x3 0 x 2 0 x 7 x5 0 x 4 0 x3 x 2 7 x2 x5 7 x3 1

x 2 7 x 18

x

x2

x2 7 x3 1

42 x 3

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.3

23. x 2 0 x 1 2 x3 8 x 2 2 x3 0 x 2 8 x 2 8 x 2 2 x3 8 x 2 3x 9 x2 1

2x 8 3x 9 2x x 9 0x 8 x 1 2x 8

29.

Polynomial and Synthetic Synth Division

3

6

0 x2 3x 2 3x 2 6 x2 9 x2 9 x2

x 4 5 x3 20 x 16 x2 x 3

x 1 x2 1

30.

x4

x

1

x 3

3

6

20 x 18 x 2 x 16 9 x 27 7 x 11

31.

27.

5

1

17

3 3

15

1

2x

0 x2 3x 2 3x 2 9 x2 6x2

x 3 0x 0 x x 0 9x 3 8x 3

3

2

10

25

2

5

0

3x 17 x 15 x 25 x 5 3

28.

3

2

5 5

18

7

32.

2

6

3

2

0

5 x3 18 x 2 7 x 6 x 3

7

12

192

2

32

199

8

9

18

8

0

18

0

9

0

18

32

18

0

32

0

16

0

9 x3 18 x 2 16 x 32 x 2

1 1

34.

6

3

75

250

10

100

250

10

25

0

0

72

18

12

72

2

12

0

3x 16 x 72 x 6

35.

4

2

5 5

36.

2

0

8

20

56

224

14

56

232

2

5 5

3x 2 2 x 12, x z 6

6

5x 6x 8 x 4 3

x 2 10 x 25, x z 10

16

3

5 x 2 14 x 56

0

6

8

10

20

52

10

26

44

5 x3 6 x 8 x 2

199 x 6

9 x 2 16, x z 2

0

x 3 75 x 250 x 10

248 x 3

4 x 2 9, x z 2

16

9

3

5 x 2 3 x 2, x z 3

2 x 2 2 x 32

2

9

6

9

20

4 x 8 x 9 x 18 x 2

2

3x 2 2 x 5, x z 5

15

14

4

33. 10

17 x 5 x 2x 1

6 x 2 25 x 74

12

4

25

15

248

2

3

26. x 2 2 x 1 2 x3 4 x 2 15 x 5 2 x3 4 x 2 2 x 17 x 5

x

222

74

2

2x

2

75

25

2 x 14 x 20 x 7 x 6

6x2 8x 3

2 x3 4 x 2 15 x 5

18

3

7 x 11 x 6x 9 2 x x 3

x

26

2

20 x 16

2

25. x3 3x 2 3 x 1 x 4 0 x3 x 4 3x3 3x3 3x3

1

6 x 3 7 x 2 x 26 x 3

x2 6 x 9

24. x 2 x 3 x 4 5 x3 x 4 x3 6 x3 6 x3

7

6

203

232 x 4

44 x 2

INSTRUCTOR USE ONLY 5 x 2 10 x 26

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© Cengage Learning. All Rights Reserved.

204 37.

NOT FOR SALE

Chapter 2

6

Polynomial ynomial and Rational Function Functions

10

50 60

60

360

2160

10

10

60

360

1360

0

10 x 4 50 x 3 800 x 6

38.

3

1 1

10 x 3 10 x 2 60 x 360

13

0

0

120

80

3

48

144

16

48

144

432 312

936 856

x 13 x 120 x 80 x 3 5

39.

4

8

1 1

9

1

8

64

0

9

81

729

9

81

0

46.

6

12

24

48

6

12

24

48

0

0

3

0

0

0

0

6

12

24

48

6

12

24

48

1

1

4

48 x 2

3 2

3

47. f x

0

180

0

6

36

216

6

36

36

216 216

1

4 x 2 14 x 30, x z

4

0

5

9 2 1 2

3 4 3 4

9 8 49 8

3x 2

1 3 49 x 2 4 8 x 12

1

14

11

4

12

8

3

2

3

48. f x

x3 5 x 2 11x 8, k

–2

1 1

6

f 2

–6

11

1 2

4

4 x 2 3 x 2 3

43 42 14 4 11

3

5 3 x 2 x 2 x3 x 1

0

f 4

f x

3

15

30

x

5

1

7

14

f x

216 x3 6 x 2 36 x 36 x 6

3

2

x3 x 2 14 x 11, k

1

2

2

15

3x3 4 x 2 5 3 x 2

4

0

4

23

16

3

48 3 x 6 x 12 x 24 x 2 3

1

1 2

0

3 x 3 6 x 2 12 x 24

180 x x x 6

4 x3 16 x 2 23 x 15 1 x 2

3

1

1

512

0

3 x x 2

44.

64

729

3

6

8

45.

3

4

43.

512

856 x 3

x 2 9 x 81, x z 9

3 x 4 x 2

2

0

0

x 729 x 9

42.

0

0

3

2

x 4 16 x 3 48 x 2 144 x 312

x 2 8 x 64, x z 8

1

41.

1360 x 6

4

x3 512 x 8

40.

800

0

5

11

8

2

14

6

7

3

2

x

3 2

2 x 2 7 x 3 2

2 3

5 2 11 2 8 2

2

11 x 1

INSTRUCTOR USE ONLY x 2 3x 6

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.3

49. f x

23

15 x 4 10 x3 6 x 2 14, k

10

6

0

14

10

0

4

83

0

6

4

34 3

15

15

51. f x

23

50. f x 1 5

3

10

3

4

2

4

75

20

7

13 5

34 3

52. f x

53. f x 1

54. f x 2

4

f

1

5

1

x 5

2º 8 ¼

2

5

4

5

2 5 5

10

5

2 5

6

3

5

2

5 ªx 2 2 ¬

5

8

2

5 x 2 5º 6 ¼

5

2

5

5 4

6

3

6

12

4

4 4 3

10 2 3

4

2 4 3

2 2 3

0

2

3

2

3 x3 8 x 2 10 x 8, k 2

3

2

8

2 10

8

6 3 2

2 4 2

8

23 2

8 4 2

0

x 2 2 ª¬3x 2 3 2 x 8 4 2 º¼ 2 3 2 2 8 2 2 10 2 2 8

f x

8

x 1 3 ª¬4 x 2 4 3 x 2 2 3 º¼ 3 0 41 3 61 3 121 3 4

3

f 2

3 2

x3 2 x 2 5 x 4, k

f x

f x

2

2

4

f 1

6

2

4 x3 6 x 2 12 x 4, k 3

2 3 2

2

1

2

3

14

2

3

x 15 10 x 20 x 7 135 13 f 15 10 15 22 15 3 15 4 5 f x

2

x 2 ¬ªx 3 2 x 3 f 2 2 3 2 2 2 14

1 5

2

22

10

3

205

2

3

f x

2

10 x 22 x 3x 4, k 3

1 1

3

4

x3 3 x 2 2 x 14, k

2

x 23 15x 6 x 4 343 f 23 15 23 10 23 6 23 14 f x

Polynomi Polynomial al and Synthetic Synth Division

2

3

2

0

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206

NOT FOR SALE

Chapter 2

55. f x

Polynomial ynomial and Rational Functions

2 x3 7 x 3

(a) Using the Remainder Theorem: f 1

21 71 3 3

(b) Using the Remainder Theorem: f 2

2

Using synthetic division: 1

2

7

3

2

2

5

2

5

2

0

2

2x x 1 2x 0 x2 2 x3 2 x 2 2x2 2x2 3

–2

2x 5 7x 3

1 4

1 2

1

1 2 13 2

f 2

8

–2

1

1

2 2 7 2 3 3

13 4 1 4

7

2

0 4

8

2

2

4

1

5

3

4x 1 7x 3

5

Using synthetic division: 2

7

–4 –4

2

7x 8x x 3 x 2 1 (d) Using the Remainder Theorem:

7x 2x 5x 3 5x 5 2

3

0

3

0

2x2 x 2 2 x 0 x2 2 x3 4 x 2 4x2 4x2

§1· §1· 2¨ ¸ 7¨ ¸ 3 2 © ¹ © 2¹

2

7

2

3

Using synthetic division: 1 2

1

Verify using long division:

(c) Using the Remainder Theorem: §1· f¨ ¸ © 2¹

3

Using synthetic division:

Verify using long division: 2

2 2 7 2 3

3

Verify using long division: 2x2 4 x 1 x 2 2x 0 x2 7 x 3 2 x3 4 x 2 4x2 7 x 4x2 8x x 3 x 2 5 3

Verify using long division: 13 2 1 x 2 x3 0 x 2 7 x 3 2 2 x3 x 2 2x2 x

x2 7 x 1 x2 x 2 13 x 3 2 13 13 x 2 4 1 4

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© Cengage Learning. All Rights Reserved.

Section 2.3 56. g x

Polynomi Polynomial al and Synth Synthetic Division

207

2 x6 3x 4 x2 3

(a) Using the Remainder Theorem: g 2

(b) Using the Remainder Theorem:

2 2 3 2 2 3 6

4

2

g 1

175

Using synthetic division: 2

2 2

0

1

0

4

8

22

44

86

172

4

11

22

43

86

175

0

3

1

g 3

4 x 11x 22 x 43x 86 3 x 4 0 x3 x 2 0 x 3 3

2

3x 4 8x4 11x 4 0 x3 11x 4 22 x3 22 x3 x 2 22 x3 44 x 2 43 x 2 0 x 43 x 2 86 x 86 x 3 86 x 172 175

2 2

2

0

1

0

3

6

18

63

6

21

63

189 188

564 564

1692 1695

Verify using long division: 2x

0

1

0

3

2

2

5

2

5

5

5 4

4 4

4 7

2 x 4 5 x3 5 x 2 4 x 4 3x 4 0 x3 x 2 0 x 3 3x 4 2x4 5 x 4 0 x3 5 x 4 5 x3 5 x3 x 2 5 x3 5 x 2 4x2 0x 4x2 4x 4x 3 4x 4 7

2 1 3 1 1 3 6

4

2

7

4

–1

0

3

0

1

0

3

2

2

5

2

5

5

5 4

–4 –4

4 7

2 2

Verify using long division:

6 x 21x 63 x 188 x 564 4

3

2

x 3 2 x 0 x 3x 0 x x 0x 3 2 x 6 6 x5 6 x5 3 x 4 6 x 4 18 x 4 21x 4 0 x3 21x 4 63 x3 x2 63x3 3 63x 189 x 2 188 x 2 0 x 188 x 2 564 x 564 x 3 564 x 1692 1695 5

3

Using synthetic division:

3

6

7

0

2 x5 x 1 2 x 6 0 x5 2 x 6 2 x5 2 x5 2 x5

g 1

1695

0

5

2

(d) Using the Remainder Theorem:

23 33 3 3 4

Using synthetic division: 3

4

Verify using long division:

4

(c) Using the Remainder Theorem: 6

2 2

Verify using long division: 2x x 2 2 x6 0 x5 2 x6 4 x5 4 x5 4 x5

6

Using synthetic division:

3

5

21 31 1 3

3

2

2 x5 x 1 2 x 0 x5 2 x6 2 x5 2 x5 2 x5 6

2 x 4 5 x3 5 x 2 4 x 4 3 x 4 0 x3 x 2 0 x 3 3x 4 2x4 5 x 4 0 x3 5 x 4 5 x3 5 x3 x 2 5 x3 5 x 2 4x2 0x 4x2 4x 4 x 3 4 x 4 7

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208

NOT FOR SALE

Chapter 2

57. h x

Polynomial ynomial and Rational Functions

x3 5 x 2 7 x 4

(a) Using the Remainder Theorem: h3

3

(b) Using the Remainder Theorem:

53 73 4

3

2

h 2

35

Using synthetic division: 3

1 1

7

4

3

6

39

2

13

35

2

x x 3 x 5x2 x3 3x 2 2 x 2 2 x 2

h 2

2

2 x 13 7x 4

5

7

4

2

6

26

3

13

22

x2 x 2 x 5x2 x3 2 x 2 3x 2 3x 2 3

7x 6x 13 x 4 13 x 39 35

5 2 72 4

10

3 x 13 7x 4 7x 6x 13 x 4 13 x 26 22

h 5

5 3

5 5 75 4 2

211

Using synthetic division:

1

5

7

2

14

14

1

7

7

10

–5

4

Verify using long division:

1

5

7

4

5

50

215

1

10

43

211

Verify using long division:

x2 7 x 7

x 2 10 x 43

x 2 x3 5 x 2 7 x 4 x3 2 x 2 7 x 2 7 x 7 x 2 14 x 7x 4 7 x 14 10

x 5 x3 5 x 2 7 x 4 x3 5 x 2 10 x 2 7 x 10 x 2 50 x 43 x 4 43 x 215 211

58. f x

22

(d) Using the Remainder Theorem:

2

Using synthetic division: –2

2

Verify using long division:

(c) Using the Remainder Theorem: 3

1 1

Verify using long division: 3

5 2 7 2 4

Using synthetic division:

5

2

2 3

4 x 4 16 x3 7 x 2 20

(a) Using the Remainder Theorem: f 1

(b) Using the Remainder Theorem:

41 161 71 20 4

3

Using synthetic division: 1

4 4

15

f 2

4 2 16 2 7 2 20 4

3

2

240

Using synthetic division:

16

7

0

4

12

5

20 5

12

5

5

15

Verify using long division: 4 x 3 12 x 2 5 x x 1 4 x 4 16 x 3 7 x 2 0 x 4 x 4 4 x3 12 x 3 7 x 2 12 x 3 12 x 2 5 x 2 0 x 5 x 2 5 x 5 x 5 x

5 20

20 5 15

2

4 4

16

7

0

20

8

48

110

24

55

110

220 240

Verify using long division: 4 x3 24 x 2 55 x x 2 4 x 4 16 x3 7 x 2 0 x 4 x3 8 x3 24 x3 7 x 2 24 x3 48 x 2 55 x 2 0 x 55 x 2 110 x 110 x 110 x

110 20

20 220 240 24

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© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.3

(c) Using the Remainder Theorem: f 5

3

2

695

Using synthetic division: 5

4

7

0

20

20

20

135

675

4

27

135

695

10

4x

2

1 1

0

7

6

2

4

6

2

3

0

x x

x3 7 x 6

4

4

3

2

16

7

0

20

40

560

5670

56

567

5670

56,700 56,720

4 x 27 x 135

4 x3

56 x 2 567 x

4

1

62.

2 x 2 2 x 3

2 3

48

80

41

6

32

32

6

48

48

9

0

x 23 48 x 48x 9 x 23 4 x 3 12 x 3

48 x3 80 x 2 41x 6

2

2 x 3 x 1

1

3 x

0

28

48

4

16

48

4

12

0

x3 28 x 48

x x

Zeros: 23 , 34 , 63.

1 2

2 2

15

3

4 x 2 4 x 12

2

3

6

3

3 2 3

6

3

2 3

0

1

3

2

1

2

27

1

7

10

14

20

0

2 x 3 15 x 2 27 x 10

1

10

x 12 2 x 2 x

2

1 x 2 x 5

64.

Zeros: 12 , 2, 5

0

3 x

2

1

4

2

2 2 2

4

2

2 2

0

2

2

2 2

2

2 2

2

x

x 2x 2x 4 2

Zeros: 2,

3 x 2

2

1 3

2 3

2

1 2

3

3, 2

1

2

2 3

x

2

Zeros: 3,

14 x 20

3

2

x 2 x 3x 6 3

2 4 x 3 4 x 1

1 4

1

4 x 6 x 2

Zeros: 4, 2, 6 61.

5670

7 x2 0x 20 x 10 4 x 4 16 x3 4 x 4 40 x3 56 x3 7 x 2 56 x3 560 x 2 567 x 2 0x 567 x 2 5670 x 20 5670 x 5670 x 56,700 56,720

Zeros: 2, 3, 1 60.

56,720

Verify using long division:

2

x 5 4 x 4 16 x3 7 x 2 0 x 20 4 x 4 20 x3 4 x3 7 x 2 4 x 3 20 x 2 27 x 2 0 x 27 x 2 135 x 135 x 20 135 x 675 695 59.

4 10 16 10 7 10 20

4

Verify using long division: 3

f 10

Using synthetic division:

16

4

209

(d) Using the Remainder Theorem:

45 165 75 20 4

Polynomial and Synthetic Synth Division

0

2 x 2 x

2

INSTRUCTOR USE ONLY 2,

2

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210 65.

NOT FOR SALE

Chapter 2 1

Polynomial ynomial and Rational Functions 3

1

3

1 1

3

1

1

3

1

3

2

3

1

3

0

2

3

1

3

1

3

1

3

–1

3x 3 2 x 2 19 x 6;

Factors: x 3 , x 2 3

(a)

3 2

3 º¼ x 1 3 x 1 3

2

19

6

9

21

6

7

2

0

3

3

3 ºª x 1 ¼¬

x 1 x 1 3, 1

68. f x

0

ªx 1 ¬

Zeros: 1, 1

2

2

1 x3 3x 2 2

0

3

7

2

6

2

1

0

Both are factors of f x because the remainders

3

are zero. 66.

2

2

1

13

3

2

5

7 3 5

3

1

1

5

6 3 5

0

1

1

5

6 3 5

2

5

63 5

3

0

1

5

5

1 x 3 x 2 13 x 3

x 2

5 x2

(b) The remaining factor is 3x 1 . (c) f x

3x3 2 x 2 19 x 6

3 x

1 x 3 x 2

(d) Zeros: 13 , 3, 2 (e)

35

5 x 3 −4

Zeros: 2

2

−10

2 1

1

5

2

4

6

2

3

1

0

2

2 2

3

5, 3

2 x 3 x 2 5 x 2; Factors: x 2 , x 1

67. f x

(a)

5, 2

3

1

2

1

1

0

69. f x

x 4 4 x3 15 x 2 58 x 40;

Factors: x 5 , x 4 (a)

5

1 4

Both are factors of f x because the remainders

(d) Zeros: (e)

1 , 2

1 x 2 x 1

58

40

5

5

50

1

10

8

40 0

1

10

8

4

12

8

3

2

0

Both are factors of f x because the remainders

are zero.

2 x

15

1 1

(b) The remaining factor is 2 x 1 . (c) f x

4

1

are zero. (b) x 2 3 x 2

x

1 x 2

The remaining factors are x 1 and x 2

2, 1

(c) f x

7

x

1 x 2 x 5 x 4

(d) Zeros: 1, 2, 5, 4 (e) −6

6

20 −6

6

−1

−180

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.3

70. f x

71. f x

8 x 4 14 x 3 71x 2 10 x 24;

Factors: x 2 , x 4 2

(a)

8 8

4

8

211

6 x 3 41x 2 9 x 14;

Factors: 2 x 1 , 3x 2

14

71

10

16

60

22

24 24

30

11

12

0

30

11

12

32

8

12

2

3

0

8

Polynomial and Synthetic Synth Division

(a)

12

6 6

2 3

41

9

14

3

19

14

38

28

0

6

38

28

4

28

6

42

0

Both are factors of f x because the remainders

Both are factors of f x because the remainders

are zero.

are zero.

4 x

(b) 8 x 2 2 x 3

3 2 x 1

The remaining factors are 4 x 3 and 2 x 1 . f x

(c)

4 x

This shows that

3 2 x 1 x 2 x 4

(d) Zeros: 34 , 12 , 2, 4 (e)

so

40 −3

6 x 7

(b) 6 x 42

2 x

f x 1 ·§ 2· § ¨ x ¸¨ x ¸ 2 ¹© 3¹ ©

f x 1 3 x 2

6 x 7 ,

x 7.

The remaining factor is x 7 .

5

f x

(c)

x

7 2 x 1 3 x 2

1 2 (d) Zeros: 7, , 2 3

−380

(e)

320

−9

3 − 40

72. f x

10 x 3 11x 2 72 x 45;

Factors: 2 x 5 , 5 x 3 (a)

52

10 10

3 5

10 10

11

72

45

25

90

45

36

18

0

36

18

6

18

30

0

(b) 10 x 30

10 x 3

This shows that

so

2 x

f x 5 3· § ·§ ¨ x ¸¨ x ¸ 2 5 © ¹© ¹

f x 5 5 x 3

10 x 3 ,

x 3.

The remaining factor is x 3 .

Both are factors of f x because the remainders are zero. (c)

f x

(e)

x

3 2 x 5 5 x 3

5 3 (d) Zeros: 3, , 2 5

100

−4

4

INSTRUCTOR USE ONLY −80 − 80

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212

Chapter 2

73. f x

Polynomial ynomial and Rational Function Functions

(d) Zeros: r 4 3, 3

2 x3 x 2 10 x 5;

Factors: 2 x 1 , x

(a)

1 2

2

(e)

1

10

5

1

0

5

0

10

0

2 5

5

2

−8

10

2 5

10

2 5

0

75. f x

(b) 2 x 2 5

2x

This shows that

so

2 x

1 x

5

5

x

x

(c) f x

(e)

5,

(c)

2

1

f x

5 .

1 2

76. g x

−6

4 3

48

144

3

0

144

0

48

0

0

48

4 3

48

4 3

0

1

0

x

2 x 5

x

2 x

2

5 x

5

4, x | 1.414, x | 1.414.

4 is an exact zero. 1

(c)

4

4

2

8

4

0

8

0

2

0

x

4 x 2 2

x

4 x

2 x

2

t 3 2t 2 7t 2

2

1 1

h t

2, t | 3.732, t | 0.268.

2.

t

2

7

2

2

8

2

4

1

0

2 t 2 4t 1

By the Quadratic Formula, the zeros of t 2 4t 1 are 2 r 3. Thus,

Both are factors of f x because the remainders are zero.

5

(b) An exact zero is t

3

1

0

(a) The zeros of h are t

Factors: x 4 3 , x 3

1

10

(b) x

77. ht

x 3x 2 48 x 144; 3

1

0

(c)

6

3

2

(a) The zeros of g are x

f x

(a)

10

x3 4 x 2 2 x 8

14

74. f x

5

1

−6

2.

2

1

5,

5 2 x 1

5 x

(d) Zeros: 5,

2x

5.

The remaining factor is x

2 and x | r2.236.

(b) An exact zero is x

f x 1· ¸ x 2¹

§ ¨x ©

f x

5

x 3 2 x 2 5 x 10

(a) The zeros of f are x

Both are factors of f x because the remainders are zero.

8

−240

0

2

60

h t

t

2 ªt 2 ¬

3 ºªt 2 ¼¬

3 º. ¼

(b) The remaining factor is x 4 3 . (c) f x

x 4 3 x 4 3 x 3

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.3

78. f s

s 3 12 s 2 40 s 24 6, s | 0.764, s | 5.236

(a) The zeros of f are s (b) s

6 is an exact zero.

(c)

1

6

12

40

24

6

36

24

6

4

0

1

f s

81.

s

f s

6 s 6s 4

79. h x

6 ªs 3 ¬

5 º. ¼ 82.

x 7 x 10 x 14 x 24 x 5

4

3

2

(a) The zeros of h are x

0, x

So,

5 ºªs 3 ¼¬

3, x

(c)

4

1

8

4,

1

1 h x

1

10

14

24

4

12

8

24

3

2

6

0

x

80. g x

83.

2 x

3, x

3, x

2

(c)

–3

6 6

1.5,

3.

a x

84.

x x

11

51

99

27

18

87

108

27 0

29

9

36

1 1

2

1 1

2

2

0 4x2 2 x 2

2 2 x 2 x 1

2 x 2 x 1, x z

3 . 2

1

64

64

8

56

64

7

8

0

2

x 2 7 x 8, x z 8

1

1

2

1 1

x 4 6 x3 11x 2 6 x x 1 x 2

6

11

6

0

1

5

6

0

5

6

0

0

5

6

0

2

6

0

3

0

0

x 4 6 x3 11x 2 6 x x 1 x 2

x 2 3 x, x z 2, 1

3 6 x 29 x 36 x 9 3

2

3 x 3 2 x 3 3 x 1

x 4 9 x3 5 x 2 36 x 4 x2 4

2

3

1

x | 0.333. (b) An exact zero is x

3

x 4 6 x 3 11x 2 6 x x 2 3x 2

6 x 4 11x 3 51x 2 99 x 27

(a) The zeros of a are x

6

x x 64 x 64 x 8 3

4 x 4 3 x 3 2 x 2 6 x

x x 4 x 3 x

3

x 3 x 2 64 x 64 x 8

4.

7

1

4 x3 8 x 2 x 3 2x 3

x | 1.414, x | 1.414.

(b) An exact zero is x

8

4 x3 8 x 2 x 3 3 x 2

5. Thus,

s

4

4

2

213

4 x3 8 x 2 x 3 2x 3

3 2

By the Quadratic Formula, the zeros of s 2 6 s 4 are 3 r

Polynomi Polynomial al and Synth Synthetic Division

x 4 9 x3 5 x 2 36 x 4 x 2 x 2

9

5

36

4

2

22

34

4

11

17

2

0

11

17

2

2

18

2

9

1

0

x 4 9 x3 5 x 2 36 x 4 x2 4

INSTRUCTOR USE ONLY x 2 9 x 1,, x z r2

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214

NOT FOR SALE

Chapter 2

Polynomial ynomial and Rational Function Functions x2n 6 xn 9

(c)

Year

Actual Value

0

23.2

23.4

1

24.2

23.7

2

23.9

23.8

3

23.9

24.1

4

24.4

24.6

5

25.6

25.7

x 2n x n 3

6

28.0

27.4

86. x n 2 x3n 3 x 2 n 5 x n 6 x 3n 2 x 2 n x2n 5xn x2n 2 xn 3x n 6 3x n 6 0

7

29.8

30.1

85. x 3 x 9 x 27 x 27 x3n 3 x 2 n 6 x 2 n 27 x n 6 x 2 n 18 x n 9 x n 27 9 x n 27 0 n

3n

2n

x3n 9 x 2 n 27 x n 27 xn 3

x 2 n 6 x n 9, x n z 3

x 3n 3 x 2 n 5 x n 6 xn 2

(d) 2010 o t 10

1 1

4

3

c

5

45

210

9

42

c 210

1 1

0.349

1.81

22.3

0.181

2.23

45.7

92. (a) and (b) 65

0

7 25

A | 0.0576t 3 0.913t 2 0.28t 30.7

Year

Actual Value

Estimated Value

0

0

2

1

c

0

30.5

30.7

2

4

8

20

42

1

32.2

31.8

2

4

10

21

c 42

2

34.2

34.5

3

38.0

38.2

4

42.7

42.7

5

47.9

47.7

6

52.7

52.8

7

57.6

57.6

(d) 2010 o t

10

To divide evenly, c 42 must equal zero. So, c must equal 42. 91. (a) and (b) 35

0

23.4

No, because the model will approach infinity quickly.

(c) 2

0.42

In 2010, the amount of money supporting higher education is about $45.7 billion.

To divide evenly, c 210 must equal zero. So, c must equal 210. 90.

0.168

0.0349

A10 | $45.7

x 2 n x n 3, x n z 2

88. You can check polynomial division by multiplying the quotient by the divisor. This should yield the original dividend if the multiplication was performed correctly.

5

10

0.0349

87. A divisor divides evenly into a dividend if the remainder is zero.

89.

Estimated Value

n

7

10

0

0.0576

A | 0.0349t 3 0.168t 2 0.42t 23.4 0.0576

0.913

0.28

30.7

0.576

3.37

36.5

0.337

3.65

67.2

In 2010, the amount of money spent on health care is about $67.2 billion.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.4

93. False. If 7 x 4 is a factor of f , then 74 is a zero

100. If x 3 is a factor of f x

then f 3

of f .

f 3

94. True. 1 2

6 6

f x

1

92

45

184

4

48

3

2

45

0

92

48

4

90

0

184

96

0

2 x

1 x 1 x 2 x 3 3 x 2 x 4

95. True. The degree of the numerator is greater than the degree of the denominator.

0 15 5

of f x , and f k

x3 kx 2 2kx 12

k 3 2k 3 12 2

27 9k 6k 12 3k k

x2 1 x 1

0.

1 1 1 1 0

x 1, x z 1

x2 x 1

97. False.

(b) x 1 x3 0 x 2 0 x 1 x3 x 2 x2 0x x2 x x 1 x 1 0

To divide x 4 3 x 2 4 x 1 by x 2 using synthetic division, the set up would be: 2

3

215

0. 3

x 101. (a) x 1 x 2 0 x x2 x x x

k is a zero of f x , then x k is a factor

96. True. If x

Comp Complex Numbers

1

3

0

4

1

A zero must be included for the missing x3 term. 98. f x

x

(a) k

x3 1 x 1

k q x r 5, q x

2, r

x3

any quadratic ax 2 bx c

x 3, r

(b) k

2 x 2 5

1, q x

x3 2 x 2 5

any quadratic

ax 2 bx c where a 0. One example:

f x

x

3 x 2 1

99. If x 4 is a factor of f x

then f 4 f 4 0 56 7

4

x3 3x 2 1 x3 kx 2 2kx 8,

x4 1 x3 x 2 x 1, x z 1 x 1 xn 1 x n 1 x n 2 " x 1, x z 1 x 1

0. 3

k 4 2 k 4 8 2

64 16k 8k 8

102. (a) f 3

8k

x2 x 1

(c) x 1 x 4 0 x3 0 x 2 0 x 1 x 4 x3 x3 0 x 2 x3 x 2 x2 0 x x2 x x 1 x 1 0

where a ! 0. One example: f x

x 2 x 1, x z 1

0 because x 3 is a factor of f .

(b) Because f x is in factored form, it is easier to

k

evaluate directly.

Section 2.4 Complex Numbers 1. (a) iii

2.

1; 1

(b) i (c) ii

3. principal square 4. complex conjugates

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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216

NOT FOR SALE

Chapter 2

5. a bi

Polynomial ynomial and Rational Function Functions

20. 3 2i 6 13i

12 7i

a

12

b

7

6. a bi

3 11i

13

b

4

7. a 1 b 3 i

5 a

6

b 3

8 b

5

6 a

18 4 3 2i

23. 13i 14 7i

8 5i

10. 2

27

2

24.

32 52 i 53 113 i

4

13.

0.09

5

25.

27i

26.

4 5i

5i

10

14 0i

75

2

75i

2

96 15 i 6

10 6

1 10i

29. 12i1 9i

76 i

10i

5 2 1

108 12i

30. 8i9 4i

10 3i

75

75i 2

5 i

72i 32i 2 32 72i

8 4i 31.

14

10i

14

10i

1 10

14 10i 2 14 10

3

5i 7

10i

21 50 7 21 5 2 7 36 84i 49i 2

24

21 3 10i 7 5i

2

5 2

12i 108i 2

4 2i

18. 13 2i 5 6i

22 i 6

11 41i

12i 108

5 7

21 35i 6i 10i 2

4 1 2i

19. 9 i 8 i

11 i 3

21 41i 10

14 10i 1

2

3 2i 3i 2i 2

28. 7 2i 3 5i

17. 7 i 3 4i

33. 6 7i

5 3

3i 2

0.09i

16. 4i 2 2i

27. 1 i 3 2i

2i

15. 10i i

32. 3

32 52 i

50i 2

0.3i 14. 14

8 3 2i 4 3 2i

1 6

2 3 3i

12.

13i 14 7i

52

25

80

2 2 2i 5 5 2i

14 20i

0

9. 8

11.

4

6 5i

5 b

2b

50

5 8i

a 1

a 6

8 5

3 3 2i

22. 8

8. a 6 2bi

21. 2

13 4i

a

3 2i 6 13i

50i 2

10 i

5 3 10 i 5 3

34. 5 4i

2

25 40i 16i 2

36 84i 49

25 40i 16

13 84i

9 40i

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.4

35. The complex conjugate of 9 2i is 9 2i.

9

2i 9 2i

46.

81 4i 2

2

1

5i 1

5i

37. The complex conjugate of

2 5i 2 5i

20i 2

38. The complex conjugate of

6 6 39.

3 i i i

40.

41.

44.

3i i 2

14 2i 2i 2i

28i 4i 2

8 16i 2i 2i 2i

3i 5i

2

20

2 5i is 2 5i .

i3 8i 2i3 2i

i 2i 3 2i 3 8i

3 2i 3 8i

3i 8i 2 6i 4i 2 9 24i 6i 16i 2

20 6 is

4i 2 9i 9 18i 16 4 9i 25 18i 25 18i 25 18i

6.

28i 4

13 13i 1 i2

100 72i 225i 162i 2 625 324 62 297i 62 297 i 949 949 949

7i 13 13i 2

16i 32i 2 4i 2

48.

1i 3 4 i i

3i 9 40i 9 40i 9 40i

21 i 31 i

1 i 1 i

1 i 4 i i 4 i

3i

4 i 4i i 2 3i 4i i 2 5 1 4i 1 4i 1 4i 5 20i 1 16i 2 5 20 i 17 17

8 4i

3i 16 40i 25i 2

2 3 1i 1i

13 13 i 2 2

6 12i 7i 14i 2 1 4i 2 20 5i 4i 5

27i 120i 2 120 27i 81 1600 1681 120 27 i 1681 1681 45.

47.

3i

13 1 i 1 i 1 i

4

6

6

6 7i 1 2i 42. 1 2i 1 2i

43.

5i.

1 5i 2 15

i 2 i

4i 2i 2 10 5i 4 i2 12 9i 5 12 9 i 5 5

85 5i is 1

217

2i 2 i 5 2 i

2i 5 2 i 2i

81 4

36. The complex conjugate of 1

Comp Complex Numbers

49. x 2 2 x 2

0; a

2 r

2 21

x

2 r

1, b 2

2, c

2

41 2

4

2 2 r 2i 2 1ri

2 2i 3 3i 11 1 5i 2 1 5 i 2 2

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

218

NOT FOR SALE

Chapter 2

Polynomial ynomial and Rational Function Functions

50. 4 x 2 16 x 17

4, b

16, c

53. 1.4 x 2 2 x 10

17

7 x 10 x 50

2

10 r

6 2 29

6, c

9, b

54.

37

49 37

1296 18 6 r 36i 1 r 2i 18 3

4 r

58.

59.

1 i3

60.

0; a

4 r

12 r

4, c

16, b

3

55. 6i 3 i 2

416 3

57. i

1 3

1 i 1 8i 3

6

43 18

72 2r

2i

6i 2i i 2 6 1 i 1

11 i 8

2i

2

6i 1 1 6i

1 r 8 6

12 23

6

56. 4i 2 2i 3

0

12 r 6 2i 6

176 32

1 i 2i

2i

4 2 216

0 Multiply both sides by 2.

12 r

4 r 4 11i 32

2

47 50

10 r 10 15 14

1500

3 2 x 6x 9 2 3x 2 12 x 18

x

6r

t

2

5 5 15 r 7 7 0; a

52. 16t 2 4t 3

10 27

14

51. 9 x 2 6 x 37 6 r

0

10 r

x

16 r 16 8 16 r 4i 8 1 2 r i 2

x

0 Multiply both sides by 5.

2

16 44 17 2 4

16 r

x

0; a

8i 6 1 i i i

1 8i 2i

8 1 1 1

8i 2i 2i 2 i i 2

1 8i

a1

1 i 2i

61. a bi a bi

1 1 i

4 2i

i

a 2 abi abi b 2i 2 a 2 b 2 1

8i 64i 2

a 2 b2

1 i 8

which is a real number since a and b are real numbers. Thus, the product of a complex number and its conjugate is a real number. 62.

63. a1 b1i a2 b2i

1 i3

4 1 21 i

8

i

1 8i 8i 8i

3

4i 2 2i 2i

6

6

6i 6i

6i 2

6

a2 b1 b2 i

The complex conjugate of this sum is a1 a2 b1 b2 i. The sum of the complex conjugates is a1 b1i a2 b2i

a1

a2 b1 b2 i.

So, the complex conjugate of the sum of two complex numbers is the sum of their complex conjugates.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

Section 2.4 64. (a) 24

16

(b)

2

(c)

2i 4

(d)

2i

66. f x

4

65. (a) i 40

i 2

(b) i 25

i 2

(c) i 50

i 2

25

(d) i 67

i 2

33

16 24 i 4

2

4

16 1 1

16i 2i 2 4 4

161 1

2 2

i

16i i

2 x 3 4, g x 2

16 16

Complex Comp Numbers

1 20

20

12

1 i

1

1 12 i

i

219

25

i 1

1 33 i

i

2 x 3 4 2

(a) The graph of f is a parabola with vertex at the point 3, 4 .

(c) If all the zeros contain i, then the graph has no x-intercepts.

The a value is positive, so the graph opens upward. The graph of g is also a parabola with vertex at the point 3, 4 . The a value is negative, so the graph opens downward. f has an x-intercept and g does not because when g x

0,

x is a complex number. (b)

f x

2 x 3 4

0

2 x 3 4

4

2 x 3

2

x

r 2 3r

2

2

(d) If a and k have the same sign (both positive or both negative), then the graph of f has no x-intercepts and the zeros are complex. Otherwise, the graph of f has x-intercepts and the zeros are real.

2

3

2

2

x 3 x

g x

2 x 3 4

0

2 x 3 4

4

2 x 3

2 r 2 3r

2i

67. False, if b

2 2

x

3

2

2

x 3 x 0 then a bi

a bi

68. True.

a.

x 4 x 2 14

That is, if the complex number is real, the number equals its conjugate.

i 6 i 6 4

2

14

36 6 14 56

56 ?

56 ?

56 56

69. False. i 44 i150 i 74 i109 i 61

i 2

22

1 22

i 2

75

1

i 2

75

1

111i i

70. (a) z1

(b)

1 z

9 16i, z2

1 1 z1 z2

37

i 2 i i 2 i 54

37

30

1 i 1 i 54

30

1

20 10i

1 1 9 16i 20 10i

§ 340 230i ·§ 29 6i · ¨ ¸¨ ¸ © 29 6i ¹© 29 6i ¹

20 10i 9 16i 10i

9 16i 20

11,240 4630i 877

29 6i 340 230i

11,240 4630 i 877 877

INSTRUCTOR USE ONLY z

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© Cengage Learning. All Rights Reserved.

220

Chapter 2

NOT FOR SALE

Polynomial ynomial and Rational Function Functions

Section 2.5 The Fundamental Theorem of Algebra 17. f x

1. Fundamental Theorem of Algebra

x3 6 x 2 11x 6

Possible rational zeros: r1, r 2, r 3, r 6

2. Linear Factorization Theorem

6

11

6

1

5

6

5

6

0

x3 6 x 2 11x 6

x x

1

3. Rational Zero

1

4. conjugate

1 5. linear; quadratic; quadratic 6. irreducible; reals 7. f x

x x 6

2

The zeros are: x 8. f x

9. g x

2 x 4

x

5 x 8

x

4

6, x

i , x

2, x

19. g x

9

6

3

2

0

3 x 2 3 x 2 3 x 2 x 1

x3 4 x 2 x 4 x 2 x 4 1 x 4

i

3i, x

3i

x 2x x 2 3

6

3

x x

x x

t 3 t 2 t 3i t 3i 3, x

7

0

So, the rational zeros are 2, 1, and 3.

8

4 x 2 1 4 x 1 x 1

So, the rational zeros are 4, 1, and 1.

2

20. h x

x3 9 x 2 20 x 12

Possible rational zeros: r1, r 2, r 3, r 4, r 6, r12

Zeros shown on graph: 2, 1, 1

1

1

x3 4 x 2 4 x 16

Possible rational zeros: r1, r 2, r 4, r 8, r16 Zeros shown on graph: 2, 2, 4 15. f x

1

2

Possible rational zeros: r1, r 2

14. f x

3

f x

6 x i x i

The zeros are: x 13. f x

1

1, x

Possible rational zeros: r1, r 2, r 3, r 6

1

5, x

The zeros are: x 12. ht

x 2 x 3 x 1 x 1

x3 7 x 6

3

2, x

The zeros are: x 11. f x

18. f x

3, x

0, x

The zeros are: x 10. f x

6

x 2 x 3 x 2 1

x

1 x 2 x 3

So, the rational zeros are 1, 2, and 3. 0, x

The zeros are: x

1 x 2 5 x 6

1 h x

2 x 4 17 x 3 35 x 2 9 x 45

Possible rational zeros: r1, r 3, r 5, r 9, r15, r 45,

x x

9

20

12

1

8

12

8

12

0

1 x 2 8 x 12 1 x 2 x 6

So, the rational zeros are 1, 2, and 6.

r 12 , r 32 , r 52 , r 92 , r 15 , r 45 2 2

Zeros shown on graph: 1, 32 , 3, 5 16. f x

4 x5 8 x 4 5 x3 10 x 2 x 2

Possible rational zeros: r1, r 2, r 12 , r 14

INSTRUCTOR USE ONLY Zeros shown on graph: 1, 12 , 12 , 1, 2

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.5

21. ht

25. f x

t 3 8t 2 13t 6

Possible rational zeros: r1, r 2, r 3, r 6

6

1

8

13

6

6

12

6

2

1

0

1

t t

t 3 8t 2 13t 6

Possible rational zeros: r1, r 2, r 3, r 4, r 6, r 8, r12, r 24,

9

6 t 1 t 1

1 f x

9

27

27

3

18

27

6

9

0

x x

3

f x

2

3 x 2 6 x 9

26. f x

3 x 3 x 3

2

0

1

2

1

1

1

1

2 x3 3 x 2 1

2

1

0

x x

1 2 x 2 x 1

x

1 2 x 1

f x

33

í9

9

í30

9

í10

3

0

x x

So, the rational zeros are 3 and 13 .

27

0

í12

0

í4

0

2 x 3 9 x 2 4 2 x 3 3 x 2 3 x 2

23

15

í25

10

í25

í10

í5

í2

5

25 0

2

í5

í2

5

2

í3

í5

2

í3

í5

0

2 2

f x

x

í3

í5

í2

5

í5

0

5 x 1 x 1 2 x 5

27. z 4 z 3 z 2 3 z 6

0

Possible rational zeros: r1, r 2, r 3, r 6 1

3 3 x 2 10 x 3 3 3 x 1 x 3

12

So, the rational zeros are 5, 1, 1 and 52 .

Possible rational zeros: r1, r 3, r 9, r 13

3

í1

2

í19

í4

í15

1 x 1 2 x 1

3x3 19 x 2 33 x 9

3

8 12

2 x 4 15 x3 23x 2 15 x 25

2

So, the rational zeros are 1 and 12 . 24. f x

í24 0

54 í4

Possible rational zeros: r1, r 5, r 25, r 12 , r 52 , r 25 2 5

3

24

So, the rational zeros are 2, 3, 23 , and 23 .

Possible rational zeros: r1, r 12

3

x x

4

í27 í27

9

2 x3 3x 2 1

1

í58

í18

9

So, the rational zero is 3. 23. C x

í9

9

6 t 2 2t 1

x3 9 x 2 27 x 27

1

9 x 4 9 x3 58 x 2 4 x 24

í2

Possible rational zeros: r1, r 3, r 9, r 27

3

221

r 13 , r 23 , r 43 , r 83 , r 19 , r 92 , r 94 , r 98

So, the rational zeros are 1 and 6. 22. p x

The Fundamental Theorem Theore of Algebra

1

1

z

1

1

3

í6

1

2

3

2

3

6

6 0

1 z 3 2 z 2 3 z 6

0

1 z 3 z 2

0

z

2

So, the real zeros are í2 and 1.

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© Cengage Learning. All Rights Reserved.

222

NOT FOR SALE

Chapter 2

Polynomial ynomial and Rational Function Functions

28. x 4 13 x 2 12 x

0

x x3 13 x 12

0

31. f x

x3 x 2 4 x 4

(a) Possible rational zeros: r1, r 2, r 4

Possible rational zeros of x3 13 x 12:

y

(b) 4

r1, r 2, r 3, r 4, r 6, r12

í1

1

2

0

í13

í12

í1

1

í1

í12

12 0

1

x –6

0

x x 1 x 4 x 3

0

–6

(c) Real zeros: 2, 1, 2

29. 2 y 4 3 y 3 16 y 2 15 y 4

2

2

0

r 12 ,

r1, r 2, r 4

3

í16

15

í4

2

5

í11

4

5

í11

4

0

32. f x

3 x3 20 x 2 36 x 16

(a) Possible rational zeros: r1, r 2, r 4, r 8, r16, r 13 , r 23 , r 43 , r 83 , r 16 3 (b)

y 10 8 6 4 2 x –4 –2

1

2

2

y y

5

í11

4

2

7

–4

7

í4

0

(c) Real zeros: 23 , 2, 4

1 y 1 2 y 1 y 4

0

1 2

and 1.

30. x5 x 4 3x3 5 x 2 2 x

0

x x x 3x 5 x 2

0

3

8 10 12

–6

0

So, the real zeros are 4,

6

–4

1 y 1 2 y 2 7 y 4

4

6

–4

The real zeros are 0, 1, 4, and 3.

1

4

–8

x x 1 x 2 x 12

Possible rational zeros:

–4

2

33. f x

4 x3 15 x 2 8 x 3

(a) Possible rational zeros: r1, r 3, r 12 , r 32 , r 14 , r 34 y

(b)

4 2 x

Possible rational zeros of x 4 x3 3x 2 5 x 2:

–6 –4 –2

6

8 10

–6

1

1 í2

4

–4

r1, r 2

1

2

1

1

í1

í3

5

í2

1

0

í3

2

0

í3

2

0

0

í3

2

í2

4

í2

í2

1

0

(c) Real zeros: 14 , 1, 3 34. f x

4 x3 12 x 2 x 15

(a) Possible rational zeros: r1, r 3, r 5, r15, r 12 , r 23 , r 52 , r 15 , r 14 , r 43 , r 54 , r 15 2 4 y

(b) 15 12

x x 1 x 2 x 2 2 x 1

0

x x 1 x 2 x 1 x 1

0

The real zeros are 2, 0, and 1.

x –9 –6 –3

6

9

12

INSTRUCTOR USE ONLY ((c)) Real zeros: 1,, 32 ,

5 2

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.5

35. f x

2 x 4 13 x3 21x 2 2 x 8

The Fundamental Theorem Theore of Algebra

39. f x

x 4 3x 2 2 r1, about r1.414

(a) Possible rational zeros: r1, r 2, r 4, r 8, r 12

(a) x

(b)

(b) An exact zero is x

1.

0

í3

0

2

1

1

í2

í2

1

í2

í2

0

16

1 −4

223

1

8

1 −8

(c)

(c) Real zeros: 12 , 1, 2, 4 36. f x

í1

4 x 4 17 x 2 4 f x

9

−8

í2

í2

í1

0

2

0

í2

0

1

(a) Possible rational zeros: r1, r 2, r 4, r 12 , r 14 (b)

1

1

40. Pt

8

x

1 x 1 x 2 2

x

1 x 1 x

2 x

2

t 4 7t 2 12 r 2, about r1.732

(a) t −15

(b) An exact zero is t

2.

í7

0

12

2

4

í6

í12

2

í3

í6

0

(c) Real zeros: 2, 12 , 12 , 2 37. f x

2

32 x3 52 x 2 17 x 3

0

1

(a) Possible rational zeros: r1, r 3, r 12 , r 23 , r 14 , r 43 ,

í2

2

í3

í6

í2

0

6

0

í3

0

1

6

1 −1

(c) Pt

3 −2

(c) Real zeros: 18 , 34 , 1 38. f x

41. h x

4 x 7 x 11x 18 3

2

(a) Possible rational zeros: r1, r 2, r 3, r 6, r 9, r18, 1 3 9 1 3 9 r ,r ,r ,r ,r ,r 2 2 2 4 4 4 (b)

x

(c) 1 r 8

t

2 t 2 t

3 t

3

x x 4 7 x3 10 x 2 14 x 24 0, 3, 4, about r1.414

4

145 8

10

14

í24

3

í12

í6

24

1

í4

í2

8

0

1

í4

í2

8

4

0

í8

0

í2

0

1 h x

3.

í7

1

8

(c) Real zeros: 2,

2 t 2 t 2 3

(b) An exact zero is x 3

−24

t

x5 7 x 4 10 x3 14 x 2 24 x

(a) h x

8 −8

2.

An exact zero is t

1,r 3,r 1,r 3 r 18 , r 83 , r 16 16 32 32

(b)

1

x x 3 x 4 x 2 2

INSTRUCTOR USE ONLY x x 3 x 4 x

2 x

2

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

224

NOT FOR SALE

Chapter 2

42. g x

Polynomial ynomial and Rational Function Functions

44. If 3i is a zero, so is its conjugate, 3i.

6 x 4 11x3 51x 2 99 x 27

f x

r 3, 1.5, about 0.333

(a) x

(b) An exact zero is x 3

6

3.

í11

í51

99

í27

18

21

í90

27

7

í30

9

0

6

í3

6

(c) g x

x x

x x

real number, has the zeros 4, 3i, and 3i.

í30

9

í18

33

í9

í11

3

0

f x

3 x 3 6 x 11x 3 3 x 3 3 x 1 2 x 3

2 x 2 10 x 26 a x3 12 x 2 46 x 52 , where a is

46. If 3 2i is a zero, so is its conjugate, 3 2i.

f x

1 x 2 25

x

5 x 3 2i x 3 2i

x

5 x 2 6 x 13

x3 11x 2 43 x 65

a x3 x 2 25 x 25 , where a is any

3 x

2 x 1 ª x 3 ¬

3 x

2 x 1 ª¬ x 3

3x 2

2 x 2 ª« x 3 ¬

a x3 11x 2 43 x 65 , where a is

Note: f x

any nonzero real number, has the zeros 5 and 3 r 2i.

2i is a zero, so is its conjugate, 3

3x 2 3x 2

x

any nonzero real number, has the zeros 2 and 5 r i.

nonzero real number, has the zeros 1 and r 5i.

f x

2 x 5 i x 5 i

Note: f x

x3 x 2 25 x 25

47. If 3

x

x3 12 x 2 46 x 52

2

1 x 5i x 5i

Note: f x

a x3 4 x 2 9 x 36 , where a is any

Note: f x

43. If 5i is a zero, so is its conjugate, 5i.

f x

4 x 2 9

45. If 5 i is a zero, so is its conjugate, 5 i.

7

6

4 x 3i x 3i

x3 4 x 2 9 x 36

3.

An exact zero is x

x x

2i.

2i ºª x 3 ¼¬ 2iºª ¼¬ x 3

2i º ¼ 2iº¼

2i º»¼ 2

x 2 x 2 6 x 9 2 x 2 x 2 6 x 11

3x 4 17 x3 25 x 2 23x 22 Note: f x 48. If 1

f x

a3 x 4 17 x3 25 x 2 23 x 22 , where a is any nonzero real number, has the zeros 23 , 1, and 3 r

3i is a zero, so is its conjugate, 1

x

x2

5 x 1 2

3i x 1

3i

49. f x

3i.

10 x 25 x 2 2 x 4

x 8 x 9 x 10 x 100 4

Note: f x

3

x 4 6 x 2 27

(a) f x

x2

9 x 2 3

(b) f x

x2

9 x

(c) f x

x

2

a x 4 8 x3 9 x 2 10 x 100 , where

a is any real number, has the zeros 5, 5, and 1 r

2i.

3 x

3i x 3i x

3

3 x

3

3i.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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NOT FOR SALE Section 2.5

50. f x

225

x 4 2 x3 3 x 2 12 x 18

x2 2x 3 x 6 x 2 x 3x 2 12 x 18 6x2 x4 3 2 x 3x 2 12 x 2 x3 12 x 2 18 3x 18 3x 2 0 2

51. f x

The Fundamental Theorem Theore of Algebra

4

3

(a) f x

x2

(b) f x

x x

(c) f x

6 x 2 2 x 3

6 x

6 x

6 x 1

6 x 2 2 x 3

2i x 1

2i

Note: Use the Quadratic Formula for (c).

x 4 4 x3 5 x 2 2 x 6

x 2x 2 x 4x x 4 2 x3 2 x3 2 x3 2

4

3

(a) f x

x2

(b) f x

x 1 x 1

(c) f x

x2 5x2 2x2 7 x2 4x2 3x 2 3x 2

2x 3 2x 6

2x 4x 6x 6 6x 6 0

2 x 2 x 2 2 x 3

3 x 1

3 x 1

3 x 1

3 x 2 2 x 3

2i x 1

2i

Note: Use the Quadratic Formula for (b) and (c). 52. f x

x 4 3 x3 x 2 12 x 20 x 2 3x 5

x 2 4 x 4 3 x3 x 2 12 x 20 4x2 x4 3 3 x 5 x 2 12 x 3 x3 12 x 5 x 2 20 5 x 2 20 0

(a) f x

x2

4 x 2 3 x 5

(b) f x

x2

§ 3 29 ·§ 3 29 · 4 ¨¨ x x ¸¨ ¸¸ ¸¨ 2 2 © ¹© ¹

(c) f x

x

§ 3 29 ·§ 3 29 · x 2i x 2i ¨¨ x ¸¨ ¸¸ ¸¨ 2 2 © ¹© ¹

Note: Use the Quadratic Formula for (b).

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226

Chapter 2

53. f x

NOT FOR SALE

Polynomial ynomial and Rational Function Functions

x3 x 2 4 x 4

Alternate Solution: Because x

Because 2i is a zero, so is í2i. 2i

1

í1

4

í4

2i

4 2i

4

1

2i 1

2i

0

2i

1

1 f x

x

2i 1

í2i

2i

2i

í1

0

x 2 0 x 4 x3 x 2 x3 0 x 2 x2 x2 f x

2 x3 3 x 2 18 x 27

Because x

3

18

27

6i

9i 18

í27

2

3 6i

9i

0

2 2

f x

x

3 6i

9i

6i

í9i

3

0

55. f x

x

2

r 3i are zeros of f x ,

3i x 3i

f x

x2

2 2

f x

x x

Because x 49

í25

í25

10i

5i 50

5i 25

25

1 10i

1 5i

í5i

0

1 10i

1 5i

í5i

10i

5i

5i

í1

í1

0

x

r 5i are zeros of f x , x 2 25 is a factor of f x .

5i x 5i

By long division, you have: 2x2 x 0 x 25 2 x x 2 x 4 0 x3 x3 x3 4

5i x 5i 2 x 2 x 1 5i x 5i 2 x 1 x 1 r5i, 12 , 1.

r 3i, 32 .

Alternate Solution:

í1

The zeros of f x are x

18 x 27 18 x 0 x 27 0 x 27 0

9 2 x 3

2

í5i

x 2 9 is a factor of f x .

The zeros of f x are x

32 .

2 x 4 x3 49 x 2 25 x 25

2

1, r 2i.

2x 3

Because 5i is a zero, so is í5i. 5i

4 4 0

4 x 1

x 2 0 x 9 2 x3 3x 2 2 x3 0 x 2 3x2 3x2

r 3i,

1 4

By long division, you have:

3i x 3i 2 x 3

The zeros of f x are x

x 4x 4x 0x 0x

Alternate Solution:

2

í3i

x2

The zeros of f x are x

1, r 2i.

Because 3i is a zero, so is 3i. 3i

x 2 4 is a factor of f x .

2i x 2i

By long division, you have:

2i x 2i x 1

The zeros of f x are x 54. f x

x

r 2i are zeros of f x ,

f x

x2

3

x 1

49 x 2 25 x 25 50 x 2 x 2 25 x 0 x 2 25 x x 2 0 x 25 x 2 0 x 25 0

25 2 x 2 x 1

The zeros of f x are x

r 5i, 12 , 1.

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© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.5

56. g x

The Fundamental Theorem Theore of Algebra

227

x3 7 x 2 x 87

Because 5 2i is a zero, so is 5 2i. 5 2i

í1

87

5 2i

í7

14 6i

í87

1

2 2i

15 6i

0

1

2 2i

15 6i

5 2i

15 6i

3

0

1

5 2i

1

3. The zeros of f x are x

The zero of x 3 is x 57. g x

4 x3 23 x 2 34 x 10

Alternate Solution Because 3 r i are zeros of g x ,

Because 3 i is a zero, so is 3 i. 3 i

3 i

23

34

í10

12 4i

37 i

10

4

11 4i

3 i

0

4

11 4i

3 i

12 4i

3i

4

í1

4 The zero of 4 x 1 is x g x are x

¬ª x 3 i ºª ¼¬ x 3 i º¼

4x 1 x 6 x 10 4 x 23 x 4 x3 24 x 2 x2 x2 2

. The zeros of

3

3 r i, 14 .

x2

2

34 x 10 40 x 6 x 10 6 x 10 0

6 x 10 4 x 1 3 r i, 14 .

3x3 4 x 2 8 x 8

Because 1

1

3

is a factor of g x . By long division, you have:

The zeros of g x are x

1

3 i 2

x 2 6 x 10

g x

58. h x

ª¬ x 3 iºª ¼¬ x 3 iº¼

x

0 1 4

3, 5 r 2i.

3i

3i

3i is a zero, so is 1

3i.

í4

8

8

3 3 3i

10 2 3i

í8

3

1 3 3i

2 2 3i

0

3

1 3 3i

2 2 3i

3 3 3i

2 2 3i

3

3 The zero of 3x 2 is x

2

0

23 . The zeros of f x are x

23 , 1 r

3i.

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228

NOT FOR SALE

Chapter 2

59. f x

Polynomial ynomial and Rational Function Functions

x 4 3 x3 5 x 2 21x 22 2i is a zero, so is 3

Because 3

ª x 3 ¬

2i ºª x 3 ¼¬

2i, and ª x 3 ¬

2i º ¼

x

3 2

2iºª ¼¬ x 3

2i

2iº¼

2

x 2 6 x 11 is a factor of f x . By long division, you have: x 2 3x 2 x 6 x 11 x 3 x 5 x 2 21x 22 2

4

3

x 4 6 x3 11x 2 3 x3 16 x 2 21x 3 x3 18 x 2 33x 2 x 2 12 x 22 2 x 2 12 x 22 0

x2 x2

f x

6 x 11 x 2 3 x 2 6 x 11 x 1 x 2

The zeros of f x are x 60. f x

3 r

2i, 1, 2. 63. h x

x3 4 x 2 14 x 20

By the Quadratic Formula, the zeros of f x are

Because 1 3i is zero, so is 1 3i. 1 3i

1 3i

1

4

14

20

1 3i

12 6i

í20

1

3 3i

2 6i

0

1

3 3i

2 6i

1 3i

2 6i

1

2

0

The zero of x 2 is x

2.

The zeros of f x are x

2, 1 r 3i.

61. f x

x 2 36

x

62. f x

x f x

64. g x

223i 2

4 68 2

2r

64 2

1 r 4i.

x 1 4i x 1 4i x

1 4i x 1 4i

x 2 10 x 17

By the Quadratic Formula, the zeros of f x are x

65. f x

1r

2r

10 r

100 68 2

10 r 2

32

5 r 2

2.

x 5 2 2 x 5 2 2 x 5 2 2 x 5 2 2

By the Quadratic Formula, the zeros of f x are 1 224 2

f x

r 6i.

x 2 x 56

1r

x

f x

6i x 6i

The zeros of f x are x

x 2 2 x 17

.

x 4 16

x2 x

4 x 2 4 2 x 2 x 2i x 2i

Zeros: r 2, r 2i

§ 1 223i ·§ 1 223i · x ¨¨ x ¸¨ ¸¸ ¸¨ 2 2 © ¹© ¹

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.5

66.

f y

70. f x

y 4 256

y2 y

The Fundamental Theorem Theore of Algebra

16 y 2 16

x3 x 2 x 39

Possible rational zeros: r1, r 3, r13, r 39

4 y 4 y 4i y 4i

3

1

Zeros: r 4, r 4i 67. f z

1

z2 2z 2

By the Quadratic Formula, the zeros of f z are z

2r

f z

48 2

1 r i.

1

4

í2

1

í2

2

í2

2

0

are x

48 2

x

69. g x

1 r i.

1 x 1 i x 1 i

1

3

1

5

1

4

5

4

5

0

x

h x

4r

16 20 2

x

16 52 2

2 r 3i

3 x 2 3i x 2 3i

1

0

1

6

2

4

6

2

3

0

2r

4 12 2

1r

2i.

2i.

x

2 ª x 1 ¬

x

2 x 1

2i x 1

2i ºª x 1 ¼¬ 2i

2i º ¼

x3 9 x 2 27 x 35

Possible rational zeros: r1, r 5, r 7, r 35 5

2ri

1 x 2 i x 2 i

1 1

9

27

35

5

20

35

4

7

0

By the Quadratic Formula, the zeros of x 2 4 x 7 are x

Zeros: 1, 2 r i g x

0

x3 x 6

72. h x

By the Quadratic Formula, the zeros of x 2 4 x 5 are: x

39

13

Zeros: 2, 1 r

x3 3x 2 x 5

1

12

4

By the Quadratic Formula, the zeros of x 2 2 x 3 are

Possible rational zeros: r1, r 5 1

x

1

Zeros: 1, 1 r i h x

f x

2

By the Quadratic Formula, the zeros of x 2 2 x 2 2r

3

Possible rational zeros: r1, r 2, r 3, r 6

Possible rational zeros: r1, r 2 í3

39

4r

71. h x

x3 3x 2 4 x 2

1

1

Zeros: 3, 2 r 3i

z 1 i z 1 i

1

1

By the Quadratic Formula, the zeros of x 2 4 x 13 are: x

ª¬ z 1 i ºª ¼¬ z 1 i º¼

68. h x

229

4 r

16 28 2

Zeros: 5, 2 r h x

x

2 r

3i.

3i

5 x 2

3i x 2

3i

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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230

NOT FOR SALE

Chapter 2

73. f x

Polynomial ynomial and Rational Function Functions

76. h x

5 x3 9 x 2 28 x 6

Possible rational zeros: r1, r 3, r 9

Possible rational zeros: 1 2 3 6 r1, r 2, r 3, r 6, r , r , r , r 5 5 5 5 15

9

28

6

1

2

6

10

30

0

5 5

3

2r

4 24 2

1 Zeros: , 1 r 5

3

1r

h x

5i ºª x 1 ¼¬

5i x 1

5i

5i º ¼

2

8

21

3

6

21

4

14

0

4r

16 112 4

2

78. f x

g x

3

9 0

1

3

0

3

0

1

0

x

r i.

3 x i x i 2

x 4 10 x 2 9 1 x 2 9 i x i x 3i x 3i

x 4 29 x 2 100

x2 x

25 x 2 4 2i x 2i x 5i x 5i

Zeros: r 2i, r 5i

4r

96 4

6i x 1

x 4 4 x3 8 x 2 16 x 16

1

4

8

2 1

2

1

2

4

8

2

0

8

0

4

0

1

3

1

Zeros: r i, r 3i

Possible rational zeros: r1, r 2, r 4, r 8, r16 2

9

3

1r

79. f x

x3 24 x 2 214 x 740

Possible rational zeros: r1, r 2, r 4, r 5, r10, r 20, r 37, r 74, r148, r185, r 370, r 740

6i.

6i

3· § ¨x ¸ x 1 2¹ ©

75. g x

3

2000

3 Zeros: , 1 r 2

f x

9

3

x

By the Quadratic Formula, the zeros of 2 x 2 4 x 14 are x

6

3

x2

Possible rational roots: 1 3 7 21 r , r1, r , r 3, r , r 7, r , r 21 2 2 2 2 1

77. f x

2 x3 x 2 8 x 21

2

10

The zeros of x 2 1 are x

5i.

5 x 1 x 1

3 2

6

Zeros: 3, r i

74. g x

1 1

5i.

ª § 1 ·º ª « x ¨ 5 ¸»5 ¬ x 1 © ¹¼ ¬

f x

1 1

By the Quadratic Formula, the zeros of 5 x 2 10 x 30 5 x 2 2 x 6 are x

x 4 6 x3 10 x 2 6 x 9

16

16

4

8

4

8

16 0

x

2 x 2 x 2 4

x

2 x 2i x 2i

6i

−20

10

−1000

Based on the graph, try x 10

1 1

10.

24

214

740

10

140

740

14

74

0

By the Quadratic Formula, the zeros of x 2 14 x 74 are x

14 r

196 296 2

The zeros of f x are x

7 r 5i.

10 and x

7 r 5i.

2

INSTRUCTOR USE ONLY 2i Zeros: 2, r 2i Zeros

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© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.5

80. f s

The Fundamental Theorem Theore of Algebra

82. f x

2 s 3 5s 2 12 s 5

1 5 Possible rational zeros: r1, r 5, r , r 2 2

9 x3 15 x 2 11x 5

1 5 1 5 Possible rational zeros: r1, r 5, r , r , r , r 3 3 9 9

10

5

−10

−5

10

5

−5

−10

Based on the graph, try s 1 2

2

5

12

1

2

5

2

4

10

0

Based on the graph, try x

1 . 2

1

2r

4 20 2

9

1.

15

11

5

9

6

5

6

5

0

By the Quadratic Formula, the zeros of 9 x 2 6 x 5

1 r 2i.

6r

are x

36 180 18

1 2 r i. 3 3

The zeros of f x are x

1 and s 2

The zeros of f s are s 81. f x

9

5

By the Quadratic Formula, the zeros of 2 s 2 2 s 5 are s

1 r 2i.

83. f x

16 x3 20 x 2 4 x 15

2 x 4 5 x3 4 x 2 5 x 2

−4

4 −5

2 and x

Based on the graph, try x −3

2

3

2

−5

Based on the graph, try x 16

16

3 . 4

20

4

15

12

24

15

32

20

0

2

x

64 80 8

1 2

2

2

5

4

5

2

4

2

4

1

2

1

2 0

1

2

1

1

0

1

0

2

0

The zeros of 2 x 2 2

By the Quadratic Formula, the zeros of 16 x 2 32 x 20 4 4 x 2 8 x 5 are 8r

1 2

20

20

3 4

1 2 r i. 3 3

1 and x

Possible rational zeros: r1, r 2, r

Possible rational zeros: 1 3 5 15 1 3 r1, r 3, r 5, r15, r , r , r , r , r , r , 2 2 2 2 4 4 5 15 1 3 5 15 1 3 5 15 r ,r ,r ,r ,r ,r ,r ,r ,r ,r 4 4 8 8 8 8 16 16 16 16

231

The zeros of f x are x

1 1 r i. 2

The zeros of f x are x

3 and x 4

1r

2 x 2 1 are x 2, x

1 . 2

r i.

1 , and x 2

r i.

1 i. 2

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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232

NOT FOR SALE

Chapter 2

84. g x

Polynomial ynomial and Rational Function Functions

87. f y

x5 8 x 4 28 x3 56 x 2 64 x 32

Possible rational zeros: r1, r 2, r 4, r 8, r16, r 32 10

4 y3 3 y 2 8 y 6

Possible rational zeros: r1, r 2, r 3, r 6, r 12 , r 32 , r 14 , r 34

34 −10

4

10

3

8

6

3

0

6

0

8

0

4 −10

Based on the graph, try x

2

2

2

1

8

28

56

64

32

1

2 6

12 16

32 24

48 16

32 0

1

6

16

24

16

1

2 4

8 8

16 8

16 0

1

4

8

8

2

4

8

2

4

0

1

4 y

are x

4 16 2

88. g x

The zeros of g x are x 85. f x

1

1r

3i.

4 x3 3x 1

4 4

0

3

1

4

4

1

4

1

0

4 x3 3x 1

1 4 x 2 4 x 1

x

1 2 x 1

12 f z

18

21

9

14

6

0

3 y 2 2

0

15

0

2

15

3 x

2 x 2 5

x4 4

2 x

25 2 x 4

9

25 x 2 36 9 x 2 4

3 2 x 3 x 2 x 2

The rational zeros are r 32 and r 2. 90. f x

2

1 2

2 x3 3x 2

23 x 12

Possible rational zeros: r1, r 2, r 3, r 4, r 6, r12, r 12 ,

r 12 ,

r 32 ,

r 92 ,

1 r 14 , r 34 , r 94 , r 16 , r 12

9

10

4 x 1 4 x2 4

x

27

10

0

1 4

4

4

2

So, the only real zero is 23 .

3

2

12 z 3 4 z 2 27 z 9

12

15

2

x 23 3x

1 4

Possible rational zeros: r1, r 3, r 9,

3 2

2

3

r 14

1 So, the real zeros are 1 and . 2

86. f z

3

89. P x

Possible rational zeros: r1,

2

3 x3 2 x 2 15 x 10

g x

3i.

2 and x

r 12 ,

8

Possible rational zeros: r1, r 2, r 5, r10, r 13 , r 23 , r 53 , r 10 3 2 3

1r

2

So, the only real zero is 34 .

By the Quadratic Formula, the zeros of x 2 2 x 4 2r

y 34 4 y y 34 4 y

4 y3 3 y2 8 y 6

2.

r 13 ,

2 f x

23

12

8

20

12

5

3

0

1 2

x

4 2 x 2 5 x 3

1 2

x

4 2 x 1 x 3

3 2

The rational zeros are 3, 12 , and 4.

2 z 32 6 z 2 7 z 3

2 z

3 3 z 1 2 z 3

INSTRUCTOR USE ONLY So, the real zeros are 32 , 13 , and 32 .

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.5

91. f x

x3 1 4

x

1 x2 4

4 x3 x 2 4 x

1 ªx 2 4¬

The Fundamental Theore Theorem of Algebra

233

97. Zeros: 2, 12 , 3

1 4

4 x 1

f x

x 2 2 x 1 x 3 2 x3 3x 2 11x 6

1 1 4 x 1 º¼

1 4

4 x

1 x 2 1

Any nonzero scalar multiple of f would have the same three zeros.

1 4

4 x

1 x 1 x 1

Let g x

The rational zeros are

and r1.

1 4

af x , a ! 0.

There are infinitely many possible functions for f. y

92. f z

1 6

6 z

3

11z 3z 2 2

8

Possible rational zeros: r1, r 2, r 12 , r 13 , r 23 , r 16 2

11

3

2

12

2

2

1

1

0

6 6

f x

x

2 6 x 2 x 1

1 6

x

2 3x 1 2 x 1

(3, 0) x

−8

1 6

( 21, 0)

(−2, 0)

98.

−4

4

8

12

y 50

(−1, 0)

The rational zeros are 2, 13 , and 12 .

10

(1, 0)

(4, 0) x

93. f x

x

x 1 3

Rational zeros: 1 x

1 x x 1 2

1

Matches (d).

Value of f(x): Positive Negative Negative Positive (a) Zeros of f x : x

x3 2

x

3

2 x2

3

2x

3

Rational zeros: 0

Irrational zeros: 1 x

3

2

Matches (a). 95. f x

x x 1 x 1

x3 x

Rational zeros: 3 x

0, r1

4

4.

(b) The graph touches the x-axis at x

1.

(c) The least possible degree of the function is 4 because there are at least four real zeros (1 is repeated) and a function can have at most the number of real zeros equal to the degree of the function. The degree cannot be odd by the behavior at r f. (d) The leading coefficient of f is positive. From the information in the table, you can conclude that the graph will eventually rise to the left and to the right.

x

2

(Any nonzero multiple of f(x) is also a solution.)

x3 2 x (f )

x x 2 2

2 x 1 x 4

x 4 4 x3 3x 2 14 x 8

Matches (b).

x x

2, x

1, x

(e) f x

Irrational zeros: 0

96. f x

5

99. Interval: f, 2 , 2, 1 , 1, 4 , 4, f

Irrational zeros: 0

94. f x

(3, 0) 4

y

(− 2, 0) 2

2 x

Rational zeros: 1 x

0

r

Irrational zeros: 2 x

2

2

−3

(1, 0)

−1 −4 −6 −8 −10

2

(4, 0) 3

x

5

Matches (c).

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234

NOT FOR SALE

Chapter 2

Polynomial ynomial and Rational Function Functions

100. (a) 2, 0, 2

(e) f x

(b) The graph touches the x-axis at x x 2.

101. (a) Combined length and width:

Volume

l wh

(f )

120 4 x

2

y

12 10 8 6

(−2, 0) −4 − 3

(2, 0) (0, 0)

1

2

3

x

4

4 x 2 30 x

13,500 4 x3 120 x 2 13,500

0

x3 30 x 2 3375

0

2

x y

15

1

30

1

4 x 30 x 2

18,000

x

0

3375

15

225

3375

15

225

0

15 x 2 15 x 225

0

Using the Quadratic Formula, x 0

2

14

(c)

x 2 120 4 x

(b)

2

x 5 2 x 4 4 x3 8 x 2

(d) The leading coefficient of f is positive. From the information in the table, you can conclude that the graph will eventually fall to the left and rise to the right.

120 y

0 ª¬ x 2 º¼ x 2

x 2 x 2 x 2

0 and at

(c) The least possible degree of f is 5 because there are at least 5 real zeros (0 and 2 are repeated) and a function can have at most the number of real zeros equal to the degree of the function. The degree cannot be even by the definition of multiplicity.

4x y

x

15,

30

15 r 15 5 . 2

0

Dimensions with maximum volume: 20 in. u 20 in. u 40 in.

The value of

15 15 5 is not possible because it is 2

negative. 102. (a)

(b) V

15

9−

x

2x

15

x −2

V 125

Volume of box

2 x 9 2 x x

Because length, width, and height must be positive, you have 0 x 92 for the domain.

x

(c)

15

x9 2 x 15 2 x

x

9

l wh

100

(d) 56

x9 2 x 15 2 x

56

135 x 48 x 2 4 x3

0

75 50

4 x3 48 x 2 135 x 56

The zeros of this polynomial are 12 , 72 , and 8.

25 x 1

2

3

4

x cannot equal 8 because it is not in the domain of V.

5

Length of sides of squares removed

[The length cannot equal 1 and the width cannot equal 7. The product of 8 1 7 56 so it

The volume is maximum when x | 1.82. The dimensions are: length | 15 21.82 width | 9 21.82 height

x | 1.82

11.36 5.36

showed up as an extraneous solution.] So, the volume is 56 cubic centimeters when x centimeter or x

7 2

1 2

centimeters.

1.82 cm u 5.36 cm u 11.36 cm

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Section 2.5

103.

The Fundamental Theorem Theore of Algebra

235

76 x3 4830 x 2 320,000, 0 d x d 60

P

76 x3 4830 x 2 320,000

2,500,000 76 x 3 4830 x 2 2,820,000

0

The zeros of this equation are x | 46.1, x | 38.4, and x | 21.0. Because 0 d x d 60, we disregard x | 21.0. The smaller remaining solution is x | 38.4. The advertising expense is $384,000. P

45 x3 2500 x 2 275,000

800,000

45 x 2500 x 275,000

0

45 x 3 2500 x 2 1,075,000

0

9 x3 500 x 2 215,000

104.

3

105. (a) Current bin: V New bin: V

2

V x

2u3u4 24 cubic feet 5 24 120 cubic feet

2 x 3 x 4

(b) x3 9 x 2 26 x 24

x

120

120

The zeros of this equation are x | 18.0, x | 31.5,

x 9 x 26 x 96

and x | 42.0. Because 0 d x d 50, disregard x | 18.02. The smaller remaining solution is x | 31.5, or an advertising expense of $315,000.

The only real zero of this polynomial is x 2. All the dimensions should be increased by 2 feet, so the new bin will have dimensions of 4 feet by 5 feet by 6 feet.

106. (a) A x

250 x 160

x

1.5 160 250

(b) 60,000

x 2 410 x 40,000

0

x 2 410 x 20,000

x

410 r

4102 4 1 20,000 21

3

2

60,000

410 r

248,100 2

410

248,100 | 44.05. 2 The new length is 250 44.05 294.05 ft and the new width is 160 44.05 so the new dimensions are 294.05 ft u 204.05 ft .

x must be positive, so x

(c) A x

250

2 x 160 x

2 x 570 x 20,000

570 r

0

5702 4 2 20,000 2 2

x must be positive, so x

204.05 ft,

60,000

2

x

0

570 r

484,900 4

570

The new length is 250 231.6

484,900 | 31.6. 4 313.2 ft and the new width is 160 31.6

191.6 ft,

so the new dimensions are 313.2 ft u 191.6 ft. 107. C

x · § 200 100¨ 2 ¸, x t 1 x 30 ¹ © x

C is minimum when 3x3 40 x 2 2400 x 36000

108. (a)

0.

The only real zero is x | 40 or 4000 units.

12

0

7 8

(b) A | 0.01676t 4 0.2152t 3 0.794t 2 0.44t 8.7 (c) The model is a good fit to the actual data. (d) A t 10 when t

3, 4, and 7, which corresponds

to the years 2003, 2004, and 2007. (e) Yes. The degree of A is even and the leading coefficient is positive, so as t increases, A will increase. This implies that attendance will continue to ggrow.

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236

Chapter 2

109. ht

NOT FOR SALE

Polynomial ynomial and Rational Function Functions

119. Because 1 i is a zero of f, so is 1 i. From the graph, 1 is also a zero.

16t 2 48t 6

Let h

64 and solve for t. 64

16t 2 48t 58

16t 2 48t 6

48 r i 1408 . 32

Because the equation yields only imaginary zeros, it is not possible for the ball to have reached a height of 64 feet. R C

P

x 3x 2 4 x 2 120. Because 1 i is a zero of f, so is 1 i. From the graph, 1 is also a zero.

f x

x

0.0001x 60 x 150,000 2

0.0001x 2 60 x 9,150,000.

60 r 60 0.0002

x 1 i x 1 i x 1

x2

xp C

0.0001x 2 60 x 150,000

Thus, 0

2 x 2 x 1

3

x140 0.0001x 80 x 150,000

9,000,000

x 1 i x 1 i x 1

x2

0

By the Quadratic Formula, t

110.

f x

2 x 2 x 1

x x2 2 3

Because the graph rises to the left and falls to the right, a 1, and f x x3 x 2 2. 121. Because f i

f 2i

0, then i and 2i are zeros of f.

Because i and 2i are zeros of f, so are i and –2i. 300,000 r10,000 15i

f x

x

Because the solutions are both complex, it is not possible to determine a price p that would yield a profit of 9 million dollars. 111. False. The most complex zeros it can have is two, and the Linear Factorization Theorem guarantees that there are three linear factors, so one zero must be real.

x

i x i x 2i x 2i 2

1 x 2 4

x4 5x2 4 122. Because f 2

0, 2 is a zero of f. Because f i

f x

1 x 2 x i x i

112. False. f does not have real coefficients.

1 x 2 x 2 1

113. g x

x3 2 x 2 x 2

f x . This function would have the same

zeros as f x , so r1 , r2 , and r3 are also zeros of g x . 114. g x

3 f x . This function has the same zeros as f

because it is a vertical stretch of f. The zeros of g are r1 , r2 , and r3. 115. g x

f x 5 . The graph of g x is a horizontal

123. Answers will vary. Some of the factoring techniques are:

1. Factor out the greatest common factor. 2

Use special product formulas. a 2 b2

a

b a b

a 2 2ab b 2

a

b

2

shift of the graph of f x five units of the right, so the

a 2 2ab b 2

zeros of g x are 5 r1 , 5 r2 , and 5 r3 .

a 3 b3

a b a b a 2 ab

a 3 b3

a

116. g x

f 2 x . Note that x is a zero of g if and only if

2x is a zero of f. The zeros of g are 117. g x

r1 r2 r , , and 3 . 2 2 2

3 f x . Because g x is a vertical shift of

the graph of f x , the zeros of g x cannot be determined. 118. g x

f x . Note that x is a zero of g if and only if

x is a zero of f. The zeros of g are r1 , r2 , and r3 .

0,

i is a zero of f. Because i is a zero of f, so is i .

2

b2

b a 2 ab b 2

3. Factor by grouping, if possible. 4. Factor general trinomials with binomial factors by “guess-and-test” or by the grouping method. 5. Use the Rational Zero Test together with synthetic division to factor a polynomial. 6. Use Descartes’s Rule of Signs to determine the number of real zeros. Then find any zeros and use them to factor the polynomial. 7. Find any upper and lower bounds for the real zeros to eliminate some of the possible rational zeros. Then test the remaining candidates by synthetic division and use any zeros to factor the polynomial.

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NOT FOR SALE Section 2.6

124. f x

x

Ration Rational Functions

237

x4 4x2 k

2

4 r

4 2 21

x

r

4 k

2r

41 k

4r 2 4 k 2

2r

4 k

(a) For there to be four distinct real roots, both 4 k

(d) For there to be four complex zeros, 2 r 4 k must be nonreal. This occurs when k ! 4. Some 5, k 6, k 7.4, etc. possible k-values are k

and 2 r 4 k must be positive. This occurs when 0 k 4. So, some possible k-values are 1 k 1, k 2, k 3, k ,k 2, etc. 2

(e) g x

No. This function is a horizontal shift of f x . Note that x is a zero of g if and only if x 2 is a zero of f; the number of real and complex zeros is not affected by a horizontal shift.

(b) For there to be two real roots, each of multiplicity 2, 4 k must equal zero. So, k 4. (c) For there to be two real zeros and two complex zeros, 2 4 k must be positive and

(f ) g x

2 4 k must be negative. This occurs when k 0. So, some possible k-values are 1 k 1, k 2, k , etc. 2

125. (a) f x

(b) f x

x

bi x

bi

126. (a) f x cannot have this graph because it also has a

x2 b

zero at x

a bi

function. Its graph is a parabola. (c) h x is the correct function. It has two real zeros,

2

x 2ax a b 2

2

0.

(b) g x cannot have this graph because it is a quadratic

ª¬ x a biºª ¼¬ x a biº¼

x

f 2 x

No. Because x is a zero of g if and only if 2x is a zero of f, the number of real and complex zeros of g is the same as the number of real and complex zeros of f.

ª¬ x a bi ºª ¼¬ x a bi º¼ 2

f x 2

2

x 2 and x 3.5, and it has a degree of four, needed to yield three turning points. (d) k x cannot have this graph because it also has a zero at x 1. In addition, because it is only of degree three, it would have at most two turning points.

Section 2.6 Rational Functions 1. rational functions

3. horizontal asymptote

2. vertical asymptote

4. slant asymptote

1 x 1

5. f x

(a)

x

f x

x

f x

x

f x

0.5

–2

1.5

2

5

0.25

0.9

–10

1.1

10

10

0. 1

0.99

–100

1.01

100

100

0.01

0.999

–1000

1.001

1000

1000

0.001

(b) The zero of the denominator is x 1, so x 1 is a vertical asymptote. The degree of the numerator is less than the degree of the denominator, so the x-axis, or y 0, is a horizontal asymptote. (c) The domain is all real numbers x except x

1.

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238

NOT FOR SALE

Chapter 2 5x x 1

6. f x

(a)

Polynomial ynomial and Rational Function Functions

f x

x

f x

x

x

(b) The zero of the denominator is x 1, so x 1 is a vertical asymptote. The degree of the numerator is equal to the degree of 5 5 the denominator, so the line y 1

f x

0.5

–5

1.5

15

5

6.25

0.9

–45

1.1

55

10

5.5

0.99

–495

1.01

505

100

5.05

is a horizontal asymptote. (c) The domain is all real numbers x except x

0.999

9. f x

f x

x

f x

x

5.005

x

f x

0.5

–1

1.5

5.4

5

3.125

0.9

–12.79

1.1

17.29

10

3.03

0.99

–147.8

1.01

152.3

100

3.0003

0.999

–1498

1.001

1502

1000

3

1.

r1, (b) The zeros of the denominator are x so both x 1 and x 1 are vertical asymptotes. The degree of the numerator equals the degree of the denominator, so 3 y 3 is a horizontal asymptote. 1 (c) The domain is all real numbers x except x r1.

4x x2 1

f x

x

f x

4.8

5

0.83

1.1

20.95

10

0.40

–199

1.01

201

100

0.04

–1999

1.001

2001

1000

0.004

x

f x

x

0.5

2.6

1.5

0.9

–18.95

0.99 0.999 4 x2

Vertical asymptote: x

0

(c) The domain is all real numbers x except x r1.

0

Domain: all real numbers x except x

ª¬Degree of N x

12. f x

4 3

Domain: all real numbers x except x Vertical asymptote: x Horizontal asymptote: y

x 5 x 5

2

3 7x 3 2x

1

degree of D x º¼ 7 x 3 2x 3

Domain: all real numbers x except x

2 Vertical asymptote: x

0

ª¬Degree of N x degree of D x º¼

Horizontal asymptote: y ª¬Degree of N x

5

5

Horizontal asymptote: y

ª¬Degree of N x degree of D x º¼ 2

5 x 5 x

Vertical asymptote: x

0

Horizontal asymptote: y

x

(b) The zeros of the denominator are x r1, so both x 1 and x 1 are vertical asymptotes. The degree of the numerator is less than the degree of the denominator, so the x-axis, or y 0, is a horizontal asymptote.

11. f x

Domain: all real numbers x except x

10. f x

1000

2

8. f x

(a)

5005

3x 2 x 1

7. f x

(a)

1.001

–4995

3 2

3 2

7 2

degree of D x º¼

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NOT FOR SALE Section 2.6

13. f x

x3 x 1

Domain: all real numbers x except x Vertical asymptotes: x

r1

Vertical asymptote: x

r1

20. f x

4 x2 x 2

Domain: all real numbers x except x

degree of D x º¼

x2 9 x 3

21. g x

x

3 x 3 x 3

The only zero of g x is x

3. x

3 makes g x

undefined.

Domain: All real numbers x. The denominator has no real zeros. [Try the Quadratic Formula on the denominator.]

10 x2 5 10 4 2 x 5 10 x2 5 10

h x

22.

Vertical asymptote: None

4

0

Horizontal asymptote: y

16. f x

1

Matches graph (b).

2

3x 2 1 2 x x 9

ª¬Degree of N x

4

Horizontal asymptote: y

2

Horizontal asymptote: None

15. f x

x 2 x 4

Vertical asymptote: x

Vertical asymptote: x

1

Matches graph (c).

ª¬Degree of N x ! degree of D x º¼

ª¬Degree of N x

4

Horizontal asymptote: y

Horizontal asymptote: None

14. f x

239

x 1 x 4

19. f x

2

Ration Rational Functions

3 4

degree of D x º¼

4 x 2 5

3x x 5 x2 1 2

4 x 2

30 15 2

x2

Domain: All real numbers x. The denominator has no real zeros. [Try the Quadratic Formula on the denominator.]

No real solution, h x has no real zeros.

Vertical asymptote: None Horizontal asymptote: y ¬ªDegree of N x

17. f x

degree of D x º¼

0

4 x 5

Vertical asymptote: x

2 x 7 x 7 x

5

Horizontal asymptote: y

0

x

Matches graph (d). 18. f x

5 x 2

Vertical asymptote: x Horizontal asymptote: y Matches graph (a).

f x

23.

3

2 0

x3 8 x2 1 x3 8

2 9

x3 8 x2 1 0 0

3

8

x

2

x

x

1

9 is a zero of f x . g x

24.

2 x 7 2 1 x 7

1

2 is a real zero of g x .

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240

Chapter 2

25. f x

NOT FOR SALE

Polynomial ynomial and Rational Function Functions

x 4 x 2 16

x 4 x 4 x 4

1 ,x z 4 x 4 r4

Domain: all real numbers x except x 4

Vertical asymptote: x

common factor of N x and

Because x 4 is a D x , x 4 is not a

Domain: all real numbers x except x

x 1 x2 1

x 1 x 1 x 1

1 , x z 1 x 1

asymptote of f x . )

vertical asymptote of f x . )

¬ªDegree of N x

x 2 25 x 4x 5

0

2 x

x 5 x 5 x 5 x 1

Domain: all real numbers x except x Vertical asymptote: x

x5 ,x z 5 x 1

5 and x

1

1 Because x 5 is a

common factor of N x and D x , x

5 is not a

vertical asymptote of f x . Horizontal asymptote: y ª¬Degree of N x

28. f x

degree of D x º¼

x2 4 2 x 3x 2 x 2 x 2

x

1

2 x 1

x 2 ,x z 2 x 1

asymptote of f x . )

3x 1 3 ,x z 3x 1 2

3 3x 1

Domain: all real numbers x except 3 1 x or x 2 3 1 (Because 2 x 3 is a 3 3 common factor of N x and D x , x is not a 2 vertical asymptote of f x . )

Vertical asymptote: x

Horizontal asymptote: y ª¬Degree of N x

Domain: all real numbers x except x 1 and x 2 Vertical asymptote: x 1 (Because x 2 is a common 2 is not a vertical factor of N x and D x , x

1

degree of D x º¼

1 x 2

31. f x

(a) Domain: all real numbers x except x § (b) y-intercept: ¨ 0, ©

2

Horizontal asymptote: y

1

degree of D x º¼

(d)

x

–4

y

1 2

2

1· ¸ 2¹

(c) Vertical asymptote: x

Horizontal asymptote: y ª¬Degree of N x

degree of D x º¼

6 x 2 11x 3 6x2 7 x 3 2 x 3 3x 1

30. f x

2

1 2

Horizontal asymptote: y

ª¬Degree of N x degree of D x º¼

27. f x

1

1 (Because x 1 is a 2 1 is not a common factor of N x and D x , x

Domain: all real numbers x except x r1 Vertical asymptote: x 1 (Because x 1 is a common factor of N x and D x , x 1 is not a vertical Horizontal asymptote: y

1 and x 2

Vertical asymptote: x

0

ª¬Degree of N x degree of D x º¼

26. f x

x 4 , x z 1 2x 1

2 x 1 x 1

vertical asymptote of f x . Horizontal asymptote: y

x 2 3x 4 2x2 x 1 x 1 x 4

29. f x

3 –1

0

1

0

1

1

1 2

1 3

y 2

(0, 12 ) –3

1 x

–1 –1

INSTRUCTOR USE S ONLY –2

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© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.6

(d)

1 x 3

32. f x

(a) Domain: all real numbers x except x

3

x

4

5

7

8

y

1 2

1

–1

1· § (b) y-intercept: ¨ 0, ¸ 3¹ ©

0

1

y

1 3

1 2

1 2

6

3

4

Horizontal asymptote: y x

241

y

(c) Vertical asymptote: x

(d)

Ration Rational Functions

2

0 2 –1

4

5

6

1

1 2

1 3

2

4

10

−4 −6

7 2x 2 x

3 2

x

−2 −2

35. C x

y

(0, 16 )

2x 7 x 2

2

(a) Domain: all real numbers x except x

1

§ 7 · (b) x-intercept: ¨ , 0 ¸ © 2 ¹

x 2

4

5

6

–1 –2

(0, − 13 )

§ 7· y-intercept: ¨ 0, ¸ © 2¹

–3

1 x 4

33. h x

2

(c) Vertical asymptote: x Horizontal asymptote: y

(a) Domain: all real numbers x except x

4

(d)

1· § (b) y-intercept: ¨ 0, ¸ 4¹ ©

x

–3

–1

1

3

y

–1

5

3

13 5

4

(c) Vertical asymptote: x

2

y

Horizontal asymptote: y (d)

0

6 5

x

–5

–3

–1

1

)0, ) 7 2

3

y

1

–1

1 3

1 5

1 x

−6 −5 −4 7 − 2, 0

)

y

−1

)

1

2

−2

4 3 1 −7 −6 −5

x −1 −2

)0, − ) 1 4

y-intercept: 0, 1

1 x 6

(a) Domain: all real numbers x except x § (b) y-intercept: ¨ 0, ©

1

§1 · (b) x-intercept: ¨ , 0 ¸ ©3 ¹

−4

1 6 x

3x 1 x 1

(a) Domain: all real numbers x except x

−3

34. g x

1 3x 1 x

36. P x

2

Horizontal asymptote: y

(c) Vertical asymptote: x Horizontal asymptote: y

1· ¸ 6¹

(c) Vertical asymptote: x

6

(d) 6

1 3 y

x

–1

0

2

3

y

2

1

5

4

6 5 4

0

(0, 1)

( 13 , 0)

INSTRUCTOR USE ONLY N x

–22

––1

2

3

4

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242

NOT FOR SALE

Chapter 2

Polynomial ynomial and Rational Function Functions

x2 x 9

37. f x

40. f x

2

(a) Domain: all real numbers x

x

±1

±2

±3

y

1 10

4 13

1 2

1

–2

–1

1

0

y

2

2t 1 t

5 2

3

7 2

–1

–4

–4

–1

4 4 9

1 4

1

3

–3

2

–4

1 2

1

2

–3

0

–1

x2 5x 4 x2 4

41. h x 3 2

x 1 x 4 x 2 x 2

(a) Domain: all real numbers x except x

r2

(b) x-intercepts: 1, 0 , 4, 0

( 12 , 0)

y-intercept: 0, 1

t 1

4 9

5 2

x

0

0

–1

–1

3 2

–2

y

–2

1 4

1

–1

Horizontal asymptote: y

y

0

(0, − 14 )

§1 · (b) t-intercept: ¨ , 0 ¸ ©2 ¹ (c) Vertical asymptote: t

–2

2

y

(a) Domain: all real numbers t except t

t

1 2

x

–1

(d)

2

(d) x

2

Horizontal asymptote: y

2

(0, 0)

1 2t t

2

(c) Vertical asymptote: x

3

38. f t

x

1· § (b) y-intercept: ¨ 0, ¸ 4¹ © y

(d)

1

(a) Domain: all real numbers x except x

(b) Intercept: 0, 0 (c) Horizontal asymptote: y

2

–1

(c) Vertical asymptotes: x Horizontal asymptote: y

2, x

2

1

–3

(d)

4s s2 4

39. g s

x

–4

3

1

y

10 3

28 5

10 3

–1

1

3

4

0

2 5

0

y

(a) Domain: all real numbers s

6

(b) Intercept: 0, 0

4 2

(c) Vertical asymptote: none Horizontal asymptote: y (d)

0

s

–2

–1

y

–1

4 5

−6 −4

0

(1, 0) x

(4, 0) 6

(0, −1)

0

1

2

0

4 5

1

y 4 3 2 1 −2

−1

(0, 0) 2 3 4

s

INSTRUCTOR T USE ONLY −2

−3 3 −4

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.6

x 4 x x 3 x

x2 2 x 8 x2 9

42. g x

2

r3

(a) Domain: all real numbers x except x

and x

x

–5

4

y

27 16

16 7 y

−6 −4

Horizontal asymptote: y

2

0

2

0

8 9

8 5

4

5

0

7 16

(d)

y

4

6

(0, ) 2 x

x

2, x

1

0

5 9

0

5 4

−2 −3 −4 −5

x x 3 x 3 x 2

x 2 3x x x6

45. f x

2

1,

3 2

x , x z 3 x2

3 and

2

Horizontal asymptote: y 1, x

2, x

Horizontal asymptote: y 2

4 5

(c) Vertical asymptote: x

(c) Vertical asymptotes: x

3 4

1 3

4

(b) Intercept: 0, 0

3· § y-intercept: ¨ 0, ¸ 2¹ ©

(a) Domain: all real numbers x except x x 2

§ 1 · (b) x-intercept: ¨ , 0 ¸, 3, 0 © 2 ¹

y

0

2

(2, 0)

1 x 3

2 x 1 x 1

(a) Domain: all real numbers x except x

–3

0

x 2 1 −3

2x2 5x 3 x 2x2 x 2

x

0

–1

5 12

3

3 2 1

(4, 0)

2

and x

1, x

y 4

2

3

9 35

(−1, 0)

−6

(d)

–4

4

−4

43. f x

x

6

−2

2, x

(c) Vertical asymptotes: x

1

x

(− 2, 0)

2,

3

(0, 89)

2

1, x

1· § y-intercept: ¨ 0, ¸ 3¹ ©

3

3, x

Horizontal asymptote: y (d)

1 x 2

(b) x-intercepts: 1, 0 , 2, 0

8· ¸ 9¹

(c) Vertical asymptotes: x

x

243

x 1 x 2 x 3

3

(a) Domain: all real numbers x except x

(b) x-intercepts: 4, 0 , 2, 0 § y-intercept: ¨ 0, ©

x2 x 2 x 2 x2 5x 6

44. f x

3

Ration Rational Functions

1

(d)

x

–1

0

1

3

4

y

1 3

0

–1

3

2

0 1.5

3

4

48 5

0

3 10

1

y 6 4

y 2

(

(

−4 −3

(0, 0)

6 3

x

−6 −4 −2

9

− 1, 0 2

(3, 0) 3

4

4

6

−4 x

−6

(0, − 32(

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244

Chapter 2

5 x 4

46. f x

NOT FOR SALE

Polynomial ynomial and Rational Function Functions 5 x 4

x 4 x 3

x x 12 2

(a) Domain: all real numbers x except x x 3

5 , x z 4 x3

4 and

5· § (b) y-intercept: ¨ 0, ¸ 3¹ ©

(c) Vertical asymptote: x

(d)

x y

–2

0

–1

5 3

3x 2 ,x z 2 2x 1

x 2 2 x 1

(a) Domain: all real numbers x except x 1 x 2

3

Horizontal asymptote: y

3x 2 8 x 4 2 x 2 3x 2 x 2 3x 2

48. f x

2 and

§2 · (b) x-intercept: ¨ , 0 ¸ ©3 ¹

0

2

5

7

–5

5 2

5 4

y-intercept: 0, – 2

(c) Vertical asymptote: x

1 2

y

3 2

Horizontal asymptote: y

6 4

(d)

2 x

(

2

)

4

6

x

–3

–1

0

2 3

3

y

11 5

5

–2

0

1

8

5 0, − 3 −4 −6

y

2 x2 5x 2 2x2 x 6 2 x 1 x 2

47. f x

2x 1 ,x z 2 2x 3

2 x 3 x 2

(a) Domain: all real numbers x except x 3 x 2 §1 · (b) x-intercept: ¨ , 0 ¸ ©2 ¹

x −4 −3 − 2 − 1

( 23 , 0) 3

4

(0, − 2)

2 and

t

t2 1 t 1

49. f t

1 t 1 t 1

t 1; t z 1

(a) Domain: all real numbers t except t

1· § y-intercept: ¨ 0, ¸ 3¹ ©

1

(b) t-intercept: 1, 0 y-intercept: 0, 1

3 2

(c) Vertical asymptote: x Horizontal asymptote: y (d)

1

(c) Vertical asymptote: none Horizontal asymptote: none

1 (d)

x

–3

–2

–1

0

y

7 3

5

–3

1 1 3

1 5

t

–1

0

2

3

y

0

1

3

4

y 4

y

3

4

2

3

(− 1, 0)

2

−4 −3

1 −5 −4 −3 − 2 1 0, − 3

)

)

)12 , 0)

x 3

1

−1

(0, 1) t 1

2

3

4

−2 −3 −4

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.6

x

x 2 36 x 6

50. f x

6 x 6 x 6

(a) Domain: all real numbers x except x

x 6; x z 6

6

y-intercept: 0, 6

(b) f x

(c) Vertical asymptote: none Horizontal asymptote: none x

1

2

3

4

y

–5

–4

–3

–2 (c)

x

8 10

−4 −6

x 2 x 2

x

x x 2

x 2x 2

2

0

1

1.5

2

2.5

3

f x

–1

Undef.

1

1.5

Undef.

2.5

3

g x

–1

0

1

1.5

2

2.5

3

x 6

x 2 x 2

0 and

–1

(6, 0) 2

x

and the denominator of f, neither x 0 nor x is a vertial asymptote of f. So, f has no vertical asymptotes.

y

−6 −4 −2

, g x

245

Because x x 2 is a factor of both the numerator

4 2

x2 2x

(a) Domain of f: All real numbers x except x x 2 Domain of g: All real numbers x

(b) x-intercept: 6, 0

(d)

x 2 x 2

52. f x

Ration Rational Functions

(0, − 6)

(d)

−10

2

−12 −2

51. f x

x2 1 , g x x 1

x 1 −2

(a) Domain of f: all real numbers x except x

1

(e) Because there are only a finite number of pixels, the utility may not attempt to evaluate the function where it does not exist.

Domain of g: all real numbers x (b) f x

x2 1 x 1 x 1 x 1

Because x 1 is a factor of both the numerator and 1 is not a vertical the denominator of f , x asymptote. So, f has no vertical asymptotes.

–3

–2

–1.5

–1

–0.5

0

1

f x

–4

–3

–2.5

Undef.

–1.5

–1

0

g x

–4

(b) f x

(d)

–2

–1.5

–1

x 2 x x 2

0

1 x

asymptote. The only vertical asymptote of f is x

0

1 −4

x 2 x2 2 x

0 and

Because x 2 is a factor of both the numerator and 2 is not a vertical the denominator of f , x (c)

–2.5

1 x

(a) Domain of f: All real numbers x except x x 2 Domain of g: All real numbers x except x

(c)

–3

x 2 , g x x2 2x

53. f x

x 1 x 1

x

4

2

x

–0.5 0

0.5 1 1.5 2

f x

–2

Undef. 2

1

g x

–2

Undef. 2

1

(d)

2 3 2 3

Undef. 1 2

0. 3 1 3 1 3

2

−3

(e) Because there are only a finite number of pixels, the utility may not attempt to evaluate the function where it does not exist.

−3

3

−2

(e) Because there are only a finite number of pixels, the utility may not attempt to evaluate the function where here it does not exist. e

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246

NOT FOR SALE

Chapter 2

Polynomial ynomial and Rational Function Functions

2x 6 , g x x 7 x 12

54. f x

2 x 4

2

(a) Domain of f: All real numbers x except x x 4 Domain of g: All real numbers x except x (b) f x

3 and

(c) Vertical asymptote: x (d)

3 x 4

f x g x (d)

0

1

1 2 1 2

2 3 2 3

2

–2

–1

1

2

y

5 2

8

–8

y=x 4

(3, 0) 4

6

x 8

−4

5 6

−6 −8

–1 Undef. Undef. 2 1

x2 5 5 x x x (a) Domain: all real numbers x except x

56. g x

–1 –2

Undef. 2 1

0

(b) No intercepts

3

−1

2

(− 3, 0) −8 − 6

4

5 2

y

4 as its only vertical

3

x

x

Because x 3 is a factor of both the numerator and 3 is not a vertical the denominator of f , x

x

0

Slant asymptote: y

asymptote of f. So, f has x asymptote.

0

(b) x-intercepts: 3, 0 , 3, 0 4

2 x 4

(c)

9 x

x

(a) Domain: all real numbers x except x

2x 6 x 2 7 x 12 2 x 3

x

x2 9 x

55. h x

(c) Vertical asymptote: x 0 Slant asymptote: y x

8

(d)

x

–3

y

−3

(e) Because there are only a finite number of pixels, the utility may not attempt to evaluate the function where it does not exist.

–2

14 3

9 2

–1

1

2

3

–6

6

9 2

14 3

y 6 4

y=x

2 −6 −4 −2 −2

x 2

4

6

−4

2 x2 1 1 2x x x (a) Domain: all real numbers x except x

57. f x

0

(d)

x

–4

y

(b) No intercepts (c) Vertical asymptote: x Slant asymptote: y

0

–2

33 4

9 2

2

4

6

9 2

33 4

73 6

2x y 6 4 2

y = 2x x

–6

–4

–2

2

4

6

INSTRUCTOR USE ONLY –6

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.6

1 x2 1 x x x (a) Domain: all real numbers x except x

x2 1 1 x x x (a) Domain: all real numbers x except x

247

59. g x

58. f x

0

(b) x-intercepts: 1, 0 , 1, 0

0

(b) No intercepts (c) Vertical asymptote: x 0 Slant asymptote: y x

(c) Vertical asymptote: x 0 Slant asymptote: y x (d)

Ration Rational Functions

(d)

x

–6

4

2

2

y

35 6

15 4

3 2

4 3 2

6 15 4

35 6

x

–4

y

2

17 4

2

4

6

5 2

17 4

37 6

5 2

y y

y = −x

6

8

4

6 4

y=x

2

2

(−1, 0)

(1, 0)

–8 –6 –4 –2

4

x x 6

–6

–4

–2

2

4

6

8

–4 –6

–6

–8

x2 x 1

60. h x

1 x 1

x 1

(a) Domain: all real numbers x except x

1

(b) Intercept: 0, 0 (c) Vertical asymptote: x

1

Slant asymptote: y

x 1

(d)

y

x

–4

y

2

16 5

4 3

2

4

6

4

16 3

36 5

8 6 4

y=x+1

2

(0, 0)

x

–4

2

4

6

8

–2 –4

61. f t

t2 1 t 5

t 5

26 t 5

(a) Domain: all real numbers t except t 1· § (b) y-intercept: ¨ 0, ¸ 5¹ ©

5

1 1 1 x 3 9 93 x 1

(a) Domain: all real numbers x except x (b) Intercept: 0, 0

5

(c) Vertical asymptote: t

t 5

Slant asymptote: y (d)

x2 3x 1

62. f x

t

–7

–6

–4

–3

0

y

25

37

–17

–5

y

1 5

1 3

(c) Vertical asymptote: x

Slant asymptote: y

1 1 x 3 9

(d) x

–3

y

25

1

–2

9 8

4 5

1 2

1 3

1 2 1 2

0

2

0

4 7

20

y=5−t

y

15

(0, − 15( −20 −15 − 10 − 5

5

1

t

1

1

y = 3x − 9

2 3

(0, 0) x

−1

1 3

2 3

1

4 3

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248

NOT FOR SALE

Chapter 2

Polynomial ynomial and Rational Function Functions

x3 4x x 2 x 4 x 4 (a) Domain: all real numbers x except x

63. f x

r2

x

1 x 1

(a) Domain: all real numbers x except x

(b) Intercept: 0, 0

1

(b) y-intercept: 0, 1 2

2 and x

(c) Vertical asymptotes: x Slant asymptote: y x

(d)

x2 x 1 x 1

65. f x

2

(c) Vertical asymptote: x 1 Slant asymptote: y x (d)

x

–3

–1

1

3

y

27 5

1 3

1 3

27 5

x

–4

–2

0

2

4

y

21 5

7 3

–1

3

13 3

y

y

8

8

6 4

6

y=x

4

(0, 0) −8 − 6 − 4

4

x 6

y=x

2

(0, −1)

8 –4

x

–2

2

4

6

8

–4

x3 2x 8

64. g x

1 4x x 2 2x2 8

2

2 x2 5x 5 3 2x 1 x 2 x 2 2 (a) Domain: all real numbers x except x

66. f x

(a) Domain: all real numbers x except x

r2

(b) Intercept: 0, 0 r2

(c) Vertical asymptote: x

1 x 2

Slant asymptote: y (d)

x y

–6

–4

–1

27 8

8 3

1 6

5· § (b) y-intercept: ¨ 0, ¸ 2¹ © (c) Vertical asymptote: x 2 Slant asymptote: y 2x 1

(d) 1

4

6

1 6

8 3

27 8

x

–6

y

–3

107 8

38 5

1

3

6

7

–2

8

47 4

68 5

y y 15 8

12

6

9

4

6

(0, 0)

y = 2x − 1

3 x

–8 –6 –4

4

6

x

8

–9 –6 –3

y = 12 x

3

6

9 12 15

(0, − ) 5 2

–9

67. f x

2 x3 x 2 2 x 1 x 2 3x 2 2 x 1 x 1 x 1

x

2 x

1 x 2

1 x 1

, x 2 2 x 2 3x 1 x 2 15 2x 7 , x 2

x z 1

x z 1 1 and x

2

INSTRUCTOR USE ONLY (a) Domain: all real numbers x except x

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.6

§ 1· (b) y-intercept: ¨ 0, ¸ © 2¹

(d)

§1 · x-intercepts: ¨ , 0 ¸, 1, 0 ©2 ¹ (c) Vertical asymptote: x 2

x

–4

y

x 2 x 1 x 2 2 x 1 , x z x 1

1 2

0

18 12

−5 −4 −3

x

−1

3

) 12 , 0)

−12 −18 −24

x2 5x 8 x 3

y = 2x − 7

x 2

2 x 3

Line: y

3

3

8

x 2 −14

x 2

10

−8

72. f x

1 and

2 x2 x x 1

2x 1

(c) Vertical asymptote: x

1

Slant asymptote: y

2x 7

1 x 1

Domain: all real numbers x except x Vertical asymptote: x

(d)

)0, 12 )

(1, 0)

Slant asymptote: y

2

Slant asymptote: y

§ 1 · x-intercepts: 2, 0 , ¨ , 0 ¸ © 2 ¹

5 4

1

Vertical asymptote: x

(b) y-intercept: 0, 2

y

0

Domain: all real numbers x except x

(a) Domain: all real numbers x except x 2 x

y

20

71. f x

2x2 5x 2 x 1 9 2x 7 ,x z 2 x 1

–3

–28

249

−30 −36

2 x3 x 2 8 x 4 x 2 3x 2 x 2 x 2 2 x 1

x

2x 7

Slant asymptote: y

68. f x

45 2

3 2

–3

Ration Rational Functions

Line: y

1

1

6

2x 1

2x 1

−12

12

−10

–2

–1

0

1 2

0

1 2

–2

–10

3 2

3

4

28

35 2

18

73. g x

1 3x 2 x3 x2

Domain: all real numbers x except x Vertical asymptote: x Slant asymptote: y

y 30

1 3 x x2

Line: y

x 3

1 x2

0

0

12

x 3

x 3

24

−12

12

18

y = 2x + 7 12 (−2, 0) −6

−2

−4

(

1 − ,0 2

2

)

4

x

74. h x

6

(0, −2)

Vertical asymptote: x

1 x 2 2

Slant asymptote: y

Domain: All real numbers except x One possibility: f x

1 2 x 1 2 4 x

Domain: all real numbers x except x

69. Domain: All real numbers

One possibility: f x

12 2 x x 2 2 4 x

2.

1 x 2

Line: y

1 x 1 2

4

4 10

1 x 1 2 −16

8

−6

(Answers are not unique). 70. An asymptote is a line to which a graph gets arbitrarily close to, but does not reach, as x or y increases

INSTRUCTOR USE ONLY without bound.

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© Cengage Learning. All Rights Reserved.

250

NOT FOR SALE

Chapter 2

Polynomial ynomial and Rational Function Functions

x 1 x 3

75. y

(a) x-intercept: 1, 0 (b)

(a)

0 1

0

x

(b) C 15

(a) x-intercept: 0, 0

0

2x x 3 2x

0

x

C 50 C 90

x x2 x

x 3

78. y

(a)

$25,000

100 50 25,00090

$225,000

100 90

1 0.04t

,t t 0

N 10

500 deer

N 25

800 deer

82. (a) 0.2550 0.75 x

2 x

0

2 x x 2 3x 2

0

x

x

1, x

x 3

C 50 x

12.50 0.75 x 4 50 x 4 50 3 x 3x 50 450 x 4 x 50

C C

(b) Domain: x t 0 and x d 1000 50

1 x 2

So, 0 d x d 950. Using interval notation, the domain is >0, 950@.

2

255 p , 0 d p 100 100 p

79. C

25,00050

(b) The herd is limited by the horizontal asymptote: 60 N 1500 deer 0.04

(a) x-intercepts: 1, 0 , 2, 0 (b) 0

| $4411.76

(a) N 5 | 333 deer

1 x x 1 x 1 r1

0

100 15

205 3t

81. N

(a) x-intercepts: 1, 0 , 1, 0 (b)

25,00015

(c) C o f as x o 100. No. The function is undefined for p 100.

1 x x

77. y

100 0

2x x 3

(b) 0

300,000

x 1 x 3 x 1

0

76. y

25,000 p , 0 d p 100 100 p

80. C

(c)

C

1.0 0.8

2,000

0.6 0.4 0.2 0

x

100

200 400 600 800 1000

0

(b) C 10 C 40 C 75

25510 100 10 255 40 100 40 25575 100 75

| 28.33 million dollars 170 million dollars

(d) As the tank is filled, the concentration increases more slowly. It approaches the horizontal asymptote 3 of C 0.75 75%. 4

765 million dollars

(c) C o f as x o 100. No. The function is undefined at p 100.

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE Section 2.6

84. (a) A

83. (a) Let t1 time from Akron to Columbus and t2 time from Columbus back to Akron.

xt1

100 t1

yt2

100 t2

50t1 t2

4

100 100 x y

4

100 y 100 x 25 y 25 x

4 y 2

30 x 4 2

y

2 x x 11

§ 2 x 22 · x¨ ¸ © x 4 ¹

xy

2 x 22 x 4

30 x 4

x 4

.

4 xy

(b) Domain: Because the margins on the left and right are each 2 inches, x ! 4. In interval notation, the domain is 4, f .

xy

(c)

25 x

xy 25 y

25 x

y x 25

200

4

40 4 0

25

Horizontal asymptote: y

x 25

5

6

7

8

9

10

y1 Area

160

102

84

76

72

70

x

11

12

13

14

15

69.143

69

69.333

70

70.999

200

y1 Area 25

65 0

(d)

30

y 2

So, A

(b) Vertical asymptote: x

(c)

x

100 x 100 y

25 x . x 25

So, y

251

xy and

200

t1 t2

Ration Rational Functions

x

30

35

40

45

50

55

60

y

150

87.5

66.7

56.3

50

45.8

42.9

(e) Sample answer: No. You might expect the average speed for the round trip to be the average of the average speeds for the two parts of the trip. (f ) No. At 20 miles per hour you would use more time in one direction than is required for the round trip at an average speed of 50 miles per hour.

The area is minimum when x | 11.75 inches and y | 5.87 inches. 85. False. Polynomial functions do not have vertical asymptotes. x crosses y x2 1 which is a horizontal asymptote.

86. False. The graph of f x

0,

87. False. A graph can have a vertical asymptote and a horizontal asymptote or a vertical asymptote and a slant asymptote, but a graph cannot have both a horizontal asymptote and a slant asymptote.

A horizontal asymptote occurs when the degree of N x is equal to the degree of D x or when the degree of N x is less than the degree of D x . A slant asymptote occurs when the degree of N x is greater than the degree of D x by one. Because the degree of a polynomial is constant, it is impossible to have both relationships at the same time. 88. (a) f x

x 1 x3 8

(b) f x

x 2 x3 1

(c) f x (d) f x

x

2 x 2 9

2 x 1

2 x 2 x 3

x

1 x 2

89. No; Yes;

2 x 2 18 x2 x 2

Not every rational function is a polynomial because 1 3 and h x are rational functions, but g x x 2 x they are not polynomials. Every polynomial f x is a rational function because it can be written as

f x 1

.

2 x 2 2 x 12 x2 x 2

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252

NOT FOR SALE

Chapter 2

Polynomial ynomial and Rational Function Functions

Review Exercises for Chapter 2 1. (a) y

2x2

2. (a) y

Vertical stretch

x2 4

Vertical shift four units downward y

y 4

3

3

2 1

2

x –4 –3

x

−4 −3 −2 −1 −1

1

2

3

–1

4

1

3

4

–2

−2 −3 −4

(b) y

–5

2 x 2

(b) y

Vertical stretch and a reflection in the x-axis y

4 x2

Reflection in the x-axis and a vertical shift four units upward y

4 3

5

2 1

3 x

−4 −3 −2 −1

1

2

3

2

4

1 x −4 −3

−3

−1 −1

−4

(c) y

1

3

4

−2 −3

x2 2

Vertical shift two units upward

(c) y

x

3

2

Horizontal shift three units to the right

y

y

4 3

5 4

1

3

x

−4 −3 −2 −1 −1

1

2

3

4

2

−2

1

−3

x

−4

(d) y

x

–2 –1 –1

2

1

2

3

4

5

–2

2

Horizontal shift two units to the left

(d) y

1 x2 2

1

Vertical shrink each y -value is multiplied by

y 4

1 2

,

and a vertical shift one unit downward y

1 −4 −3 −2 −1 −1

4

x 1

2

3

3

4

2

−2

1

−3 −4

x −4 −3 −2

2

3

4

−2 −3 −4

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NOT FOR SALE

Review Exercises ffor Chapter 2

3. g x

6. h x

x2 2 x

3 4x x2

x2 2 x 1 1

x 2 4 x 3

x

x 2 4 x 4 4 3

1 1 2

Vertex: 1, 1

2 ª x 2 7º ¬ ¼

Axis of symmetry: x

1

x 2 7

8 6 4

Axis of symmetry: x

y 7 6 5 4 3

0

3 4x x

0

x2 4x 3

4r

−2

2

28

x-intercepts: 2 r

x 6 x 9 9

x 2

4

6

8

10

41 3

2r

2 6x x2

2 −2

4 21

x 2 3 4 5 6

2

2

4 r

x

4. f x

10

Vertex: 2, 7

x-intercepts: 0, 0 , 2, 0

−3 −2 −1

y

2

x x 2

x2 2x

0

253

7

7, 0

2

7. f t

x 3 9 2

2t 2 4t 1 2t 2 2t 1 1 1

Vertex: 3, 9 Axis of symmetry: x

2t 1 3 2

x6 x

6x x2

0

2 2 ªt 1 1º 1 ¬ ¼

3

Vertex: 1, 3

x-intercepts: 0, 0 , 6, 0

Axis of symmetry: t

y 10

0

8

2t 1

6

2

1

2t 1 3 2

y

3

6 5 4 3

4

t 1

2 −2

3 2

r

x −2

4

2

8

t 5. f x

x 2 8 x 10 4 6

8. f x

4

2

x 4 x

t −3 −2 −1

1

2 3 4 5

6

6 · , 0 ¸¸ 2 ¹

x

x 2 8 x 16 16 12

4 6

4 4 2

y

Vertex: 4, 4

8 6

2

r 6

x-intercepts: 4 r

x

y

2

6

4 r

x 2 8 x 12

4

Axis of symmetry: x

x

1r

2

Vertex: 4, 6

0

6 2

§ t-intercepts: ¨¨1 r ©

x 2 8 x 16 16 10

x

2 1

10

Axis of symmetry: x

4

4

x −8

6 6, 0

−4 −2

0

x 2 8 x 12

−4

0

x

2

−6

2 x 6

x-intercepts: 2, 0 , 6, 0

2 x −2

−2

4

8

10

−4

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254

NOT FOR SALE

Chapter 2

9. h x

Polynomial ynomial and Rational Functions

11. h x

4 x 2 4 x 13 4 x 2 x 13

x2 5x

1 1 13 4 4 4 x x 14 1 13 4 x 12 12 Vertex: 12 , 12

4 x2 x

2

2

5· 41 § ¨x ¸ 2 4 © ¹

2

4 x

0

x 12

2

1 2

20

12

15 10

3

No real zeros −3

10. f x

−2

−1

2

3

x

x 6x 1 2

Axis of symmetry: x

6r

12. f x

3

6 2 21

32

52 41 4 21

41 1

x-intercepts: 3 r 2 2, 0 y

2

4 x2 4x 5 1 1 5· § 4¨ x 2 x ¸ 4 4 4¹ © 2 ª§ º 1· 4 «¨ x ¸ 1» 2¹ «¬© »¼

3r 2 2

2

5 r

5 r 41 2

x2 6x 1 6 r

−2 −4

4

8

2

§ 1 · Vertex: ¨ , 4 ¸ © 2 ¹

10

0

−6 −8

8

2

Axis of symmetry: x 2

y 12

1· § 4¨ x ¸ 4 2¹ ©

x −2

2 −2

§ 5 r 41 · x-intercepts: ¨¨ , 0 ¸¸ 2 © ¹

3 8

Vertex: 3, 8

x

−2

−10

x2 6 x 9 9 1

0

−4

x2 5x 4

2

x

−6

−4

x 1

x −8

Axis of symmetry: 5 x 2 0

5

x-intercepts: none

y

§ 5 41 · Vertex: ¨ , ¸ 4¹ © 2

y

12 2

25 25 4 4 4

5· 25 16 § ¨x ¸ 2¹ 4 4 ©

2

Axis of symmetry: x

x2 5x 4

x

−8 −6 −4 −2 −2

x 2

4

6

1 2

4 x2 4 x 5

4 r

42 4 4 5 2 4

4 r

64

8 4 r 8i 8 1 ri 2 The equation has no real zeros. x-intercepts: none

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NOT FOR SALE

Review Exercises ffor Chapter 2

15. Vertex: 4, 1 f x

1 2 x 5 x 4 3 1§ 2 25 25 · 4¸ ¨ x 5x 3© 4 4 ¹

13. f x

1 ª§ 5· 41º «¨ x ¸ » 3 «¬© 2¹ 4 »¼

f x

2

1§ 5· 41 ¨x ¸ 3© 2¹ 12

2

4a

12

a

12 x 4 1 2

5 2

f x

52 41 4

5 r

x

a 2 4 1

2

Point: 0, 3 3

x2 5x 4

0

2

16. Vertex: 2, 2 f x

§ 5 41 · Vertex: ¨ , ¸ © 2 12 ¹ Axis of symmetry: x

a x 4 1

Point: 2, 1 1

2

21

5 r 41 2

1 4

x

a x 2 2 2

a0 2 2 2

3

4a 2

1

4a

1 4

a

2 2 2

17. Vertex: 1, 4 f x Point: 2, 3 3

§ 5 r 41 · , 0 ¸¸ x-intercepts: ¨¨ 2 © ¹

f x

x

a x 1 4 2

a 2 1 4 2

1

y

255

a

1 4 2

4

18. Vertex: 2, 3 f x

2 x −8

−6

14. f x

−4

−2

Point: 1, 6 6

2

a x 2 3 2

a 1 2 3 2

−4

6

9a 3

−6

3

9a

1 3

a

1 2 6 x 24 x 22 2 3 x 2 12 x 11

f x

3 x 2 4 x 4 4 11

1 3

x

2 3 2

19. (a) y

3 x 2 3 4 11 2

3 x 2 1 2

x

Vertex: 2, 1 Axis of symmetry: x 0 x

(b) x x y y 2x 2 y y

2

3x 2 12 x 11 12 r 12 r 12 6 2 r

3 3

12 2 23

A

43 11

1000 500 x

xy x500 x 500 x x 2

y

(c) A

14

500 x x 2 x 2 500 x 62,500 62,500

12 10

x 250 62,500

8

2

6 4

x-intercepts:

§ 3 · , 0 ¸¸ ¨¨ 2 r 3 © ¹

P

2 x –6 –4 –2

4

6

8 10

The maximum area occurs at the vertex when 500 250 250. x 250 and y The dimensions with the maximum area are x 250 meters and y 25 meters. 250

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256

NOT FOR SALE

Chapter 2

Polynomial ynomial and Rational Functions

10 p 2 800 p

20. R

(a) R 20

$12,000

R 25

$13,750

R30

$15,000

y

2 1

800 2 10

b 2a

R 40

−3

−2

−1

x 1

−1

2

3

−2 −3

$40

Transformation: Reflection in the x-axis and a vertical stretch

$16,000

The revenue is maximum when the price is $40 per unit.

x 4 , f x

25. y

The maximum revenue is $16,000.

6 x4

y

Any price greater or less than $40 per unit will not yield as much revenue.

7 5 4

70,000 120 x 0.055 x 2

21. C

4 x3

3

(b) The maximum revenue occurs at the vertex of the parabola.

x 3 , f x

24. y

3 2

The minimum cost occurs at the vertex of the parabola.

120 | 1091 units 20.055

b Vertex: 2a

About 1091 units should be produced each day to yield a minimum cost.

1

0.107 x 5.68 x 48.5

0

0.107 x 2 5.68 x 74.5

1

2

3

4

Transformation: Reflection in the x-axis and a vertical shift six units upward x 4 , f x

26. y 22. 26

x

−4 −3 − 2

2 x 8

4

2

5.68 2 4 0.107 74.5 2 0.107

5.68 r

x

y 10 8 6 4

x | 23.7, 29.4

The age of the bride is about 24 years when the age of the groom is 26 years.

x

−2 −2

2

4

6

8

10

Transformation: Horizontal shift eight units to the right and a vertical stretch

y

Age of groom

27 26

24

x

5

5

y

23 4

22

3

x

2

20 21 22 23 24 25

Age of bride

23. y

x5 , f x

27. y

25

x 3 , f x

1

x 2

x

−2

1

2

3

5

6

3 −3 −4

y 4

Transformation: Horizontal shift five units to the right

3 2 1 x

− 4 −3 − 2

1

2

4

−2 −3 −4

INSTRUCTOR USE ONLY Transformation: Reflection in the x-axis and a horizontal shift two units to the right

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE

Review Exercises ffor Chapter 2

x5 , f x

28. y

1 x5 2

3

36. f x

x3 8 x 2

0

x3 8 x 2

0

x 2 x 8

y 8

257

10 − 10

10

6

−6 −4

− 80

Zeros: x 0 of multiplicity 2 (even multiplicity)

4

x

−2

2

4

x 8 of multiplicity 1 (odd multiplicity)

6

Turning points: 2

Transformation: Vertical shrink and a vertical shift three units upward 29. f x

37. f x

18 x3 12 x 2

0

18 x3 12 x 2

0

6 x 2 3 x 2

2 x 2 5 x 12

The degree is even and the leading coefficient is negative. The graph falls to the left and falls to the right. 30. f x

2x

1 x3 2

x

The degree is odd and the leading coefficient is positive. The graph falls to the left and rises to the right. 31. g x

3 4

x4

3 x 2 2

2

3

−2

38. g x

x 4 x3 12 x 2

0

x 4 x3 12 x 2

0

x 2 x 2 x 12

0

x 2 x 4 x 3

The degree is odd and the leading coefficient is negative. The graph rises to the left and falls to the right. 3x 2 20 x 32

0

3x 2 20 x 32

0

3 x

Zeros: x

40

−12

4

4 x 8 4 3

and x

8,

Zeros: x

0 of multiplicity 2 (even multiplicity)

x

4 of multiplicity 1 (odd multiplicity)

x

3 of multiplicity 1 (odd multiplicity)

− 80

Turning points: 3

both of multiplicity 1 (odd multiplicity) Turning points: 1

39. f x

34. f x

x x 3

2

0

x x 3

2

0 of multiplicity 2 (even multiplicity)

−3

x7 8 x 2 8 x

33. f x

of multiplicity 1 (odd multiplicity)

Turning points: 2

The degree is even and the leading coefficient is positive. The graph rises to the left and rises to the right. 32. f x

2 3

Zeros: x

3

−6

Zeros: x 0 of multiplicity 1 (odd multiplicity)

6

x3 x 2 2

(a) The degree is odd and the leading coefficient is negative. The graph rises to the left and falls to the right. 1

(b) Zero: x −5

(c)

x 3 of multiplicity 2 (even multiplicity)

x f x

–3

–2

1

0

1

2

34

10

0

–2

–2

–6

Turning points: 2 35. f t

t 3 3t

0

t 3 3t

0 Zeros: t

y 4 3

−5

t t 3

2

4

2

0, r

(d)

3

(− 1, 0)

1 x

−4 −3 −2

3, all of

multiplicity 1 (odd multiplicity)

1

2

3

4

−3

−3 −4

INSTRUCTOR USE ONLY Turning points: 2

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258

NOT FOR SALE

Chapter 2

40. g x

Polynomial ynomial and Rational Functions

2 x3 4 x 2

(c)

(a) The degree is odd and the leading coefficient, 2, is positive. The graph falls to the left and rises to the right. (b) g x

2 x3 4 x 2

0

2 x3 4 x 2

0

2 x 2 x 2

–2

1

0

1

2

h x

–4

2

0

2

–4

(d)

y 4 2

(−

3, 0)

(

3, 0)

−4 −3

x x 2

x −1 −1

1

3

4

−2 −3

2, 0

Zeros: x

(0, 0)

3

2

0

(c)

x

−4

x

–3

2

1

0

1

g x

–18

0

2

0

6

6x 3 43. 5 x 3 30 x 2 3x 8 30 x 2 18 x

y

(d)

15 x 8

4

15 x 9

3 2

−4 −3

17

(0, 0)

(− 2, 0)

x −1 −1

1

2

3

30 x 2 3 x 8 5x 3

4

−2 −3

6x 3

17 5x 3

−4

4 3 44. 3x 2 4 x 7

x x x 5 x 3

41. f x

3

2

(a) The degree is even and the leading coefficient is positive. The graph rises to the left and rises to the right.

x f x

(d)

–4

3

2

–1

0

1

2

3

100

0

–18

–8

0

0

10

72

(− 3, 0)

4x 7 3x 2

4 29 3 33 x 2

5x 4 45. x 2 5 x 1 5 x3 21x 2 25 x 4

y

−4

8 3 29 3

0, 1, 3

(b) Zeros: x (c)

4x

3

(1, 0)

5 x 3 25 x 2 5 x

x

−2 −1

1

2

3

4

(0, 0)

4 x 2 20 x 4 4 x 2 20 x 4

−15 −18

0

−21

42. h x

3x x 2

5 x3 21x 2 25 x 4 x2 5x 1

4

(a) The degree is even and the leading coefficient, 1, is negative. The graph falls to the left and falls to the right. (b) g x

3x 2 x 4

0

3x 2 x 4

0

x 2 3 x 2

Zeros: x

0, r

3

5 x 4, x z

3x 2

5 r 2

29 2

3

46. x 1 3 x 0 x 0 x 2 0 x 0 2

4

3x 4

3

3x 2 3x 2

0

3x 2

3 3

4

3x x2 1

3x 2 3

3 x2 1

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© Cengage Learning. All Rights Reserved.

NOT FOR SALE

Review Exercises ffor Chapter 2

259

x 2 3x 2

47. x 0 x 2 x 3 x 4 x 2 6 x 3 2

4

3

x 4 0 x3 2 x 2 3x3 2 x 2 6 x 3x3 0 x 2 6 x 2x2 0 x 3 2x2 0 x 4 1 x 3x 4 x 6 x 3 x2 2 4

3

2

x 2 3x 2

1 x2 2

3x 2 5 x 8 48. 2 x 0 x 1 6 x 10 x 13 x 2 5 x 2 2

4

3

6 x 4 0 x3 3x 2 10 x3 16 x 2 5 x 10 x3 0 x 2 5 x 16 x 2 0 x 2 16 x 2 0 x 8 10 6 x 4 10 x3 13 x 2 5 x 2 2 x2 1

49.

2

6

–4

6

– 27

18

12

16

– 22

–8

8

–11

–4

–8

0.1 0.1

51. 8

6 x3 8 x 2 11x 4

0.3

0

0.5

0.5

4

20

0.8

4

19.5

0.1x3 0.3 x 2 0.5 x 5

(a)

0.1x 2 0.8 x 4

– 25 16

66 – 72

48 – 48

2

–9

–6

0

2 x 25 x 66 x 48 x 8

52.

–4

2

5 5

19.5 x 5

–8

–20

– 52

8

13

–2

0

–1

20

9 – 20

14 11

–3 3

0 0

20

–11

–3

0

0

1 is a zero of f.

3 4

Yes, x

x z 8

0

(d)

1

14

–3

0

15

18

3

0

24

4

0

3 4

0

is a zero of f.

20

9 0

14 0

–3 0

0 0

20

9

–14

–3

0

Yes, x

5 x 2 13 x 2, x z 4

9

20 20

(c)

50

20 x 4 9 x3 14 x 2 3 x

Yes, x (b)

2 x 2 9 x 6,

33

5 x3 33 x 2 50 x 8 x 4

8 x 2

53. f x

2

3

10 2 x2 1

0

6 x 4 4 x3 27 x 2 18 x x 2

50. 5

3x 2 5 x 8

0 is a zero of f.

20

9 20

14 29

–3 15

0 12

20

29

15

12

12

INSTRUCTOR USE ONLY No, x

1 is not a zero off ff.

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260

NOT FOR SALE

Chapter 2

54. f x

(a)

Polynomial ynomial and Rational Functions

4

3

–8

– 20

16

12

16

–16

4

–4

0

3 Yes, x (b)

(c)

2 3

16

–12

80

– 240

3

– 20

60

– 224

Yes, x

–8

– 20

16

2

–4

–16

–6

– 24

0

103 243 20 3 44 3

2

– 20

1

10

– 24

20

44

1

–3 7

– 21 – 45

135 155

– 465 – 421

So, f 3

421.

(b) Remainder Theorem: f 1

1

4

101 24 1 20 1 44 3

2

9

is a zero of f.

Synthetic Division:

3

–8 –3

11

9

3

–11

–9

25

16

–1

1

10

– 24

20

44

–1

–9

33

– 53

9

–33

53

–9

1

1 is not a zero of f.

No, x 56. g t

–3

4 is not a zero of f.

2 3

4

Synthetic Division:

– 20

3

3

421

–8

3

–1

f 3

3

No, x

x 4 10 x3 24 x 2 20 x 44

(a) Remainder Theorem:

4 is a zero of f.

–4

(d)

55. f x

3 x3 8 x 2 20 x 16

So, f 1

9.

2t 5 5t 4 8t 20

(a) Remainder Theorem: g 4

2 4 5 4 84 20 5

4

3276

Synthetic Division:

–4

2

–5

0

0

–8

20

2

–8 –13

52 52

–208 –208

832 824

– 3296 – 3276

So, g 4

3276.

(b) Remainder Theorem:

g

2

2

2

5

2

5

4

2 20

8

0

Synthetic Division: 2

So, g

2

–5

0

0

–8

20

5 2 4 5 2 4

10 4 2 10 4 2

10 2 8 10 2

–20

2

2 2 5 2 2

2

0

0.

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NOT FOR SALE

Review Exercises ffor Chapter 2

x 3 4 x 2 25 x 28; Factor: x 4

57. f x

(a)

4

1 1

4

25

28

4

32

28

8

7

0

(b) x 2 3 x 4

x

(b) x 8 x 7

(c) f x

x

(e)

40

7 x 1 −3

5

x3 4 x 2 25 x 28

x

−10

7 x 1 x 4

60. f x

(d) Zeros: 7, 1, 4

(a)

(e)

1 x 4 x 2 x 3

(d) Zeros: 2, 1, 3, 4

The remaining factors are x 7 and x 1 . (c) f x

1 x 4

The remaining factors are x 1 and x 4 .

Yes, x 4 is a factor of f x . 2

x

261

80

x 4 11x 3 41x 2 61x 30

2

1

–11

41

– 61

30

2

–18

– 30

–9

23

46 –15

1

−8

0

5

5

1

–9

23

–15

5

–20

15

–4

3

0

−60

2 x 3 11x 2 21x 90; Factor: x 6

58. f x

(a)

–6

11

21

– 90

–12

6

90

–1

–15

0

2 2

1

Yes, x 2 and x 5 are both factors of f x . (b) x 2 4 x 3

(b) 2 x x 15

2 x

(c) f x

5 x 3

2 x

(e)

5 x 3 x 6

(d) Zeros: x

52 ,

(e)

50

x

(d) Zeros: x

The remaining factors are 2 x 5 and x 3 . (c) f x

1 x 3

The remaining factors are x 1 and x 3 .

Yes, x 6 is a factor of f x . 2

x

1 x 3 x 2 x 5 1, 2, 3, 5

4

−6

12

3, 6 −8

−7

61. A | 0.0022t 3 0.044t 2 0.17t 2.3

5

62.

8

−100

59. f x

x 4 4 x 3 7 x 2 22 x 24

Factors: x 2 , x 3 (a)

–2

3

1 1

0

12 0

1

–4

7

22

24

–2

12

–10

–24

1

–6

5

12

0

–6

5

12

3

–9

–12

–3

–4

0

The model is a good fit to the actual data.

INSTRUCTOR USE ONLY Yes, x 2 and x 3 are both factors of f x .

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262 63.

Chapter 2

NOT FOR SALE

Polynomial ynomial and Rational Functions

t

A, actual

A, cubic model

0

2.3

2.3

1

2.4

2.5

2

2.9

2.8

3

3.2

3.1

4

3.6

3.5

5

4.0

4.0

6

4.2

7

64. 18

–0.0022

66. 5

49

4.9

4.9

67. i 2 3i

8

5.4

5.3

68. 5i i 2

9

5.8

5.8

10

6.4

6.2

11

6.5

6.6

12

6.9

6.9

72. 1 6i 5 2i

2 2 i 2 2

2 2

2 i 2

6.7856

1 5i

§ 2 · 2¨¨ i ¸¸ © 2 ¹

7

4 5i 2i

3 7i

2i

74. i6 i 3 2i

63 77i

i18 12i 3i 2i 2 i18 9i 2 i 20 9i 20i 9i 2 9 20i

20 30i 16i 24i 2 20 46i 24

75. 8 5i

2

64 80i 25i 2 64 80i 25

4 46i 2

0.2492

1 3i

17 28i

2

0.0044

5 7i

5 2i 30i 12i 2

76. 4 7i 4 7i

4.4856

8 10i

5 28i 12

73. 10 8i 2 3i

0.0792

69. 7 5i 4 2i

2 · i¸ 2 ¸¹

77i 63i 2

–0.0396

No, the model falls to the right as t increases since the degree is odd and the leading coefficient is negative.

4.4

71. 7i11 9i

2.3

A18 | 6.8 million

100

2 · § 2 i¸ ¨ 2 ¸¹ ¨© 2

0.17

–0.0022

65. 8

§ 2 70. ¨¨ © 2

0.044

39 80i

16 56i 49i 2 16 56i 49i 2 32 98i 2 66

77.

6i 4 i

6i 4i 4i 4i

79.

4 2 2 3i 1 i

24 10i i 2 16 1 23 10i 17 23 10 i 17 17 8 5i i 78. i i

8i 5i 2 i 2

5 8i 1

5 8i

4 2 3i 2 1i 2 3i 2 3i 1 i 1 i 8 12i 2 2i 4 9 11 8 12 i 1i 13 13 §8 · § 12 · ¨ 1¸ ¨ i i ¸ © 13 ¹ © 13 ¹ 21 1 i 13 13

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE

Review Exercises ffor Chapter 2

80.

1 4i 52 i 2 i 1 4i

1 5 2 i 1 4i

1 4i 10 5i 2 8i i 4i 2 9 i 2 9i 2 9i 2 9i 18 81i 2i 9i 2 4 81i 2 9 83i 9 83 i 85 85 85 81. 5 x 2 2

5x2

2

x2

x

r

2 i 5

x

r

10 i 5

2 5

8x

2

x

2

14

x

r 12 i

83. x 2 2 x 10

x x

x

4 x 6 x 2i x 2i 4, 6, 2i, 2i

x

8 x 5 x 3 i x 3 i 2

5, 8, 3 r i

4 x 3 8 x 2 3 x 15

3x 4 4 x3 5 x 2 8

93. f x

x

1 r 3i

–2

1 1

3 r

32 46 27

60

–2

–2

60

1

–30

0

x x

The zeros of f x are x

26

94. f x

639

2 x 6 x 5 2, x

6, and x

5.

4 x 3 27 x 2 11x 42

1 r 4

71 i 4

r 2, r 3, r 72 , r 21 , r 6, r 7, r 21 , r14, r 21, r 42 4 2 –1

4

2

4

–27

11

42

–4

31

– 42

–31

42

0

0, 3

4 x3 27 x 2 11x 42 4 x 9 9, 4

2 x 2 x 30

Possible rational zeros: r 14 , r 12 , r 34 , r1, r 32 , r 74 ,

12

3 r 3i 71 12

28

3

x3 3 x 2 28 x 60

0

b 2 4ac 2a

4 x x 3

x 3 3 x 2 28 x 60

Possible rational zeros: r1, r 2, r 3, r 4, r 5, r 6, r10, r12, r15, r 20, r 30, r 60

9 r 9

x

Possible rational zeros: r1, r 2, r 4, r 8, r 13 , r 23 , r 43 , r 83

b r

86. f x

10i

Possible rational zeros: r1, r 3, r 5, r15, r 12 , r 32 , r 52 , r 15 , r 14 , r 34 , r 54 , r 15 2 4

x 1

Zeros: x

10i x

0, r 10i

Zeros: x

92. f x

84. 6 x 2 3x 27

85. f x

x x 2 10

91. f x

10 1

2

3 r

x3 10 x

Zeros: x

0

x2 2 x 1

x

88. f x

2 x 9 2, 9

Zeros: x

90. f x

0

1

x

Zeros: x

2

x

x 2 11x 18

89. f x

0

82. 2 8 x 2

87. f x

263

2

The zeros of f x are x

x x

1 4 x 2 31x 42 1 x 6 4 x 7

INSTRUCTOR USE ONLY Zeros: x

1, x

7 , 4

and x

6.

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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264

NOT FOR SALE

Chapter 2

95. f x

Polynomial ynomial and Rational Function Functions

x 3 10 x 2 17 x 8

Possible rational zeros: r1, r 2, r 4, r 8 1

1

–10

17

–8

1

–9

8

–9

8

0

1

The zeros of f x are x 96. f x

1 x 2 9 x 8

x

x3 10 x 2 17 x 8

1 and x

x

1 x 1 x 8

97. f x

9

1

24

3

20

–5

–20

–20

4

4

0

x3 9 x 2 24 x 20

x 4 x3 11x 2 x 12

1 1

x

5 x 2 4 x 4

x

5 x 2 .

The zeros of f x are x

2

Possible rational zeros: r1, r 2, r 3, r 4, r 6, r12

Possible rational zeros: r1, r 2, r 4, r 5, r10, r 20 1

1 x 8

8.

x 3 9 x 2 24 x 20

–5

x

–4

1

–11

1

–12

3

12

3

12

4

1

4

0

1

4

1

4

–4

0

–4

0

1

0

2

5 and x

2.

1

x

x 4 x3 11x 2 x 12 The zeros of f x are x 98. f x

3 x 4 x 2 1

3 and x

4.

25 x 4 25 x3 154 x 2 4 x 24

Possible rational zeros: r1, r 2, r 3, r 4, r 6, r 8, r12, r 24, r 15 , r 52 , r 53 , r 54 , r 56 , r 85 , r 12 , 5 1 , r 2 , r 3 , r 4 , r 6 , r 8 , r 12 , r 24 r 24 , r 25 5 25 25 25 25 25 25 25

–3

25 25

2

25

25

–154

–4

24

–75

150

12

–24

– 50

–4

8

0

– 50

–4

8

50

0

–8

0

–4

0

25

x

25 x 4 25 x3 154 x 2 4 x 24 The zeros of f x are x 99. f x

3, x

2, and x

x3 4 x 2 x 4, Zero: i

Because i is a zero, so is i. i

i

–4

1

–4

i

1 4i

4

1

4 i

4i

0

4 i

4i

i

4i

–4

0

1 f x

x

x

3 x 2 5 x 2 5 x 2 .

2 r . 5

100. h x

x3 2 x 2 16 x 32

Because 4i is a zero, so is 4i.

1

1

3 x 2 25 x 2 4

4i

4i

–1

2

16

32

4i

16 8i

–32

–1

2 4i

8i

0

–1

2 4i

8i

4i

8i

–1

i x i x 4

h x

ri, 4

Zeros: x

x

2

0

4i x 4i x 2

INSTRUCTOR USE ONLY Zeros: Zeros x

r4i, 2

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE

Review Exercises ffor Chapter 2

101. g x

105. g x

2 x 4 3 x3 13 x 2 37 x 15, Zero: 2 i

Because 2 i is a zero, so is 2 i. 2i

2 2

13

37

–15

4 2i

5i

15

1 2i

13 5i

31 3i 6 3i

1

2

1 2i

13 5i

6 3i

4 2i

10 5i 3

6 3i

2

5

0

2 ª¬ x 2 i ºª ¼¬ x 2 i º¼ 2 x 5 x 3

g x

x Zeros: x 102. f x

–4

0

12

–208

0

–3

52

0

1

0

3

52

–4

16

–52

–4

13

0

1

g x

x

2 r 3i.

g x

2

0. Because 1 i is a zero, so is 1 i. 11

14

–6

4 4i

11 3i

4

7 4i

3 3i

6 0

4

7 4i

3 3i

4 4i

3 3i 0

3

4

i, 1 i

g x

8

–72

–153

33

123

153

11

41

51

0

1

11

41

51

–3

–24

–51

8

17

By the Quadratic Formula, the zeros of x 2 8 x 17 are

f x

x3 7 x 2 36

8

8 r

2

41 17

21

8 r

4

4 r i.

2

x

3 x 3 x 4 i x 4 i 3i is a zero, so is 3i.

3x

2 3

x 4 x

3i x

–7

0

36

3 x

2 x 4 x 3

–2

18

–36

3x 2

14 x 8 x 2 3

–9

18

0

3i

2

3x 4 14 x3 17 x 2 42 x 24

x

3 x 6 are

3, 6. The zeros of g x are x

x

1

8

Multiply by 3 to clear the fraction.

The zeros of x 2 9 x 18 x

2

3

107. Because

0, 5, 1

1

4 x 2 3i x 2 3i

x 4 8 x3 8 x 2 72 x 153

f x

x x 5 x 1

1

x

2

The zeros of f x are 3, 3, 4 i, 4 i.

x x 2 4 x 5

–2

4 ª¬ x 2 3i ºª ¼¬ x 2 3i º¼

1

x

x3 4 x 2 5 x

104. g x

3

x

1

x x 1 i x 1 i 4 x 3 3 ,1 4

106. f x

–3

x ª¬ x 1 i ºª ¼¬ x 1 i º¼ 4 x 3

Zeros: x

2

x

4

103. f x

4 x 2 4 x 13

2 r i, 12 , 3

One zero is x

Zeros: 0,

208

1

x 4 x 11x 14 x 6

f x

40

2 i x 2 i 2 x 1 x 3

3

1i

3

By the Quadratic Formula the zeros of x 2 4 x 13 4 and are x 2 r 3i. The zeros of g x are x

4 x 4 11x3 14 x 2 6 x

1i

4

0 –4

2i

4

x 4 4 x3 3 x 2 40 x 208, Zero: x

–4

3

265

2 x 3 x 6

2, 3, 6.

Note: f x

a3 x 4 14 x3 17 x 2 42 x 24 ,

where a is any real nonzero number, has zeros 23 , 4, and r 3i.

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266

NOT FOR SALE

Chapter 2

Polynomial ynomial and Rational Function Functions

108. Because 1 2i is a zero, so is 1 2i.

f x

x

2 x 3 x 1 2i x 1 2i

x2 x2

2 x 6 ª x 1 4º ¬ ¼

x 6 x 2 2 x 5

2 x 1

2, x

1

3 2x2

117. f x

(a) Domain: all real numbers x except x

3x x 10

4 x3 2 5x

x 2 x 4

0

(b) No intercepts (c) Vertical asymptote: x

Domain: all real numbers x except x 110. f x

x

2

Vertical asymptotes: x

x 4 x3 3 x 2 17 x 30 109. f x

x3 4 x 2 x 3x 2

116. h x

10

Horizontal asymptote: y (d)

4 x3 5x 2

Domain: all real numbers x except x

0

2 5

x

–1

y

3 2

1 2

–6

0

1 2

1

6

3 2

y

111. f x

8 x 2 10 x 24 8 x 4 x 6

1

Domain: all real numbers x except x 112. f x

x

−4 −3

x2 x 2 x2 4

4 and x

1

−1

3

4

6

4 x

118. f x

Domain: all real numbers x

(a) Domain: all real numbers x except x 113. f x

4 x 3

Vertical asymptote: x

(b) No intercepts (c) Vertical asymptote: x

3

Horizontal asymptote: y 2 x 5x 3 x2 2 2

x y

Horizontal asymptote: y

0

Horizontal asymptote: y

0 (d)

114. f x

0

0

–3

–2

–1

1

2

4 3

–2

–4

4

2

3 4 3

2 y

115. f x

5 x 20 x 2 2 x 24 5 x 4

x

4 3 2 1 x

6 x 4

–3 –2 –1

Horizontal asymptote: y

2

3

4

–2

5 ; x z 4 x 6

Vertical asymptote: x

1

–3

6 0

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE

Review Exercises ffor Chapter 2

2 x 1 x

119. g x

x 2 x 1

(d)

(a) Domain: all real numbers x except x

1

x

–2

–1

0

1

2

y

20 17

1

0

1

20 17

(b) x-intercept: 2, 0

y

y-intercept: 0, 2

2

(c) Vertical asymptote: x

1

x

–1

y

1 2

0

2

1

1

Horizontal asymptote: y (d)

267

−2

x

(0, 0)1

−1

2

−1

3

−2

2

5 2

–4

2x x2 4

122. f x

y 6

(a) Domain: all real numbers x

4

(0, 2) (−2, 0)

2

(b) Intercept: 0, 0

x

(c) Horizontal asymptote: y

0

–4

(d)

–6

x

–2

–1

0

2 5

0

1

2

–8

y

x 4 x 7

120. f x

1 2

2 5

1 2

y

(a) Domain: all real numbers x except x

7

2

(b) x-intercept: 4, 0

1

(0, 0)

§ 4· y-intercept: ¨ 0, ¸ © 7¹

–2

–1

y

2 3

5 8

2

–1 –2

7

Horizontal asymptote: y x

x 1

(c) Vertical asymptote: x

(d)

0

1 1

4 7

2

1 2

2 5

y

x x2 1

123. f x

(a) Domain: all real numbers x (b) Intercept: 0, 0 (c) Horizontal asymptote: y

0

8

(d)

6

(0, 47) 4

x

−4 − 2

x

–2

y

(4, 0) 2

4

10 12

−4

–1

2 5

−6

1 2

0

1

2

0

1 2

2 5

y

−8 2

121. f x

2

5x 4x2 1

1

(0, 0) x 1

(a) Domain: all real numbers x

–1

(b) Intercept: 0, 0

–2

(c) Horizontal asymptote: y

2

5 4

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268

NOT FOR SALE

Chapter 2

Polynomial ynomial and Rational Function Functions

9

124. h x

x

3

(a) Domain: all real numbers x except x (b) y-intercept: 0, 1 (c) Vertical asymptote: x

3

2

(a) Domain: all real numbers x except x

2, x

Horizontal asymptote: y

0

x

–3

–2

–1

0

1

2

y

1 4

9 25

9 16

1

9 4

9

r2

(b) Intercept: 0, 0 (c) Vertical asymptotes: x

3

Horizontal asymptote: y (d)

2x2 x 4

126. y

2

(d)

2

2

x

r5

r4

r3

r1

0

y

50 21

8 3

18 5

2 3

0

y

y

8

6

6

4

4

(0, 0) x

2

–6 –4

(0, 1) −2

4

6

x

2

4

6

8

−2

6 x 2 x2 1

125. f x

6 x 2 11x 3 3x 2 x 3x 1 2 x 3

127. f x

(a) Domain: all real numbers x (b) Intercept: 0, 0

x3 x 1

(c) Horizontal asymptote: y (d)

x

r3

y

r2

27 5

2x 3 1 ,x z 3 x

24 5

6

r1

0

–3

0

(a) Domain: all real numbers x except x 1 x 3

0 and

§3 · (b) x-intercept: ¨ , 0 ¸ ©2 ¹

y

(c) Vertical asymptote: x

4

0

Horizontal asymptote: y

2

2

(0, 0) −6

−4

x

−2

2

−8

4

6

(d)

x

–2

–1

1

2

3

4

y

7 2

5

–1

1 2

1

5 4

y

2 −8 −6 −4

x −2 −4

4 6 3 ,0 2

( (

8

−6 −8

INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

© Cengage Learning. All Rights Reserved.

NOT FOR SALE

Review Exercises ffor Chapter 2

6 x2 7 x 2 4x2 1 2 x 1 3x 2

128. f x

2 x

1

(c) Vertical asymptote: x

x 1

Slant asymptote: y 3x 2 1 ,x z 2x 1 2

1 2 x 1

(a) Domain: all real numbers x except x

(d)

r

1 2

x

–6

y

3 2 13 2

–2

37 5

§2 · (b) x-intercept: ¨ , 0 ¸ ©3 ¹

269

–5

1 2

5 2

0

4

1

17 5

y 4

(0, 1)

y-intercept: 0, 2 x −6 −4

1 2

(c) Vertical asymptote: x

(d) x

–3

–2

–1

0

2 3

1

2

y

11 5

8 3

5

–2

0

1 3

4 5

3 x

3 x

x

−3 −2 −1

2

(23, 0)

2 x3 x2 1

2 x 1

1· § y-intercept: ¨ 0, ¸ 2¹ ©

(b) Intercept: 0, 0 (c) Slant asymptote: y

1 and

§2 · (b) x-intercepts: 1, 0 , ¨ , 0 ¸ ©3 ¹

2x 2x 2 x 1

(a) Domain: all real numbers x

4 3

(c) Vertical asymptote: x

2x

–2

–1

0

1

2

16 5

–1

0

1

16 5

4 x 1

(a) Domain: all real numbers x except x 4 x 3

3

(0, −2)

y

6

3x 4 1 23 x , x z 1 3 3x 4

2

x

4

3x3 2 x 2 3x 2 3x 2 x 4 3x 2 x 1 x 1

131. f x

y

(d)

2

3 2

Horizontal asymptote: y

129. f x

−2

x

Slant asymptote: y (d)

x

y

y

–3

3

–2

44 13

0

12 5

1 2

1 3 1

2

3

0

2

14 5

2

y 1

(0, 0) −3

130. f x

−2

−1

4

x 1

2

3

3

−2

2

−3

(0, − 12 ( 1

(1, 0)

−2 −1

2

x2 1 x 1

x 1

(23 , 0( 3

x 4

−2

2 x 1

(a) Domain: all real numbers x except x

1

(b) y-intercept: 0, 1

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© Cengage Learning. All Rights Reserved.

270

NOT FOR SALE

Chapter 2

Polynomial ynomial and Rational Function Functions

3x 3 4 x 2 12 x 16 3x 2 5 x 2 x 2 x 2 3x 4

132. f x

2 3x 4 5 , x z 2 3x 1

2 and (a)

§4 · (b) x-intercepts: ¨ , 0 ¸, 2, 0 ©3 ¹

4000

0

(c) Vertical asymptote: x

1 3

Slant asymptote: y

x 3

–4

–1

96 13

y

100 0

y-intercept: 0, 8

x

0.5

528 p , 0 d p 100 100 p

134. C

(a) Domain: all real numbers x except x 1 x 3

(d)

0.5 1

As x increases, the average cost per unit approaches the 0.5 $0.50. horizontal asymptote, C

3x 1 x 3

0.5 x 500 ,0 x x

Horizontal asymptote: C

x 2 3x 1

x

C x

133. C

0

21 4

–8

1 1 2

2 0

4 16 11

528 25

(b) When p

25, C

100 25

When p

50, C

100 50

When p

75, C

100 75

52850 52875

$176 million. $528 million. $1584 million.

(c) As p o 100, C o f. No, it is not possible.

y 4

(43, 0)

2

x

− 6 − 4 −2 −2

4

6

(2, 0)

−6

(0, − 8)

−8

Chapter Test for Chapter 2 1. f x

x2

(a) g x

3. (a) y

1 x 2 60 x 900 900 5 20

2 x2

Reflection in the x-axis followed by a vertical shift two units upward

x

(b) g x

3 2

Horizontal shift

2

3 2

a x 3 6 2

Point on the graph: 0, 3 3

a0 3 6

9

9a a

1 ª x 30 2 900º 5 20 ¬ ¼ 1 x 30 20 50 2

Vertex: 30, 50 units to the right

2. Vertex: 3, 6

y

1 x 2 3x 5 20

The maximum height is 50 feet. (b) The constant term, c 5, determines the height at which the ball was thrown. Changing this constant results in a vertical translation of the graph, and, therefore, changes the maximum height.

2

x

1

3 6. 2

INSTRUCTOR USE ONLY So, y

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Chapter Test ffor Chapter 2

4. ht

34 t 5 2t 2

10. Because 2 i is a zero, so is 2 i.

y

−4 − 3 − 2 −1

2 3 4 5

−2 −3 −4 −5

x 4 7 x3 17 x 2 15 x any non-zero real number, has the zeros 0, 3, and 2 i. 11. Because 1

x 1

6. 2

2 2

So,

2

x 1 3i x 2 x 2 3i x 1 3i x 2 x 2

–5

0

–3

x 2 2 x 4 x 2 4 x 4

4

8

6

12

x 4 6 x3 16 x 2 24 x 16

4

3

6

9

–5

–6

15

5

0

–15

0

–6

0

9 . x 2

a is any non-zero real number, has the zeros 1

3i, 2,

and 2. 12. f x

3x3 14 x 2 7 x 10

Possible rational zeros: r 13 , r 23 , r1, r 53 , r 2, r 10 , r 5, r10 3 1

x 52 2 x 6 2 x 52 x 3 2 x x 3 x

2 x 3 5 x 2 6 x 15

a x 4 6 x3 16 x 2 24 x 16 , where

Note: f x

2 x3 4 x 2 3 x 6

3i.

3i

x 1

2 x3 5 x 2 6 x 15

2

x 1

f x

x 1 . x2 1

3i is a zero, so is 1

0

2 x4 5x2 3 x 2

7. f x 5 2

3x

a x 4 7 x3 17 x 2 15 x , where a is

Note: f x

2

3x3 4 x 1 x2 1

i

x x3 7 x 2 17 x 15

3x 0 x 3x

So,

2 i

x x 3 x 2 4 x 5

t

x 1 3x 2 x 1 5. x 2 0 x 1 3 x3 0 x 2 4 x 1 3

x 0 x 3 x 2 i x x x 3 x 2 i x 2

f x

5 4 3

The degree is odd and the leading coefficient is negative. The graph rises to the left and falls to the right.

271

3

2

3

14

–7

10

3

17

10

17

10

0

2

5 , 2

Zeros: x

r

x

8. (a) 10i 3

25

3

2

3i 2

3i

10i 3 5i

13. f x

5 2 i

5 2 i 2i 2i 5 2 i 41 2i

1,

23 ,

1 3 x 2 17 x 10 1 3x 2 x 5

5

x 4 9 x 2 22 x 24

4 3i 2

Possible rational zeros: r1, r 2, r 3, r 4, r 6, r 8, r12, r 24

43

–2

1 4

0

9

22

–24

–2

4

10

–2

–5

–12

24 0

1

7 9.

x x

3x3 14 x 2 7 x 10

Zeros: x

3 5i

(b)

3

1

–2

–5

12

4

8

12

1

2

3

0

f x

x

2 x 4 x 2 2 x 3

By the Quadratic Formula, the zeros of x 2 2 x 3 are x

1 r

1 r

2i.

2i. The zeros of f are: x

2, 4,

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272

NOT FOR SALE

Chapter 2

Polynomial ynomial and Rational Function Functions

4 1 x2

14. h x

y

4 x2 x2 2 x 2 x

x2 2 x 1

16. g x

x 1

3 x 1

y-intercept: 0, 2

4 3 2

x2

Vertical asymptote: x

1

(− 2, 0)

(2, 0)

−1

x-intercepts: r 2, 0

1

x

1 x 1

Slant asymptote: y

2

−2

y 10

Vertical asymptote: x

0

8 6

1

Horizontal asymptote: y

4 2

2 x 5 x 12 x 2 16 2 x 3 x 4 2

15. f x

x

x

−8 −6 −4

y

2 −4

8

4

6

8

(0, −2)

−6

6

(− 32, 0( (0, 34 (

4 x 4

2x 3 ,x z 4 x 4

−8

−6 − 4

x −2

2

−4

§ 3 · x-intercept: ¨ , 0 ¸ © 2 ¹

18.47 x 2.96 ,0 x 0.23 x 1

17. y

2

4

The limiting amount of CO2 uptake is determined by the horizontal asymptote. 18.47 | 80.3 mg dm 2 hr. 0.23

y

§ 3· y-intercept: ¨ 0, ¸ © 4¹

90

4

Vertical asymptote: x Horizontal asymptote: y

2 0

100 0

Problem Solving for Chapter 2 x 2 x 3

f x

20

x 3 x 20 3

2. False. Because f x

x 2 x 3

l wh

1. V

2

d x

0

Possible rational zeros: r1, r 2, r 4, r 5, r10, r 20 2

1 1

x x

3

0

20

2

10

20

5

10

0

2 x 2 5 x 10 2 or x

x+3

0 x

5 r

r x

d x

.

The statement should be corrected to read f 1 because

x

q x

d x q x r x , you have

f x x 1

q x

f 1 x 1

2

.

3. If h 0 and k 0, then a 1 produces a stretch that is reflected in the x-axis, and 1 a 0 produces a shrink that is reflected in the x-axis.

15i 2

Choosing the real positive value for x we have: x 2 and x 3 5. The dimensions of the mold are 2 inches u 2 inches u 5 inches.

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Problem Solving ffor Chapter 2

273

13 x 2 1 is decreasing. 5

4. (a) y1

3 5

y2

x

2 3 is increasing. 5

8

−12

12

y1

y2 −8

(b) The graph is either always increasing or always decreasing. The behavior is determined by a. If a ! 0, g x will always be increasing. If a 0, g x will always be decreasing. (c) H x

x5 3x3 2 x 1

Since H x is not always increasing or always decreasing, H x z a x h k . 5

6

−9

9

−6

ax

5. f x

x

b

2

(a) b z 0 x

b is a vertical asymptote. a causes a vertical stretch if a ! 1 and a vertical shrink if 0 a 1.

For a ! 1, the graph becomes wider as a increases. When a is negative, the graph is reflected about the x-axis. (b) a z 0. Varying the value of b varies the vertical asymptote of the graph of f. For b ! 0, the graph is translated to the right. For b 0, the graph is reflected in the x-axis and is translated to the left 6. G

0.003t 3 0.137t 2 0.458t 0.839, 2 d t d 34

7. f x

60

(a)

(a)

2 x2 x 1 x 1 6

−9

−10

9

45 −5

(b) The tree is growing most rapidly when it is approximately 15.2 years old.

The graph has a “hole” when x vertical asymptotes.

0.009t 2 0.274t 0.458

(c) y

−6

b 2a

0.274 | 15.2222 2 0.009

y15.2222 | 2.5434

(b)

2 x2 x 1 x 1

(c) As x o 1,

Vertex: 15.2222, 2.5434

2 x

1 x 1 x 1

1. There are no 2 x 1, x z 1

2x2 x 1 o 3 x 1

(d) In both (b) and (c) the point of diminishing returns occured when t | 15.2.

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274

NOT FOR SALE

Chapter 2

8. Let x

Polynomial ynomial and Rational Function Functions

length of the wire used to form the square.

Then 100 x (a) let s Let r

length of wire used to form the circle.

the side of the square. Then 4 s

x s

the radius of the circle. Then 2S r

x and the area of the square is s 2 4

100 x r

2

§ x· ¨ ¸ . ©4¹

100 x and the area of the circle 2S

2

is S r 2

§ 100 x · ¸ . © 2S ¹

S¨

The combined area is: A x

2

§ x· § 100 x · ¨ ¸ S¨ ¸ © 4¹ © 2S ¹

2

§ 10,000 200 x x 2 · x2 S¨ ¸ 16 4S 2 © ¹ 2500 50 x x2 x2 S S 16 4S 1 · 2 50 x 2500 §1 ¨ ¸x S S 4S ¹ © 16 50 2500 §S 4· 2 x ¨ ¸x S S © 16S ¹ (b) Domain: Since the wire is 100 cm, 0 d x d 100. (c) A x

50 2500 §S 4· 2 x ¨ ¸x 16 S S S © ¹ 800 · 2500 § S 4 ·§ 2 x ¨ ¸¨ x S 4 ¹¸ S © 16S ¹© 2 2 800 §S 4·ª 2 § 400 · § 400 · º 2500 x x « ¨ ¸ ¨ ¸ ¨ ¸ » S 4 S © 16S ¹ «¬ ©S 4¹ © S 4 ¹ »¼ 2

2500 §S 4·ª § 400 ·º § S 4 ·§ 400 · ¨ ¸ «x ¨ ¸» ¨ ¸¨ ¸ S © 16S ¹ ¬ © S 4 ¹¼ © 16S ¹© S 4 ¹ 2

2

10,000 2500 §S 4·ª § 400 ·º ¨ ¸ «x ¨ ¸» S S 4 S © 16S ¹ ¬ © S 4 ¹¼ 2

2500 §S 4·ª § 400 ·º ¨ ¸ «x ¨ ¸» S 4 © 16S ¹ ¬ © S 4 ¹¼

The minimum occurs at the vertex when x

400

S 4

| 56 cm and A x | 350 cm 2 .

The maximum occurs at one of the endpoints of the domain. When x

0, A x | 796 cm 2 .

When x

100, A x

625 cm 2 .

Thus, the area is maximum when x

0 cm.

(d) Answers will vary. Graph A x to see where the minimum and maximum values occur.

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NOT FOR SALE

Problem Solving ffor Chapter 2

1 z

9. (a) zm

1 1i 1i 1i 1 1 i 2 2

1 1i 1i 2 10. y

(a)

1 z

(b) zm

1 3i 3i 10

1 z

1 2 8i 1 2 8i 2 8i 2 8i 1 2 2 8i i 68 34 17

(c) zm 1 3i 3i 3i 3 1 i 10 10

ax 2 bx c

0, 4 : 1, 0 : 2, 2 :

4

(b)

c

0

a b c a b

2

4a 2b c 4a 2b

4a 2b a

6

b

6

0

–4

1

0

3

2

2

4 a b

4

4

0

6

– 10

1

1 b

4

b

5

Use the “Quad Reg” feature of your graphing utility to obtain y x2 5x 4

L2

2a b a

x 2 5 x 4.

Thus, y Check:

L1

4

Solve the system of equations:

4, 0 : 6, 10 :

11. (a) Slope

0

4 5 4 4

10

6 56 4

2

2

9 4 3 2

5. Slope of tangent line is less

4 1 2 1

3. Slope of tangent line is greater

12. (a) x 2 y

than 5. (b) Slope

A x

4.41 4 (c) Slope 2.1 2 less than 4.1.

(d) Slope

(b) 4.1. Slope of tangent line is

2 h

1000 800 600

2

400 200

h 4 2

x 20

h

3

41

5

40

60

80

100

Maximum of 1250 m 2 at x A x

(c)

4 h, h z 0

4 1 4 0.1

A

1200

4h h 2 h 4 h, h z 0 Slope

xy

1400

f 2 h f 2

2

100 x 2 x2 § 100 x · 50 x x¨ ¸ 2 2 © ¹

100 y

Domain: 0 x 100

than 3.

(e)

275

A50

4.1

The results are the same as in (a)–(c).

x

50 m, y

25 m

1 2 x 100 x 2 1 x 2 100 x 2500 1250 2 1 2 x 50 1250 2 1250 m 2 is the maximum.

50 m, y

25 m

(f ) Letting h get closer and closer to 0, the slope approaches 4. Hence, the slope at 2, 4 is 4.

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Chapter 2

NOT FOR SALE

Polynomial ynomial and Rational Function Functions

ax b cx d

13. f x

f x has a vertical asymptote at x

d and a horizontal asymptote at y c

a . c

(i) a ! 0

(ii) a ! 0

b 0

b ! 0

c ! 0

c 0

d 0

d 0 d is positive. c

x

y

a is positive. c

d is negative. c

x

y

a is negative. c

Both asymptotes are negative on graph (b).

Both asymptotes are positive on graph (d). (iii) a 0

(iv) a ! 0

b ! 0

b 0

c ! 0

c ! 0

d 0

d ! 0

x

d is positive. c

x

y

a is negative. c

y

a is positive. c

The vertical asymptote is positive and the horizontal asymptote is negative on graph (a).

d is negative. c

The vertical asymptote is negative and the horizontal asymptote is positive on graph (c).

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