Calculus AP Edition 9th Edition Larson Solutions Manual Full Download: http://alibabadownload.com/product/calculus-ap-edition-9th-edition-larson-solutions-manual/
NOT FOR SALE CHAPTER P Preparation for Calculus
Section P.1
Graphs and Models.................................................................................2
Section P.2
Linear Models and Rates of Change....................................................11
Section P.3
Functions and Their Graphs.................................................................23
Section P.4
Fitting Models to Data..........................................................................33
Review Exercises ..........................................................................................................35 Problem Solving ...........................................................................................................41
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
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NOT FOR SALE C H A P T E R P Preparation for Calculus Section P.1 Graphs and Models 1. y
32 x 3
4 x2
7. y
x-intercept: (2, 0) y-intercept: (0, 3) Matches graph (b).
x
3
2
0
2
3
y
5
0
4
0
5
y
9 x2
2. y
6
x-intercepts: 3, 0 , 3, 0
(0, 4) 2
(− 2, 0)
y-intercept: (0, 3) −6
Matches graph (d).
(2, 0) x
−4
4
(− 3, − 5)
3. y
6
−2
(3, − 5)
−4
3 x2
−6
x-intercepts:
3, 0 , 3, 0
8. y
x 3 2
y-intercept: (0, 3)
Matches graph (a).
x
0
1
2
3
4
5
6
x3 x
y
9
4
1
0
1
4
9
4. y
x-intercepts: 0, 0 , 1, 0 , 1, 0
y 10
y-intercept: (0, 0)
8
Matches graph (c).
6
(0, 9)
4
5. y
2
1x 2
2
(6, 9)
(5, 4)
(1, 4) (2, 1)
(4, 1) x −6
x
4
2
0
2
4
y
0
1
2
3
4
9. y
y 6
(4, 4) 4
−4 − 2
−2
4
2
6
(3, 0)
x2
x
5
4
3
2
1
0
1
y
3
2
1
0
1
2
3
(2, 3)
(0, 2)
y
(−2, 1) 6
x −4
−2
(−4, 0)
2
4
−2
4
(− 5, 3)
(− 4, 2) 2
6. y
5 2x
(− 1, 1)
(− 3, 1) −6
x
1
0
1
2
5 2
y
7
5
3
1
0
3
4
1
3
−4
(1, 3) (0, 2) x
(− 2, 0)
2
−2
y 8
(−1, 7) (0, 5) 4 2
(1, 3) (2, 1)
INSTRUCTOR ST USE ONLY − −66 − −44 − −22 − −22 − −44
( ( 5,0 2
(3, − 1)
x
(4, (4 −3) 3)
© 2010 Brooks/Cole, Cengage Learning
2
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section P.1
x 1
10. y
3
2
1
0
1
2
3
y
2
1
0
1
0
1
2
x
6
4
3
2
1
y
14
12
1
Undef.
1
y
5 4 3 2
3
(3, 2)
2
(− 2, 1) 1
−1
(− 1, 0)
(−1, 1)
2
15.
x
0
1
4
9
16
y
6
5
4
3
2
2
8
12
(9, −3)
16
0 and y
0 when
(16, −2)
−4
(4, −4) (1, −5) −6 (0, −6)
16.
Xmin 20 Xmax 30 Xscl 5 Ymin 10 Ymax 40 Yscl 5
−8
x2
12. y
3 when x
Note that y 1. x
x 4
1 2 3
Xmin 5 Xmax 4 Xscl 1 Ymin 5 Ymax 8 Yscl 1
y
−2
1 4
−2 −3 −4 −5
(− 3, − 1)
x 6
−4
1 2
x −1
(− 6, − 14 ) (− 4, − 12 )
3
(1, 0) (0, − 1)
−2
11. y
2
(0, 12 ) (2, 14 )
(2, 1) x
−3 −2
0
y
4
(−3, 2)
3
1 x 2
14. y
x
Graphs and Models Graph
x
2
1
y
0
1
0 2
2
7
14
2
3
4
16 when x
Note that y
y
17. y
0 or 16.
5 x
5 5 4
(14, 4)
3
(−4.00, 3) (2, 1.73)
(7, 3)
(− 1, 1)
−6
(2, 2) (0, 2 )
2
6
−3
x
(− 2, 0)
13. y
5
10
15
20
3 x
x
3
2
1
0
1
2
3
y
1
32
3
Undef.
3
3 2
1
y
(a)
2, y
2, 1.73
(b)
x, 3
4, 3
18. y
(2, 32 (
x
−2
9
(1, −4)
1
−3 −2 −1 −1
1
2
(− 2, − 32 (
6
−6
(3, 1)
(−3, −1)
5 4
3 | 1.73
(−0.5, 2.47) −9
2
3
52
x5 5 x
(1, 3)
3
y
(a)
0.5, y
(b)
x, 4
3
0.5, 2.47 1.65, 4 and x, 4
1, 4
INSTRUCTOR S USE ONLY (−1, (−1, 1, −3) − 3))
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
4
NOT FOR SALE
Chapter P
Preparation paration for Calc Calculus
2x 5
19. y
5; 0, 5
25. y
2 x 5x
y-intercept: y
2 0 5
x-intercept: 0
2x 5
y-intercept: None. x cannot equal 0.
5
2x
x-intercept: 0
x
5; 2
52 , 0
4x2 3
20. y
y-intercept: y
4 0 3
x-intercept: 0
4 x2 3
2
3
3; 0, 3
4 x2
26. y
2 x 5x
0
2
x
4; 4, 0
x 2 3x
3 x
1
2
02 3 0
y-intercept: y
ª¬3 0 1º¼
None. y cannot equal 0.
y-intercept: y
02 0 2
y
2; 0, 2
22. y
2
0
0
x
2, 1; 2, 0 , 1, 0
x
x
2 x 1
x 4x 0; 0, 0
x-intercepts: 0
x3 4 x
x
0, r 2; 0, 0 , r 2, 0
x 24. y
x
1
y-intercept: y y x-intercept: 0 x
0; 0, 0
x 16 x 2
28. y
2x
x-intercept:
2 0
0
1; 1, 0
2x x2 1
3x 2
1
x2
1 3
x
r
1
x
x2 1 Note: x
02 1
x2 1
x2 1
4x2
x2 1
1
0; 0, 0
1; 0, 1
2x
4 x 4 x 0, 4, 4; 0, 0 , 4, 0 , 4, 0
x
0
x2 1
y
x
0 1 02 1; 0, 1
0; 0, 0
x
y-intercept: y
x 16 x 2 0 16 02
0
x-intercept: x 2 0 x 2 4 0
x x 2 x 2
0
0
y
0
x-intercepts: 0
2
y-intercept: 02 y 02 4 y
03 4 0
y
y-intercept: y
3x 1 x x 3 2 3x 1 0, 3; 0, 0 , 3, 0
27. x 2 y x 2 4 y
3
y-intercept: y 2
23. y
x 2 3x
x-intercepts: 0
x2 x 2
x-intercepts: 0
2
0; 0, 0
y
x2 x 2
21. y
x
3 3 3 § 3 ;¨ , 3 ¨© 3
· 0 ¸¸ ¹
3 3 is an extraneous solution.
29. Symmetric with respect to the y-axis because
y
x 2
6
x 2 6.
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
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NOT FOR SALE Section P.1
x2 x
30. y
31. Symmetric with respect to the x-axis because
y 2
Symmetry: None y
32. Symmetric with respect to the origin because
y
3
1
x x
−1
2
32 x 6
42. y
Intercepts: 0, 6 , 4, 0 Symmetry: None
10.
xy 2
4
y 8
x 3
6
No symmetry with respect to either axis or the origin.
4 x
xy
2
4 x2
2
(4, 0) −2
0 0.
4
2
(8, 0) x
−2 −2
x2 is symmetric with respect to the y-axis x 1
x 2 x 1
x2 . 2 x 1
x 3 x
8
10
(0, − 4) −8
44. y
x x is symmetric with respect to the y-axis x
4
− 10
3
3
2
−6
2
x
x 8
y
x . x2 1
because y
6
Symmetry: none
2
39. y
4
Intercepts: 8, 0 , 0, 4
1
because y
1x 2
43. y
x
y
2 −2
37. Symmetric with respect to the origin because
x 2
(0, 6)
4
36. Symmetric with respect to the origin because
x y
3
−1
34. Symmetric with respect to the x-axis because
x y
x 2
x3 x.
33. Symmetric with respect to the origin because x y xy 4.
38. y
( 23 , 0(
3
y
y
(0, 2)
2
x x
y
35. y
23 , 0
Intercepts: 0, 2 ,
x3 8 x.
y2
5
2 3x
41. y
No symmetry with respect to either axis or the origin.
Graphs and Models Graph
x3 x .
2x 3
1
Intercepts: 0, 1 , 32 , 0 Symmetry: none y
40. y x
3 is symmetric with respect to the x-axis
2
because y x
3
y x
3.
(0, 1)
( 32 , 0)
x −1
1
2
−1 −2
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
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6
NOT FOR SALE
Chapter P
Preparation paration for Calculus Calc
9 x2
45. y
x3 2
49. y
Intercepts: 0, 9 , 3, 0 , 3, 0
Intercepts: 3 2, 0 , 0, 2
Symmetry: y-axis
Symmetry: none
y 10
y
(0, 9)
5 4
6
3
4
(0, 2)
2
(−3, 0) −6 −4 −2
(3, 0) x 2
−2
4
6
x
−3 −2
x2 3
46. y
1
(− 3 2, 0)
1
−1
2
3
x3 4 x
50. y
Intercepts: 0, 0 , 2, 0 , 2, 0
Intercept: (0, 3) Symmetry: y-axis
Symmetry: origin
y
y 12
3
9
(−2, 0)
(0, 0)
−3
−1
(0, 3) 3
x
47. y
3
−1
x
3
−2
x
−3
−6
(2, 0)
1
−3
6
2
51. y
x 5
x
Intercepts: 3, 0 , 0, 9
Intercepts: 0, 0 , 5, 0
Symmetry: none
Symmetry: none
y
y
12
3
10 8
2
(0, 9)
(− 5, 0)
(0, 0) x
−4 −3 −2 −1
1
2
2 − 10 − 8 −6
x
(−3, 0)
−2
2
−4
x 2 x 1
2x2 x
48. y
−3
4
Intercepts: 0, 0 ,
12 ,
0
Symmetry: none
52. y
25 x 2
Intercepts: 0, 5 , 5, 0 , 5, 0 Symmetry: y-axis
y
y 5
7 6
4 3 2
(− 12 , 0)
1
(− 5, 0)
(0, 0) x
−3
−2
−1
1
2
3
4 3 2 1
(0, 5)
(5, 0) x
−4 −3 −2 −1
1 2 3 4 5
−2 −3
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
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NOT FOR SALE Section P.1
53. x
y3
Graphs and Models Graph
7
6 x
57. y
Intercepts: 0, 6 , 6, 0 , 6, 0
Intercept: (0, 0) Symmetry: origin
Symmetry: y-axis
y
y
4 3
8
2
6
(0, 0)
4 x
− 4 − 3 −2 −1
1
2
3
2
(−6, 0)
4
(0, 6) (6, 0) x
−8
−2
−4 −2 −2
−3
−4
−4
−6
4
2
6
8
−8
54. x
y2 4
Intercepts: 0, 2 , 0, 2 , 4, 0
58. y
6 x
Intercepts: (0, 6), (6, 0)
Symmetry: x-axis
Symmetry: none
y
y 3
8
(0, 2)
(0, 6)
(−4, 0) −5
−2
x
−1
1
4
(0, −2)
2
(6, 0)
−3
x 2
55. y
8 x
59. y 2 x y2
Intercepts: none
y
Symmetry: origin
4
6
8
9 x 9 r
x9
Intercepts: 0, 3 , 0, 3 , 9, 0
y 8
Symmetry: x-axis
6 4
4
(0, 3)
2 x −2
2
4
6
(− 9, 0)
8
−11
1
(0, −3) −4
56. y
10 x2 1
60. x 2 4 y 2
Symmetry: y-axis
4 x2 2
Symmetry: origin and both axes
y
Domain: >2, 2@
12
(0, 10)
2
(0, 1)
(− 2, 0) −3
(2, 0) 3
2 − 6 − 4 −2
r
Intercepts: 2, 0 , 2, 0 , 0, 1 , 0, 1
Intercept: (0, 10)
10
4 y
(0, −1) x 2
4
6
−2
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
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Chapter P
8
61. x 3 y 2
Preparation paration for Calculus Calc
6
65. x 2 y
6 y
6 x2
x y
4 y
4 x
6 x
3y2
6 x
6 x 3
r
y
Intercepts: 6, 0 , 0,
2 , 0,
2
Symmetry: x-axis
4 x
0
x2 x 2
0
x
x
2, 1
2 x 1
The corresponding y-values are y
3
( 0,
2)
y (6, 0)
−1
8
( 0, −
66.
62. 3x 4 y 2
8 3x 8
4 y2
r
y Intercept:
3x 4
2
8, 3
0
Symmetry: x-axis
x
3 y2 y2
y
x 1
3 x
x
3 x
x2 2x 1
( 83 , 0)
x y
8 y
8 x
4x y
7 y
4x 7
x
1 or x
The corresponding y-value is y
5.
Point of intersection: (3, 5) 4 y 10 y
2 for x
1
2 .
4 x 10 2 4 x 10
5 x2
1 y 1
x 1
x
5 x
x 2x 1
0
2x2 2 x 4
x
1 or x
2
2
2 x 1 x 2
2
1 for x
1
2 .
Points of intersection: 1, 2 , 2, 1 68. x 2 y 2
25 y 2
25 x 2
3 x y
15 y
25 x
2
3 x 15
3 x
25 x
2
9 x 90 x 225
15
2
2
7x
14
0
10 x 2 90 x 200
x
2
0
x 2 9 x 20
0
x
x
4 or x
The corresponding y-value is y
2 for x
The corresponding y-values are y and y
3x 4 2 4 x 10 2
5 y2
2
x
3x 4 2 3x 4
2
5 x2
5x
4x 2 y
x 1 x 2
1 for x
x y
4x 7
64. 3x 2 y
2
x2 x 2
67. x 2 y 2
−6
3
3 x
Points of intersection: 1, 2 , 2, 1
12
15
1
0
and y
−6
2 and
1 .
The corresponding y-values are y
6
8 x
5 for x
2 for x
Points of intersection: 2, 2 , 1, 5
2)
−3
63.
2
1.
Point of intersection: 2, 1
5 x 4 5
The corresponding y-values are y and y
0 for x
3 for x
4
5 .
Points of intersection: 4, 3 , 5, 0
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section P.1
Graphs Graph and Models
72. Analytically,
y
x3
y
x
y
x4 2 x2 1
x3
x
y
1 x2
x3 x
0
1 x2
69.
x x 1 x 1 x
0
1, or x
0, x
1
The corresponding y-values are 1 for x y 0 for x 0 , y y
1 for x
1 , and
1 .
x4 2 x2 1
0
x4 x2
0
x 2 x 1 x 1
x
1, 0, 1.
Points of intersection: 1, 0 , 0, 1 , 1, 0 y = x 4 − 2x 2 + 1
Points of intersection: 0, 0 , 1, 1 , 1, 1
y
x 4x
y
x 2
x 4x
x 2
70.
x
x3 3x 2
0
1 x 2
0
2
(0, 1) (−1, 0)
3
(1, 0)
−2
y = 1 − x2
x 6
73. y
x2 4 x
y
2
1 or x
2
−3
3
3
x
9
4
The corresponding y-values are y 3 for x 1 and y 0 for x
2 .
Points of intersection: 1, 3 , 2, 0
y=
x+6
(3,
3)
(− 2, 2)
−7
2
− x 2 − 4x
y=
−2
71. Analytically,
y
x3 2 x 2 x 1
y
x 2 3x 1
x3 2 x 2 x 1
x 2 3x 1
x3 x 2 2 x
0
x x 2 x 1
0
x
x6
Analytically,
x6 3 x 2
0
x
y = x3 − 2x2 + x − 1
74. y
y
x2 4 x 0
1, 0, 2.
3 | 3, 1.732
x2 4x
x2 5x 6
x
Points of intersection: 1, 5 , 0, 1 , 2, 1 4
Points of intersection: 2, 2 , 3,
3, 2.
2x 3 6 6 x 7
−4
(2, 1)
(0, −1)
6
y=6−x
(1, 5)
(−1, −5)
(3, 3) −8
−4
8
y = − x2 + 3x − 1
−1
y = −⏐2x − 3⏐+ 6
Points of intersection: (3, 3), (1, 5) Analytically, 2 x 3 6 2x 3
2x 3 x
x or 2 x 3 3 or
x
6 x x
x 1.
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
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10
Chapter P
NOT FOR SALE
Preparation paration for Calculus Calc
75. (a) Using a graphing utility, you obtain
76. (a) Using a graphing utility, you obtain
0.027t 5.73t 26.9. 2
y (b)
y (b)
250
−5
0.77t 2 2.1t 4
240
35
18
0
−50
0
The model is a good fit for the data. (c) For 2010, t 40 and y 212.9. C
77.
5.5 x 10,000
5.5 x
2
30.25 x 0
The model is a good fit for the data. (c) For 2015, t
25 and y | 538 million subscribers.
R 3.29 x
3.29 x
10,000
2
10.8241x 2 65,800 x 100,000,000 10.8241x 2 65,830.25 x 100,000,000 Use the Quadratic Formula.
x | 3133 units
The other root, x | 2949, does not satisfy the equation R
C.
This problem can also be solved by using a graphing utility and finding the intersection of the graphs of C and R. 78. y
81. (a) If (x, y) is on the graph, then so is x, y by y-axis
10,770 0.37 x2
symmetry. Because x, y is on the graph, then so
400
is x, y by x-axis symmetry. So, the graph is
0
symmetric with respect to the origin. The converse is x3 has origin symmetry not true. For example, y but is not symmetric with respect to either the x-axis or the y-axis.
100 0
If the diameter is doubled, the resistance is changed by approximately a factor of 14 . For instance,
(b) Assume that the graph has x-axis and origin symmetry. If (x, y) is on the graph, so is x, y by
x-axis symmetry. Because x, y is on the graph,
y 20 | 26.555 and y 40 | 6.36125.
then so is x, y
79. Answers may vary. Sample answer:
y
x
x
4, x
symmetry. Therefore, the graph is symmetric with respect to the y-axis. The argument is similar for y-axis and origin symmetry.
4 x 3 x 8 has intercepts at 3, and x
x, y by origin
8.
80. Answers may vary. Sample answer:
y
x 32 x 4 x 52 has intercepts at
x
32 , x
4, and x
82. (a) v ªBecause y ¬
(b) i ª¬Because y
5. 2
3 x 3 2
3x3 3x
3 x 2 3º ¼ 3 x x 1 x 1 has x-intercepts 0, 0 , 1, 0 , 1, 0 º¼
(c) None of the equations are symmetric with respect to the x-axis 2 (d) ii ªBecause 2 3 ¬
1º and vi ª¬Because ¼
3 (e) i ªBecause 3 x 3 x ¬ 3 i ªBecause 3 0 3 0 (f) ( ¬
3 x 3 3 x
2 3
1º¼
yº and iv ª¬Because ¼
0º and iv ª¬Because ¼
3
x
3 x
yº¼
0º¼
INSTRUCTOR USE ONLY 3
0
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section P.2
83. False. x-axis symmetry means that if 4, 5 is on the
Linear Models and Rate Rates of Change y
87. 4
graph, then 4, 5 is also on the graph. For example,
4, 5 is not on the graph of 4, 5 is on the graph. 84. True. f 4
(0, 3)
y 2 29, whereas
x
11
2
(x, y)
1 −2
f 4 .
x
−1
(0, 0)
2
3
−1
§ b r 85. True. The x-intercepts are ¨ ¨ © § b 86. True. The x-intercept is ¨ , © 2a
b 2 4ac , 2a
· 0 ¸. ¸ ¹
x
2
· 0 ¸. ¹
0 y 3 2
x
2
0 y 0 2
2 4 ª x 2 y 3 º ¬ ¼
x2 y 2
4 x 2 4 y 2 24 y 36
x2 y 2
3 x 2 3 y 2 24 y 36
0
x y 8 y 12
0
2
2
x y 4 2
2
2
4
Circle of radius 2 and center (0, 4). 88. Distance from the origin x2 y 2
K u Distance from 2, 0 K
x2 y 2
x
2 y 2 , K z 1 2
K 2 x2 4 x 4 y2
1 K 2 x 2 1 K 2 y 2 4 K 2 x 4K 2
0
Note: This is the equation of a circle!
Section P.2 Linear Models and Rates of Change 1. m
1
y
8. m = −3
2. m
2
3. m
0
4. m
1
1
m=3
(−2, 5)
6
m=0
4
m=3 x −6
−2
2
4
−2
5. m
12
6. m
40 3
2 4 53
9. m
6 2
3
y
7.
m = −2
y 3
m is undefined.
(5, 2) x
−1
m=1
8
(3, 4)
4
2
3
5
6
7
−3 −4
2 x −2
1
−2
6
−6 −4
2 1
m=−3 2
2
4
8 10
(3, −4)
−5
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
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12
NOT FOR SALE
Chapter P
Preparation paration for Calculus Calc
7 1 2 1
10. m
6 3
2
14. m
y
(−2, 7)
7
§ 3· § 1· ¨ ¸ ¨ ¸ © 4¹ © 4¹ §7· §5· ¨ ¸¨ ¸ ©8¹ © 4¹
6
1 3 8
8 3
y
5 3
3 2
2
(1, 1)
1
x
−4 −3 −2 − 1
1
3
1
4
( 78 , 34 ) x
−2
16 44
11. m
5 , undefined. 0
1
−1
( 54 , − 14 )
15. Because the slope is 0, the line is horizontal and its equation is y 2. Therefore, three additional points are
The line is vertical y
0, 2 , 1, 2 , 5, 2 .
7
(4, 6)
6
−1
5
16. Because the slope is undefined, the line is vertical and its 4. Therefore, three additional points equation is x
4 3 2
are 4, 0 , 4, 1 , 4, 2 .
(4, 1)
1
x −2 −1
1
2
3
5
6
17. The equation of this line is 5 5 53
12. m
0 2
The line is horizontal 1 2
3
4
5
y
3x 10.
18. The equation of this line is
x 1
3 x 1
Therefore, three additional points are (0, 10), (2, 4), and (3, 1).
y
−1 −1
y 7 0
6
y 2
−2
y
−3 −4
2 x 2.
Therefore, three additional points are 3, 4 , 1, 0 ,
(3, −5) (5, − 5)
−6
2 x 2
and (0, 2). 2 1 3 6 1 § 3· ¨ ¸ 2 © 4¹
13. m
1 2 1 4
19. (a) Slope
2
'y 'x
(b) x 10 ft
y
30 ft
3 2
(− 12 , 23 ) −3
−2
By the Pythagorean Theorem,
(− 34 , 16 ) x 1
−1 −2 −3
1 3
2
x2
3
x
302 102
1000
10 10 | 31.623 feet.
20. (a) m 800 indicates that the revenues increase by 800 in one day.
(b) m 250 indicates that the revenues increase by 250 in one day. (c) m 0 indicates that the revenues do not change from one day to the next.
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
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Section P.2
25. x 5 y
y
Population (in millions)
21. (a)
Linear Models and Rat Rate Rates of Change
300
20 15 x 4
y
295 290
Therefore, the slope is m
285 280
t 1
2
3
4
5
Year (0 ↔ 2000)
26. 6 x 5 y
15
y
6x 5
(b) The slopes are: 285.3 282.4 10 288.2 285.3 2 1 291.1 288.2 32 293.9 291.1 43 296.6 293.9 54
3
Therefore, the slope is m
2.9
6 5
and the y-intercept is
0, 3 .
2.9
27. x
4
The line is vertical. Therefore, the slope is undefined and there is no y-intercept.
2.9 2.8
1
28. y
The line is horizontal. Therefore, the slope is m the y-intercept is 0, 1 .
2.7
The population increased least rapidly from 2004 to 2005.
29.
3x 4
y
0 and
3
3x 12
4y
r
3x 4 y 12
0
100 80
y
60
5
40
4
20
(0, 3) 2
t 5
10 15 20 25 30
1 x
(b) The slopes are: 74 57 10 5 85 74 15 10 84 85 20 15 61 84 25 20 43 61 30 25
−4 −3 −2 −1
1
3.4
30. The slope is undefined so the line is vertical. 2.2
x x 5
0.2
5 0 y 1
4.6
x −4 −3 −2 −1
3.6
(−5, −2)
−1
1
−2 −3
The rate changed most rapidly between 20 and 25 seconds. The change is 4.6 mi h sec. 23. y
15 and the y-intercept is
(0, 4).
275
22. (a)
13
−4 −5
4x 3
the slope is m 24. x y
y
4 and the y-intercept is 0, 3 .
1 x 1
The slope is m
1 and the y-intercept is (0, 1).
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
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14
31.
NOT FOR SALE
Chapter P y
2x 3
3y
2x
Preparation paration for Calculus Calc
y 0
2x 3y
0
80 2 40 2 x 0
m
35.
y
y
2x
0
2x y
4 y
3
(4, 8)
8
2
6
(0, 0)
4
x 1
2
3
4
2
−1
(0, 0) −4
y
4
y 4
0
32.
2
y 0
5
6
5 x
y
(0, 4)
5x y
3
4
50 5 1 0 5 x 0
m
36.
y
x
−2
0
2
y
1
(−1, 5) −3
−2
1
4
2
3 x 3
33. y 2 y 2
1
3x 9
x
3x 11
0
3x y 11
3 2 1 x 1
2
3
4
5
6
2
3
2
y 1
2 x 2
y 1
2x 4 2x y 3
0
(3, − 2)
1
1 3 20
37. m
y
−2
(0, 0)
−3 −2 −1 −1
y
−2 −1 −1
5
x
−1
−3
y
−4 −5
2
(2, 1)
1
53 x 2
y 4
34.
5 y 20
3 x 6
3x 5 y 14
x −2 −1 −1
2
3
4
5
−2 −3
(0, − 3)
0
y 5 4
(−2, 4) 2 1 x −3
−2
−1
1
2
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section P.2
7 2 1 2
38. m
9 3
3
3 x 2
y 2
3 x 2
y 2 y
3x 4
0
3x y 4
Linear Models and Rat Rate Rates of Change 83 66
41. m
15
5 , undefined 0
The line is horizontal. x
6
x6
0
y
(6, 8)
8
y
6
8
4
(1, 7)
(6, 3) 2
6 4
x
−2
2
4
8
−2
x −6 −4
2
(−2, −2)
4
6
−4
80 25
39. m
8 3
y 8 x 3 y 40
y 2
x
−1
1
43. m
(5, 0)
x
6 7 8 9
62 3 1
y
4 4
1
y 2
1 x 1
y 2
x 1
(1, −2)
4
(3, −2)
7 3 2 4 1 0 2 3 4
11 4 1 2
11 2
11 x 0 2 11 3 x 2 4 22 x 4 y 3
y 0
y
0 4
y
3
( 12 , 72 )
2
7
1
6
(− 3, 6)
3
−4
(2, 8)
x y 3
2
−1
−2
40. m
0
1
−3
1 2 3 4
0
y
y
−1
0 2
2
y
8 x 5 3 8 40 x 3 3 0
y 0
9 8 7 6 5 4 3 2 1
2 2 31
42. m
( 0, 34 ) x
−4 −3 −2 −1
5
1
2
3
4
3
(1, 2)
2 1
x −4 −3 −2 −1
1
2
3
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
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16
NOT FOR SALE
Chapter P
Preparation paration for Calculus Calc
§ 3· § 1· ¨ ¸ ¨ ¸ © 4¹ © 4¹ §7· §5· ¨ ¸¨ ¸ ©8¹ © 4¹
44. m
1 3 8
8 3
8 § 5· ¨x ¸ 3© 4¹ 32 x 40
1 y 4 12 y 3 32 x 12 y 37
0
3 2
( 78 , 34 )
1
x −2
−1
1
x
3
x 3
0
45.
50.
y 2 1
(3, 0) 1
x y a a 1 2 a a 3 a a
2
x
2 0
1 1
x y a a 3 4 a a 1 a a
3 x y
3
x y 3
0
1 1 1 1 x y x y 1
4
−1
1 0
3
51. y
−2
1
1
( 54 , − 14 )
−1
1
3x y 2
49.
y
x y 2 2 3 y 3x 2 2 3x y
48.
y
2
46. m
b a
y b x y a x y a b
1 x
−3 −2 −1
1
2
3
4
5
−2
b xb a
−4 −5 −6
b 52. x
1
4 y
y 3 2
(0, b) 1 x 1
2
3
5
−1
(a, 0)
−2 x
53. y 47.
x y 2 3 3x 2 y 6
2 x 1 y
1
3
0 1 x −2
−1
1
2
INSTRUCTOR USE S ONLY −1 −1
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section P.2
1
1x 3
54. y
Linear Models and Rat Rate Rates of Change
59. (a)
17
5
y
−5
5
2 1 x −3 −2 − 1
−5
3
(0, −1)
The lines do not appear perpendicular.
−2
4
(b)
−3 −4
−6
55. y 2
3 2
x
3x 2
y
6
1 −4
1 2
The lines appear perpendicular.
y
The lines are perpendicular because their slopes 2 and 12 are negative reciprocals of each other. You must use
4 3 2 1 x − 4 − 3 −2
1
2
3
4
−2
a square setting in order for perpendicular lines to appear perpendicular. Answers depend on calculator used. 60. ax by
−3 −4
56. y 1
3 x 4
y
3 x 13 y
4
(a) The line is parallel to the x-axis if a b z 0.
0 and
(b) The line is parallel to the y-axis if b a z 0.
0 and
(c) Answers will vary. Sample answer: a b 8.
16 12
5 x 8 y
4
y
1 8
x − 16 − 12 − 8
4
−4
8
2x 3
5x 2 y
4
y
1 2
y
(e) a
1 x −2
−1
5 2
and b
−1
5x 2
3y
4
−2
5x 6 y
8
2
5x 8
1 2
Answers will vary. Sample answer: a b 2.
0
y
4
(d) The slope must be 52 .
−8
57. 2 x y 3
5 x
5 and
3
5 x 4
5 and
52 x 2
3.
−3
61. The given line is vertical. 58. x 2 y 6
y
0 12 x
3
(a) x
7, or x 7
0
(b) y
2, or y 2
0
62. The given line is horizontal.
y 4 2
− 10
−8
−6
(a) y
0
(b) x
1, or x 1
0
x
−2
−4
INSTRUCTOR T USE ONLY −6
© 2010 Brooks/Cole, Cengage Learning
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18
NOT FOR SALE
Chapter P
63. 4 x 2 y
Preparation paration for Calculus Calc
3 2x
y m (a) y 1
2 2 x 2
y 1
V
x 2
x 2y 4
m
(2, 4)
−3
1 x 3
y 2
x 3
You can use the graphing utility to determine that the points of intersection are (0, 0) and (2, 4). Analytically,
y 2 0
65. 5 x 3 y
0
0 0, 0
x
2 y
4 2, 4 .
The slope of the line joining (0, 0) and (2, 4) is m 4 0 2 0 2. So, an equation of the line is y 0 5 3
24 y 21
x 43 72. y
53 x
7 8
40 y 35
3 4
y
34 x
m
34 y 5
34 x 3
4 y 20
3 x 12
0
x
4x 3
9
4
x2 2x 3
2 x2 6 x
0
2 x x 3
0
x
0 y
3 0, 3
x
3 y
0 3, 0 .
The slope of the line joining (0, 3) and (3, 0) is m 0 3 3 0 1. So, an equation of the line is
0 4 3
(3, 0)
x2 4 x 3
34 x 4
3x 4 y 8
(0, 3) −9
−6
7 4
y 5
x2 2x 3
You can use the graphing utility to determine that the points of intersection are (0, 3) and (3, 0). Analytically,
3 x 7
3 y 15
x 2 4 x 3, y
0
7
y 5
2 x.
6
24 x 18
24 x 40 y 53
(b) y 5
2 x 0
40 x 24 y 9 y
4y
y
40 x 30
0
66. 3x 4 y
0 0 y
7 8
(b)
2 x x 2 x
5 3
y
0
x y 5
5x 3
m
2x2 4x
x 3
0
y
4x x2
x2
1 x 3
(b) y 2
6
(0, 0) −1
1 y 2
x y 1
4.
5600t 267,400
5
0
x 7
(a)
8.
1600t 30,000
245,000 when t
5600 t 4 245,000
71.
7
y
17,200 when t
1600 t 8 17,200
70. The slope is 5600. V
12 x 2
y 1 2y 2
(a)
V
2x y 3
(b)
(a)
69. The slope is 1600. V
2x 4
0
64. x y
68. The slope is 4.5. V 156 when t 4. V 4.5 t 4 156 4.5t 138
3 2
y 3
16 3
y
1 x 0 x 3.
4 x 16 4 x 3 y 31
67. The slope is 250. V 1850 when t 8. V 250 t 8 1850 250t 150.
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section P.2
10 2 1
73. m1
2 3
m1 z m2 The points are not collinear. 6 4 70 11 4 m2 5 0 m1 z m2
10 7 7 5
74. m1
19
77. Equations of altitudes: a b y x a c x b a b y x a c Solving simultaneously, the point of intersection is § a 2 b2 · ¨ b, ¸. c © ¹
1
2 0 2 1
m2
Linear Models and Rate Rat Rates of Change
y
(b, c)
The points are not collinear. (a, 0)
75. Equations of perpendicular bisectors:
c y 2
a b§ a b· ¨x ¸ c © 2 ¹
c 2
a b§ b a· ¨x ¸ c © 2 ¹
y
x
(− a, 0)
§b c· 78. The slope of the line segment from ¨ , ¸ to © 3 3¹
Setting the right-hand sides of the two equations equal and solving for x yields x 0. Letting x 0 in either equation gives the point of intersection: § a 2 b 2 c 2 · ¨ 0, ¸. 2c © ¹
§ a2 b2 · ¨ b, ¸ is: c © ¹
m1
3a 2 3b2 c 2 3c 2b
This point lies on the third perpendicular bisector, x 0.
§b c· The slope of the line segment from ¨ , ¸ to © 3 3¹
(b, c)
m2
( a +2 b , 2c )
b 3
(a, 0)
m1 76. Equations of medians:
y y y
ª a 2 b 2 c 2 2c º c 3 ¬ ¼ 0 b 3
3a 2 3b2 3c 2 2c 2 6c
x
(−a, 0)
3a 2 3b 2 c 2 2bc
m2
Therefore, the points are collinear.
c x b
79. Find the equation of the line through the points (0, 32) and (100, 212).
c x a 3a b c x a 3a b
m F 32 F
§b c· Solving simultaneously, the point of intersection is ¨ , ¸. © 3 3¹ y
( b −2 a , 2c )
3
3a 2 3b 2 c 2 2bc
§ a 2 b2 c 2 · ¨ 0, ¸ is: 2c © ¹
y
( b −2 a , 2c )
ª a 2 b 2 cº c 3 ¬ ¼ b b 3
180 9 100 5 9 C 0 5 9 C 32 5
or C
1 9
5 F
160
5F 9C 160 0 72q, C | 22.2q. For F (b, c)
80. C
(
a + b, c 2 2 x
)
For x
0.48 x 175 137, C
0.48 137 175
$240.76.
INSTRUCTOR NSTR NST N NS ST S TR USE ONLY ((− −a a,, 0) 0
(0, 0) (a, (a, 0) 0
© 2010 Brooks/Cole, Cengage Learning
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20
NOT FOR SALE
Chapter P
81. (a) W1
W2 (b)
Preparation paration for Calculus Calc
0.75 x 14.50
84. (a) y
18.91 3.97 x
1.30 x 11.20
x
quiz score, y
25
(b)
test score
100
(6, 19)
−2
10
0
0
0.75 x 14.50 3.3 6
1.30 x 11.20 0.55 x
(c) If x
19.
(c) When six units are produced, the wage for both options is $19.00 per hour. Choose option 1 if you think you will produce less than six units per hour, and choose option 2 if you think you will produce more than six.
86.4.
(d) The slope shows the average increase in exam score for each unit increase in quiz score. (e) The points would shift vertically upward 4 units. The new regression line would have a y-intercept 4 greater than before: y 22.91 3.97 x. 85. The tangent line is perpendicular to the line joining the point (5, 12) and the center (0, 0). y
(5, 12) 8 4 −8 −4
82. (a) Depreciation per year: 875 5
18.91 3.97 17
17, y
x
1.30 6 11.20
y
20 0
Using a graphing utility, the point of intersection is (6, 19) Analytically, W1 W2
x
(0, 0) 8
16
−8
$175 −16
875 175 x
y
Slope of the line joining (5, 12) and (0, 0) is
where 0 d x d 5.
12 . 5
The equation of the tangent line is
1000
5 12
y 12
5 x 12
y 0
5 x 12 y 169
6
x 5
169 12
0.
0
875 175 2
(b) y (c)
200 175 x
86. The tangent line is perpendicular to the line joining the point 4, 3 and the center of the circle, (1, 1).
$525
875 175 x 675
y
x | 3.86 years
4 2
83. (a) Two points are (50, 780) and (47, 825). The slope is 825 780 45 m 15. 47 50 3 p 780 p
−6
(b)
15 x 1530
4
Slope of the line joining (1, 1) and 4, 3 is
13 1 4
4 . 3
Tangent line: 1 1530 p 15
y 3
50
y 0
0
1600
855, then x
45 units. 1 1530 795 5 15
3 x 4 4 3 x6 4 3 x 4 y 24
87. 4 x 3 y 10
0
If p
2
(4, −3)
or x
x −2
−6
15 x 50 15 x 750 780
−2
(1, 1)
0 d
4 0 3 0 10 42 32 10 5
INSTRUCTOR USE ONLY (c) If p
795, then x
49 units.
2
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section P.2
88. 4 x 3 y 10
89. x y 2
4 2 3 3 10
0 d
0 d
42 32
93. If A
If B
1 0 1 1 5
1 0 2
d
7
2
3 4
A2 B 2
Ay Ay1 Bx1 Ay1
Bx Ay
2 2.
1 is 1, 1 . The
A2 B 2
3 4 10 5
9 . 5
C A. The distance to x1 , y1 is
0 is the vertical line x Ax1 By1 C
2
0 is
.
.
The equation of the line through x1 , y1 perpendicular to Ax By C B x x1 A Bx Bx1
4 2
C B. The distance to x1 , y1 is
(Note that A and B cannot both be zero.) The slope of the line Ax By C
y y1
2
3 1 4 1 10 2
Ax1 By1 C
Ax1 C A
2
distance from the point 1, 1 to 3x 4 y 10
0 is the horizontal line y
By1 C B
15
1 1 2
92. A point on the line 3x 4 y
5 2 2
1 6 0 2 1
0, then Ax C § C · x1 ¨ ¸ © A ¹
d
12 12
0, then By C § C · y1 ¨ ¸ © B ¹
d
d
0 d
21
91. A point on the line x y 1 is (0, 1). The distance from the point (0, 1) to x y 5 0 is
7 5
1 2 1 1 2 5 2
90. x 1
Linear Models and Rat Rate Rates of Change
0 is A B.
0 is:
The point of intersection of these two lines is: Ax By Bx Ay
B x1 ABy1
B x
1 2
AC B 2 x1 ABy1
By adding equations 1 and 2
x
AC B x1 ABy1 A2 B 2
A2 x ABy Bx1 Ay1 B 2 x ABy
AC
C
A
2
2
2
2
Ax By Bx Ay
C
ABx B 2 y Bx1 Ay1 ABx A2 y
A2 B 2 y y
3 4
BC ABx1 A2 y1
BC ABx1 A2 y1 By adding equations 3 and 4 BC ABx1 A2 y1 A2 B 2
§ AC B 2 x1 ABy1 BC ABx1 A2 y1 · , ¨ ¸ point of intersection A2 B 2 A2 B 2 © ¹
The distance between x1 , y1 and this point gives you the distance between x1 , y1 and the line Ax By C 2
d
ª AC B 2 x1 ABy1 º ª BC ABx1 A2 y1 º x1 » « y1 » « 2 2 2 2 A B A B ¬ ¼ ¬ ¼ 2
ª AC ABy1 A2 x1 º ª BC ABx1 B 2 y1 º « » « » 2 2 A B A2 B 2 ¬ ¼ ¬ ¼ ª A C By1 Ax1 º ª B C Ax1 By1 º « » « » 2 2 A B A2 B 2 ¬ ¼ ¬ ¼ 2
2
0.
2
2
A2 B 2 C Ax1 By1 2 2 A2 B 2
Ax1 By1 C A2 B 2
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
22
NOT FOR SALE
Chapter P
Preparation paration for Calculus Calc
mx 4 mx 1 y 4
94. y
m3 1 1 4
Ax1 By1 C
d
0
m 1
A2 B 2
2
3m 3
m2 1
2
1. In this case, the line y
The distance is 0 when m
x 4 contains the point (3, 1).
8
−9
9
(−1, 0) −4
95. For simplicity, let the vertices of the rhombus be (0, 0), (a, 0), (b, c), and a b, c , as shown in the figure. The
slopes of the diagonals are then m1
c and a b
c . Because the sides of the rhombus are ba
m2
y2* y1* x2* x1*
y2 y1 x2 x1
y
b 2 c 2 , and you have
equal, a 2
c2 2 b a2
c c a b ba
m1m2
97. Consider the figure below in which the four points are collinear. Because the triangles are similar, the result immediately follows.
c2 c 2
(x 2 , y2 )
1.
Therefore, the diagonals are perpendicular.
(x *2 , y*2 )
(x1, y1 ) (x *1, y*1 )
y x
(b, c)
1 m2 , then m1m2
98. If m1
(a + b, c)
1. Let L3 be a line with
slope m3 that is perpendicular to L1. Then m1m3 x
(0, 0)
m3 L 2 and L 3 are parallel. Therefore,
So, m2
(a , 0)
1.
L 2 and L1 are also perpendicular. 96. For simplicity, let the vertices of the quadrilateral be (0, 0), (a, 0), (b, c), and (d, e), as shown in the figure. The midpoints of the sides are §a · §a b c · §b d c e· §d e· , ¸, ¨ , ¨ , 0 ¸, ¨ ¸, and ¨ , ¸. 2¹ © 2 2 ¹ ©2 ¹ © 2 © 2 2¹
The slope of the opposite sides are equal: c 0 2 a b a 2 2 e 0 2 a d a 2 2
c e 2 b d 2 c ce 2 2 b b 2 2
e 2 d 2
d
c b
99. True. ax by
c1 y
bx ay
c2 y
m2
a c x 1 m1 b b b c2 m2 x a a
a b
b a
1 m1
100. False; if m1 is positive, then m2
1 m1 is negative.
e a d
Therefore, the figure is a parallelogram. y
(d, e)
( b +2 d ,
c+e 2
)
(b, c)
( d2 , 2e )
(a +2 b , 2c )
INSTRUCTOR S USE ONLY x
(0, 0)
( a2 , 0)
((a, a, 0)
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section P.3
Functions and T Their Graphs
23
Section P.3 Functions and Their Graphs 1. (a) Domain of f: 4 d x d 4 >4, 4@
4. (a) f 4
Range of f: 3 d y d 5 >3, 5@
(b) f 11
Domain of g: 3 d x d 3 >3, 3@
(c) f 8
Range of g: 4 d y d 4 >4, 4@
(d) f x ' x
(b) f 2 g 3
1
g x for x
(d) f x
2 for x
1
(c) g 2
(e) g x
0 for x
1, 1 and 2
(d) g t 1
(b) g
1
6. (a) g 4
Range of g: 4 d y d 2 > 4, 2@
(c) g c
5 02
5
5
5
5 2
(d) f x
2 for x
4, 4
(e) g x
0 for x
1
3. (a) f 0
7 0 4
(b) f 3 (c) f b
2 and x
7 b 4
(d) f x 1
f x 'x f x 'x
10.
f x f 1 x 1
11.
f x f 2 x2
7b 4 7 x 11
3
3 x 1 3 1 x 1
1
3 x 1 x 1
c 3 4c 2
4 t 4 4
t
4 t
2 2
§ § S ·· cos¨ 2¨ ¸ ¸ © © 4 ¹¹ § § S ·· cos¨ 2¨ ¸ ¸ © © 3 ¹¹
1
§ S· cos¨ ¸ © 2¹ cos
2S 3
0
1 2
0
§ 5S · sin ¨ ¸ © 4 ¹
§ 2S · (c) f ¨ ¸ © 3 ¹
§ 2S · sin ¨ ¸ © 3 ¹
3
t 3 8t 2 16t cos 0
§ 5S · (b) f ¨ ¸ © 4 ¹
2
45 8
t
sin S
x 3 3 x 2 ' x 3 x 2 'x ' x x 3 'x
52
2 2
3 2
3 x 2 3 x'x 'x , 'x z 0 2
3, x z 1
x 1 1 x2
1 x 1 1 2 x 1 1
x f x f 1 x 1
5 t 2 2t 1
2
9 4
cos 2 0
8. (a) f S
'x x3 'x
1
0
c 2 c 4
§S · (c) f ¨ ¸ ©3¹ 25
x
0
54
2
§ S· (b) f ¨ ¸ © 4¹
7 x 1 4
55
2
4 2 4 4
7. (a) f 0
4
4
7 3 4
2
5 t 1
2
g x for x
4
x 'x 5
(d) g t 4
2
(c) f x
16
3, undefined
32 32 32 4
(b) g
g 3
1
4 2t t 2
Domain of g: 4 d x d 5 >4, 5@ (b) f 2
1
8 5
5
(c) f x
Range of f: 4 d y d 4 > 4, 4@
12.
11 5
5. (a) g 0
4
2. (a) Domain of f: 5 d x d 5 >5, 5@
9.
4 5
x3 x 0 x 1
x 1 x 1
x x 1 x 1 x 1
2 x
x 2
x 11
x 1
1
x 11
x 1
, x z 2
x x 1 , x z 1
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
24 Chapter P 13. f x
Preparation aration for Calculus Calcu
22. f x
4x2
Domain: f, f
x 2 3x 2 t 0
x
Range: >0, f 14. g x
2 x 1 t 0
Domain: x t 2 or x d 1 x2 5
Domain: f, 1@ >2, f
Domain: f, f
15. g x
1 cos x z 0
6x
cos x z 1
Domain: 6 x t 0
Domain: all x z 2nS , n an integer
x t 0 >0, f
1 z 0 2 1 sin x z 2
x 3
sin x
Domain: x 3 t 0 >3, f Range: f, 0@ 17. f t
St 4
z
sec
2n
Domain: all x z
St 4
Range: f, 1@ >1, f
2nS ,
5S 2nS , n integer 6
x3 z 0 x3 z 0 Domain: f, 3 3, f
nS , n an integer
1 x 4
26. g x
Range: f, f 19. f x
6
Domain: all x z 3
cot t
Domain: all t
S
1 x3
25. f x
1 S t z 4n 2 2
Domain: all t z 4n 2, n an integer
18. h t
1 sin x 1 2
24. h x
Range: >0, f
2 1 cos x
23. g x
Range: >5, f
16. h x
x 2 3x 2
2
x2 4 z 0
3 x
x
2 x 2 z 0
Domain: all x z 0 f, 0 0, f
Domain: all x z r 2
Range: f, 0 0, f
Domain: f, 2 2, f
20. g x
27. f x
2 x 1
2 x 1, x 0 ® ¯2 x 2, x t 0
Domain: f, 1 1, f
(a) f 1
Range: f, 0 0, f
(b) f 0
2 0 2
2
(c) f 2
2 2 2
6
21. f x
x
1 x
2 1 1
1
x t 0 and 1 x t 0
(d) f t 2 1
2 t 2 1 2
x t 0 and x d 1
(Note: t 2 1 t 0 for all t)
Domain: 0 d x d 1 >0, 1@
Domain: f, f
2t 2 4
INSTRUCTOR USE ONLY Range: f f, 1 >2, 2 f
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section P.3
28. f x
2 ° x 2, x d 1 ® 2 °¯2 x 2, x ! 1
(a) f 2
2 2
2
(b) f 0
02 2
2
(c) f 1
12 2
(d) f s 2 2
Functions and T Their Graphs
25
4 x
32. g x
Domain: f, 0 0, f
6
Range: f, 0 0, f y
3
6 4
2 s 2 2 2 2
2 s 4 8s 2 10
2
x 2
(Note: s 2 2 ! 1 for all s)
4
6
Domain: f, f Range: >2, f 29. f x
33. h x
° x 1, x 1 ® °¯ x 1, x t 1
(a) f 3
y 3
3 1
4
(b) f 1
1 1
0
(c) f 3
3 1
2
(d) f b 2 1
2 1 x 3
b 2 1 1
b 2
Domain: f, f
12
Domain: x 6 t 0
34. f x
3
1 x3 4 y
3 4
1
1
5 4
(b) f 0
0 4
2
(c) f 5
5 4
3
(d) f 10
10 5 2
2
Domain: >4, f
1 x
25
−3
−1 −1
1
2
3
Domain: f, f
Range: >0, f 31. f x
9
Range: >0, f
° x 4, x d 5 ® 2 °¯ x 5 , x ! 5
(a) f 3
6
x t 6 >6, f
Range: f, 0@ >1, f 30. f x
x 6
Range: f, f
4 x
35. f x
Domain: f, f
9 x2
Domain: >3, 3@
Range: f, f
Range: >0, 3@
y
y
8 5
6
4
4
2 1
2
x −4 −3 −2 −1
x −4
−2
2
4
1
2
3
4
−2
INSTRUCTOR USE ONLY −3
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
26
NOT FOR SALE
Chapter P
36. f x
Preparation paration for Calculus Calc
x
27
Domain: >2, 2@
18
Range: ª¬2, 2 2 º¼ | >2, 2.83@
9
y-intercept: 0, 2
x-intercept:
2, 0
t1
4
2, 0( x 1
−1
2
3
0 y
x2 4 y
42.
4
−2
t
r
x
0 y
x2 4
y is a function of x. Vertical lines intersect the graph at most once.
−3 −4
37. g t
t3
y is not a function of x. Some vertical lines intersect the graph twice.
3
(0, 2)
−4 −3 −2
t2
41. x y 2
y
(−
d
40.
4 x2
43. y is a function of x. Vertical lines intersect the graph at most once.
3 sin S t y 3
44. x 2 y 2
4
y
r
2 1 t 1
y is not a function of x. Some vertical lines intersect the graph twice.
3
45. x 2 y 2
Domain: f, f
5 cos
46. x 2 y
T
r 16 x 2
16 y
16 x 2
y is a function of x because there is one value of y for each x.
2
Domain: f, f
47. y 2
Range: >5, 5@
x2 1 y
r
x2 1
y is not a function of x because there are two values of y for some x.
y 5 4 3 2 1
−2 π
16 y
y is not a function of x because there are two values of y for some x.
Range: >3, 3@ 38. h T
4 x2
48. x 2 y x 2 4 y 2π
θ
0 y
x2 x 4 2
y is a function of x because there is one value of y for each x.
−5
49. y
20 1 mi min during the first 4 40 2 minutes. The student is stationary for the next 2 minutes. 62 Finally, the student travels 1 mi min during 10 6 the final 4 minutes.
f x 5 is a horizontal shift 5 units to the left.
Matches d.
39. The student travels
50. y
f x 5 is a vertical shift 5 units downward.
Matches b. 51. y
f x 2 is a reflection in the y-axis, a
reflection in the x-axis, and a vertical shift downward 2 units. Matches c. 52. y
f x 4 is a horizontal shift 4 units to the right,
INSTRUCTOR USE ONLY followed ollowed by a reflection in the x-axis. x-axis. Matches a.
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section P.3
53. y
f x 6 2 is a horizontal shift to the left 6
Functions and T Their Graphs
(f) The graph is stretched vertically by a factor of 14 .
units, and a vertical shift upward 2 units. Matches e. 54. y
27
y
f x 1 3 is a horizontal shift to the right 1 unit,
4 2
and a vertical shift upward 3 units. Matches g.
x −4
−2
2
4
6
55. (a) the graph is shifted 3 units to the left. y −6
4
56. (a) g x
x −6
−4
−2
2
4
f x 4
−2
g 6
f 2
−4
g 0
f 4
−6
1 3
Shift f right 4 units y
(b) The graph is shifted 1 unit to the right. 4
y
3
4
2
(6, 1)
1
2
x −1
x −2
2
4
6
8
1
2
3
5
6
−2
−2
(0, − 3)
−4
−4 −6
(b) g x
(c) The graph is shifted 2 units upward.
f x 2
Shift f left 2 units
y
y
6
4 3
4
2
(0, 1)
2
x
x −4
−2
2
4
−7 −6 −5 −4 −3
6
−1
−2
1
−2 −3
(− 6, − 3)
(d) The graph is shifted 4 units downward. y x −4
7
−2
2
4
(c) g x
−4
f x 4
Vertical shift upwards 4 units
6
−2
y
−4
6
(2, 5)
5
−6
4
−8 2 1
(− 4, 1)
x −5 −4 −3 −2 −1
(e) The graph is stretched vertically by a factor of 3. y x −4
−2
4
1
3
−2
(d) g x
f x 1
Vertical shift down 1 unit
6
y
−2 2
−4
1
−6 −8
2
(2, 0) x
5
4
3
−10
2
1
2
3
−3 −4
INSTRUCTOR USE E ONLY (−4, −4)
−5 −6
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
28
Chapter P
NOT FOR SALE
Preparation paration for Calculus Calc
(e) g x
2 f x
g 2
2 f 2
g 4
sin x S 2 1 is a horizontal shift
58. (a) h x
S 2 units to the left, followed by a vertical shift 1 unit upwards.
2
2 f 4
6
(b) h x
y
(2, 2)
sin x 1 is a horizontal shift 1 unit to the
right followed by a reflection about the x-axis.
2 1 x − 5 − 4 − 3 −2 −1
1
2
3
−3 −4 −5
(− 4, −6)
(f ) g x
1 2
f x
g 2
1 2
f 2
g 4
1 2
1 2
f 4 2
32
x 3
1
−1
2
−3
g 1
0
(c) g f 0
g 0
−5 −6
f x 2 1
(f) g f x
g
4
sin x, g x
§S · f¨ ¸ ©2¹ g 0
x 3
4
Vertical shift 2 units upward
(b) y
(e) f g x
f S x
(f) g f x
g sin x
61. f x
x
f
y
x 2 , g x
D g x
1 x
2
3
0
§S · sin ¨ ¸ ©2¹
1
§
2· ¸¸ 2 © ¹
S
S ¨¨
2 2
sin S x
S sin x
x
f g x f
1
sin 2S
§ § S ·· g ¨ sin ¨ ¸ ¸ © © 4 ¹¹
2
2
x 1, x t 0
1
0
§ 2· g ¨¨ ¸¸ © 2 ¹
1
2
Sx
f 2S
3
1
x2 1
x x
§ § S ·· (d) g ¨ f ¨ ¸ ¸ © © 4 ¹¹
y
15
(e) f g x
(c) g f 0
x 2
1
f 15
§ § 1 ·· (b) f ¨ g ¨ ¸ ¸ © © 2 ¹¹
−4
57. (a) y
(b) g f 1
(a) f g 2
3
−2
( − 4, − 32 )
0
60. f x
(2, 12 )
1 4
f 0
(d) f g 4
y
5
59. (a) f g 1
x x
2
x, x t 0
Domain: >0, f
4
−1
g
−2 −3
D f x
g f x
g x2
x2
x
No. Their domains are different. f D g
g
Domain: f, f
Reflection about the x-axis x 2
(c) y
D f for
x t 0.
y
4 3 2 1
x −1
1
2
3
4
5
6
−2
INSTRUCTOR USE ONLY Horizontal shift 2 units to the right
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section P.3
62. f x
f
x 2 1, g x
cos x
f g x
D g x
64.
f cos x
cos 2 x 1
g x 2 1
D f x
g
cos x 2 1
Domain: f, f
f
3 , g x x
§1· g¨ ¸ © x¹
D f x
+
x2 1 f x 2 1
2
§ 3· g¨ ¸ © x¹
+
−2
1 x 2
1 2x x
1 2 x
§ 3· ¨ ¸ 1 © x¹
65. (a)
9 x2 x2
9 1 x2
+ − − +
−1 − 1 2
+
0
+
+
x
1
2
Domain: f, 12 º¼, 0, f
3 x2 1
g f x
D f x
intervals where 1 2x and x are both positive, or both
Domain: all x z r1 f, 1 1, 1 1, f
g
x 2
negative.
f g x
D g x
f
You can find the domain of g D f by determining the
No, f D g z g D f . 63. f x
D g x
29
Domain: 2, f
Domain: f, f
g
f
Functions and T Their Graphs
Domain: all x z 0 f, 0 0, f No, f D g z g D f .
f
f g 3
D g 3
4
(b) g f 2
g 1
(c) g f 5
g 5 , which is undefined
2
(d)
f
D g 3
f g 3
f 2
(e)
g
D f 1
g f 1
g 4
(f ) f g 1 66.
f 1
A D r t
3 2
f 4 , which is undefined
A r t
A 0.6t
S 0.6t
2
0.36S t 2
A D r t represents the area of the circle at time t. 67. F x
2x 2
Let h x
2 x, g x
x 2 and f x f g 2 x
Then, f D g D h x
x.
f 2 x 2
2 x
2
2x 2
F x .
[Other answers possible] 68. F x
4 sin 1 x 4 x, g x
Let f x
f
sin x and h x
f g 1 x
D g D h x
1 x. Then,
f sin 1 x
4 sin 1 x
F x .
[Other answers possible] 69. f x
x
2
4 x 2
x 2 4 x 2
f x
Even 70. f x
3
x
3 x
f x
Odd 71. f x
x cos x
x cos x
f x
Odd 72. f x
sin 2 x
sin x sin x
sin x sin x
sin 2 x
Even
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
30 Chapter P
Preparation aration for Calculus Calcu
32 , 4 is on the graph. f is odd, then 32 , 4 is on the graph.
73. (a) If f is even, then
(b) If
74. (a) If f is even, then 4, 9 is on the graph.
(b) If f is odd, then 4, 9 is on the graph. 75. f is even because the graph is symmetric about the y-axis. g is neither even nor odd. h is odd because the graph is symmetric about the origin.
81 7 53 2 7 y 1 x 3 2 7 21 y 1 x 2 2 7 19 y x 2 2 For the line segment, you must restrict the domain. 7 19 f x x , 3 d x d 5 2 2
78. Slope
76. (a) If f is even, then the graph is symmetric about the y-axis.
y
(5, 8)
8
y 6 6
f
4
4
2
(3, 1)
2
x x
−6 −4 −2 −2
2
4
−2
6
2
4
6
8
−2
−4
79. x y 2
−6
y 6
f
x
y
x
f x
x, x d 0
y
(b) If f is odd, then the graph is symmetric about the origin.
0
2
y
4 3
2
2
x
−6 −4 −2 −2
2
4
6
1
−4
x
−5 −4 −3 −2 −1
1
−6 −2 −3
4 6 2 0
77. Slope
10 2
y 4
5 x 2
y 4
5 x 10
5
80. x 2 y 2
y
5 x 6
y
36
2
36 x 2
y
36 x 2 , 6 d x d 6
y
For the line segment, you must restrict the domain. f x
5 x 6, 2 d x d 0
4 2
y x
(−2, 4)
6
−4 −2 −2
4
−4
2
4
2 x 2
−4 −6
(0, −6)
4
6
81. Answers will vary. Sample answer: Speed begins and ends at 0. The speed might be constant in the middle: y
Speed (in miles per hour) S
−6 − 4 −2
INSTRUCTOR U USE E ONLY x
Time (in hours)
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section P.3
82. Answers will vary. Sample answer: Height begins a few feet above 0, and ends at 0.
Functions and T Their Graphs
31
88. (a) For each time t, there corresponds a depth d.
(b) Domain: 0 d t d 5
y
Range: 0 d d d 30
Height
(c)
d 30 25 20 15
x
10
Distance 5
83. Answers will vary. Sample answer: In general , as the price decreases, the store will sell more. y
t 1
2
3
4
5
6
(d) d 4 | 18. At time 4 seconds, the depth is
Number of sneakers sold
approximately 18 cm. A
Average number of acres per farm
89. (a)
x
Price (in dollars)
500 400 300 200 100 t
84. Answers will vary. Sample answer: As time goes on, the value of the car will decrease
5 15 25 35 45 55
Year (5 ↔ 1955)
(b) A 20 | 384 acres farm
y
90. (a) Value
25
t
0
§ x · (b) H ¨ ¸ © 1.6 ¹
c x2
y
85.
c x2
y2 x2 y 2
25.
86. For the domain to be the set of all real numbers, you must require that x 2 3cx 6 z 0. So, the discriminant must be less than zero: 2
4 6 0
23
8 3
8 3
87. (a) T 4
(b) If H t
x x2
If x 0, then f x
x x 2
If 0 d x 2, then f x If x t 2, then f x
f x
c
6 c
91. f x
2 x 2.
x x 2
x x 2
2.
2 x 2.
So,
9c 2 24 c2
2
§ x · § x · 0.002¨ ¸ 0.005¨ ¸ 0.029 © 1.6 ¹ © 1.6 ¹ 0.00078125 x 2 0.003125 x 0.029
c, a circle.
For the domain to be >5, 5@, c
3c
100
0
8
8 3 2 3
2 x 2, x d 0 ° 0 x 2. ®2, °2 x 2, x t 2 ¯
6
16q, T 15 | 23q T t 1 , then the changes in temperature
will occur 1 hour later. (c) If H t
T t 1, then the overall temperature
INSTRUCTOR USE ONLY would be 1 degree lower.
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
32
NOT FOR SALE
Chapter P
92. p1 x
Preparation paration for Calculus Calc
x3 x 1 has one zero. p2 x
93. f x
x3 x has
three zeros. Every cubic polynomial has at least one zero. Given p x Ax3 Bx 2 Cx D, you have
a2 n 1 x
2 n 1
" a3 x a1 x 3
ª¬a2 n 1 x 2 n 1 " a3 x3 a1xº¼ f x
p o f as x o f and p o f as x o f if Odd
A ! 0. Furthermore, p o f as x o f and p o f as x o f if A 0. Because the graph has no breaks, the graph must cross the x-axis at least one time. 94. f x
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 Even 95. Let F x
f x g x where f and g are even. Then F x
So, F x is even. Let F x F x
f x g x
f x g x
f x g x
F x .
f x g x where f and g are odd. Then
ª ¬ f x ºª ¼¬ g x º¼
f x g x
F x .
So, F x is even. 96. Let F x F x
f x g x where f is even and g is odd. Then f x g x
f x ª ¬ g x º¼
f x g x
F x .
So, F x is odd. x 24 2 x
97. (a) V
2
Domain: 0 x 12 1100
(b)
−1
12 −100
Maximum volume occurs at x (c)
4. So, the dimensions of the box would be 4 u 16 u 16 cm.
x
length and width
volume
1
24 2 1
1¬ª24 2 1 º¼
2
24 2 2
2 ª¬24 2 2 º¼
3
24 2 3
3¬ª24 2 3 º¼
4
24 2 4
4 ª¬24 2 4 º¼
5
24 2 5
5¬ª24 2 5 º¼
2
6
24 2 6
6 ª¬24 2 6 º¼
2
2 2
2
2
484 800 972 1024 980 864
The dimensions of the box that yield a maximum volume appear to be 4 u 16 u 16 cm.
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section P.4
y 2 03
98. By equating slopes,
02 x 3 6 x 3 6 2 x 3
y 2 y
y
2
2x , x 3
2
99. False. If f x
x 2 , then f 3
f 3
33
103. First consider the portion of R in the first quadrant: x t 0, 0 d y d 1 and x y d 1; shown below.
§ 2x · x2 ¨ ¸ . © x 3¹
x2 y2
L
Fitting Mod Models Mo to Data
1
−1
(0, 1)
(2, 1)
x
(0, 0)
(1, 0) 2
−1
9, but
3 z 3.
The area of this region is 1
1 2
3. 2
By symmetry, you obtain the entire region R: y
100. True 2
101. True. The function is even. 102. False. If f x
(−2, 1)
x 2 then, f 3x
3 x 2
9 x 2 and
3x 2 . So, 3 f x z f 3x .
3 f x
(2, 1) x
−2
1
2
(2, −1)
(−2, − 1) −2
32
The area of R is 4 104. Let g x
c be constant polynomial.
Then f g x So, f c
6.
f c and g f x
c.
c. Because this is true for all real numbers c,
f is the identity function: f x
x.
Section P.4 Fitting Models to Data 1. Trigonometric function
6. (a)
20
2. Quadratic function 3. No relationship 0
20 0
4. Linear function
No, the relationship does not appear to be linear. y
5. (a), (b)
(b) Quiz scores are dependent on several variables such as study time, class attendance, etc. These variables may change from one quiz to the next.
250 200 150
7. (a) d
100
x 3
6
9
12
d = 0.066F
15
Yes. The cancer mortality increases linearly with increased exposure to the carcinogenic substance. (c) If x
3, then y | 136.
0.066 F
10
(b)
50
0
110 0
The model fits the data well. (c) If F
55, then d | 0.066 55
3.63 cm.
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
34
Chapter P
NOT FOR SALE
Preparation paration for Calculus Calc
9.7t 0.4
8. (a) s (b)
(b)
−1
0
100 0
The model fits the data well. 24.65 meters second. (c) If t 2.5, s 9. (a) Using a graphing utility, y 0.151x 0.10
(c) According to the model, the times required to attain speeds of less than 20 miles per hour are all about the same. (d) Adding (0, 0) to the data produces 0.002 s 2 0.02 s 0.1
t
The correlation coefficient is r | 0.880.
(e) No. From the graph in part (b), you can see that the model from part (a) follows the data more closely than the model from part (d).
35
y = 0.151x + 0.10
0
14. (a) N1
200 0
(c) Greater per capita energy consumption by a country tends to correspond to greater per capita gross national product. The four countries that differ most from the linear model are Venezuela, South Korea, Hong Kong and United Kingdom. (d) Using a graphing utility, y 0.155 x 0.22 and r | 0.984. 0.3323t 612.9333
10. (a) Linear model: H
(b)
20
5 −5
(b)
0.002s 2 0.04 s 1.9
13. (a) t
45
N3
3.72t 31.6 0.0932t 3 1.735t 2 3.77t 35.1
90
(b)
N1
N3 0
16 30
(c) The cubic model is better. (d) N 2
0.221t 2 6.81t 24.9
90
600
N1 N2
0
0
1300
16 30
0
The model fits the data well. (c) When t 500, H 11. (a) y1
0.3323 500 612.9333 | 446.78. 0.04040t 3 0.3695t 2 1.123t 5.88
y2
0.264t 3.35
y3
0.01439t 3 0.1886t 2 0.476t 1.59
The model does not fit the data well. (e) For 2007, t 17, and N1 | 94.8 million and N 3 | 14.5 million. Neither seem accurate. The linear model’s estimate is too high and the cubic model’s estimate is too low. (f ) Answers will vary 15. (a) y
18
(b)
y1 + y2 + y3
(b)
y1
1.806 x3 14.58 x 2 16.4 x 10
300
y2 y3
0
8
0
0
For year 12, y1 y2 y3 | 47.5 cents mile.
7 0
(c) If x
4.5, y | 214 horsepower.
12. (a) S 180.89 x 2 205.79 x 272 25000 (b)
14
0 0
INSTRUCTOR USE ONLY (c) When x
ounds. 2, S | 583.98 pounds.
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Review Exercises for fo f Chapter P
2.9856 u 104 p 3 0.0641 p 2 5.282 p 143.1
16. (a) T 350
(b)
18. (a) S t
(b)
35
56.37 25.47 sin 0.5080t 2.07
100
M(t)
110
0 150
0
300q F , p | 68.29 lb in.2 .
(c) For T
The model is a good fit.
(d) The model is based on data up to 100 pounds per square inch.
(c)
17. (a) Yes, y is a function of t. At each time t, there is one and only one displacement y.
2.35 1.65
2
S(t)
The model is a good fit. (d) The average is the constant term in each model. 83.70qF for Miami and 56.37qF for Syracuse.
The period is approximately 0.5.
(e) The period for Miami is 2S 0.4912 | 12.8. The
0.35 sin 4S t 2.
(c) One model is y
13 0
0.35.
2 0.375 0.125
100
0
(b) The amplitude is approximately
(d)
13 0
period for Syracuse is 2S 0.5080 | 12.4. In both
4
cases the period is approximately 12, or one year. (f ) Syracuse has greater variability because 25.47 ! 7.46.
(0.125, 2.35) (0.375, 1.65) 0
19. Answers will vary.
0.9 0
The model appears to fit the data.
20. Answers will vary.
21. Yes, A1 d A2 . To see this, consider the two triangles of areas A1 and A2 : T2 T1 a1
γ1
b1
β1
α1
γ2
a2
b2
β2
c1
α2 c2
1, 2, the angles satisfy D i E i J i
For i
S . At least one of D1 d D 2 , E1 d E 2 , J 1 d J 2 must hold.
Assume D1 d D 2 . Because D 2 d S 2 (acute triangle), and the sine function increases on >0, S 2@, you have A1
1b c 2 1 1
sin D1 d d
1 2
b2c2 sin D1
1b c 2 2 2
sin D 2
A2
Review Exercises for Chapter P 1. y
5x 8
x
0: y
5 0 8
y
0: 0
5x 8 x
2. y
x
8 0, 8 y-intercept 8 5
85 , 0 x-intercept
2 x 6
x
0: y
0 2 0 6
y
0: 0
x 2 x
12 0, 12 y-intercept
6 x
2, 6 2, 0 , 6, 0 x-intercepts
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
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36
NOT FOR SALE
Chapter P
Preparation paration for Calculus Calc
x 3 x4
3. y
9. 13 x
x
0: y
03 04
y
0: 0
x 3 x x4
4. xy
52 x
3 § · ¨ 0, ¸ y-intercept 4 © 4¹ 3 3, 0 x-intercept.
5y 6
1
y
6 5
Slope:
6 5
2 5
y-intercept:
6 5 y
4
x
2x 5
y
0 and y
0 are both impossible. No intercepts.
3 2
5. Symmetric with respect to y-axis because
x 2 y
x 4 y
0
x y x 4y
0.
2
2
−3
2
x 4
x 3
y
x x 3.
0.25
2 x 15 y
2
12 x
7. y
1
10. 0.02 x 0.15 y
2
4
x
−1 −1
6. Symmetric with respect to y-axis because
y
−2
25 2x 15
y
5 3
2 Slope: 15
3 2
53
y-intercept: 0,
Slope: 12
y
3 2
y-intercept:
3
y 3
1
2
x
−4
1
4
8
12
−1
x
−1
1
2
−2
3
−1 −2
8. 6 x 3 y
3 y
12
y-intercept: 0, 9
6 x 12
x-intercepts: 1, 0 , 9, 0
2x 4
y
x 1 x 9
9 8x x2
11. y
y
Slope: 2
28 24
y-intercept: 0, 4
12
y
8
4
4 x
2
− 12
−6 −4 − 2
2
4
x −6 −4 −2
−2 −4
−8
2
4
6
x 6 x
12. y
y-intercept: (0, 0) x-intercepts: (0, 0), (6, 0) y 10 8 6 4
INSTRUCTOR USE USE ONLY 2
−2
x
2
4
8
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Review Exercises for fo f Chapter P
18. x y 1
2 4 x
13. y
Domain: f, 4@
y x
y
2
0 y
x 1
7 y
x2 7
37
14
5
3 − 12
2
12 −2
1 −1
x 1
−1
2
3
4
x 1
y
5
x 1 x 2 x4 4
14. y
7 x2 x 6
0
y
No real solution.
6
No points of intersection.
4
x −2 −2
2
4
6
8
10
You need factors x 4 and x 4 .
−6
Multiply by x to obtain origin symmetry.
4 x 2 25
x x 4 x 4
y
Xmin 5 Xmax 5 Xscl 1 Ymin 30 Ymax 10 Yscl 5 16. y
x 2 7 do not
19. Answers will vary. Sample answer:
−4
15. y
x 1 and y
The graphs of y intersect.
2
kx3
20. y
83 x 6
(a) 4
k 1 k
(b) 1
k 2 k
(c) 0
k 0 and k will do!
3
21.
4 and y
3
4 x3
18 and y
18 x 3
3
k 1 k 3
(d) 1
Xmin 40 Xmax 40 Xscl 10 Ymin 40 Ymax 40 Yscl 10
x3 16 x
1 y
x3
y 5 4
( 5, 52 )
3 2
17. 5 x 3 y
1 y
x y
1 3
5 y
5 x
1
1
( 32 , 1 ) 1
x5
2
Slope 3 −1
Using a graphing utility, the lines intersect at 2, 3 . Analytically, 1 3
5 x
1
5 x 1 16
3 2 7 2
3 7
y
(−1, 8) 6
3 x 15
4 2
8x
x. 2 2, y x5
5
22. The line is horizontal and has slope 0.
(−7, 8)
x5
4
§ 5· ¨ ¸ 1 © 2¹ § 3· 5¨ ¸ © 2¹
6
−5
x
3
−8
−6
−4
x
−2
2
INSTRUCTOR USE S ONLY For x
2 5
3.
−22
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
38
23.
NOT FOR SALE
Chapter P
Preparation paration for Calculus Calc
t 5 0 8
1 5 2 8
t 5 8 t 5 8 5t 25
28. Because m
5t
0 x 5
y 4
6 10 3 5 24
0, the line is horizontal.
4 or y 4
y
0
y 8 6
(5, 4)
1 2
1 5
t
24.
3 1 3 t 4 3 t 44
x −4
2
7 4
y 5
2 4
6
53 x 3 5 x 15
5x 3 y
0 3
x x 3
−6 −8
(0, − 414)
0
23 x 2
y 4
30. (a)
3 y 12 26. Because m is undefined the line is vertical.
8 or x 8
x
0
y 4
6 4 2
(−8, 1) x −4
−2
2 −2
(c) m
−4
23 x 2
y 2x 3y 6
0
1 x 2 x 2
0
x y 2
4 1 26 y 4
4 y 16 3x 4 y 22
3 4 3 x 2 4 3 x 6
0
(d) Because the line is horizontal the slope is 0.
y 3 2 1
(−3, 0)
0
y
23 x 3
y 0
27.
2 x 4
2 x 3 y 16
(b) x y 0 has slope 1. Slope of the perpendicular line is 1.
y
−6
5 3
3 y 15
8
−4
−10
5 x 3 y 30
(d) Slope is undefined so the line is vertical.
x 2
3
5 x 15
y 5
7 x 4 y 41
x
50 3 0
(c) m
y
−8 −6 −4 −2
5 3
0
7 x 21
0
3 has slope 53 .
3 y 15
21 4
3
7 x 16 y 101
(b) 5 x 3 y
x 3
x
7 x 21
0
y 5
7x 4
4 y 20
7 16
16 y 80
53 3
25. y 5
6
y 5
29. (a)
3t
t
4
−2
36 3 8 3 11 9 3t
53
−2
y
4
y 4
0
x
INSTRUCTOR ST USE ONLY −44 − 3
−1 1
1
2
3
−3 −4
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Review Exercises for fo f Chapter P
31. The slope is 850. V 850t 12,500.
V 3
9 y2
36. x
850 3 12,500
39
Not a function of x since there are two values of y for some x.
$9950
y
9.25t 13.50t 36,500
32. (a) C
(b) R
22.75t 36,500
4
30t
2
22.75t 36,500
30t
(c)
7.25t
1
x − 12 − 9 − 6 − 3 −1
36,500
3
6
12
−2
t | 5034.48 hours to break even
−4
33. x y
2
6
y
r
1 x
37. f x
x6
Not a function because there are two values of y for some x.
(a) f 0 does not exist.
y 4 3
1 1 1 'x 1 'x
f 1 'x f 1 'x
(b)
2
1 , 'x z 1, 0 1 'x
1 x 2
−1
4
1 1 'x 'x 'x
1
8 10 12 14
−2 −3
34. x 2 y
0
Function of x because there is one value for y for each x. y 6 5 4
(a) f 4
4 2
2
(b) f 0
02
2
(c) f 1
12
1
(b) Domain: all x z 5 or f, 5 5, f
x −1
1
2
3
Range: all y z 0 or f, 0 0, f
x 2 x 2
35. y
(c) Domain: all x or f, f
y is a function of x because there is one value of y for each x. y
Range: all y or f, f 40. f x
4 3 1 x 1
3
4
5
6
(b)
−2 −3 −4
1 x 2 and g x
2x 1
1 x 2 2 x 1 x 2 2 x f x g x 1 x 2 2 x 1 2 x3 x 2 2 x 1 g f x g 1 x 2 2 1 x 2 1 3 2 x 2
(a) f x g x
2
−2 −1
1
Range: >0, 6@
2
−2
18 because 4 0
39. (a) Domain: 36 x 2 t 0 6 d x d 6 or >6, 6@
3
−3
2 °x 2, x 0 ® °¯ x 2 , x t 0
38. f x
−4
(c)
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
40
NOT FOR SALE
Chapter P
41. (a) f x
Preparation paration for Calculus Calc x 3 c, c y
43. (a) Odd powers: f x
2, 0, 2
x 3 , h x
x5
g
2
c=0
x, g x
h 3 −3
c = −2
1
f
x −3 −2
2
3
−2
The graphs of f, g, and h all rise to the right and fall to the left. As the degree increases, the graph rises and falls more steeply. All three graphs pass through the points (0, 0), (1, 1), and 1, 1 and are
c=2
(b) f x
x
c , c 3
y
3
2, 0, 2
c=0
symmetric with respect to the origin.
c = −2
Even powers: f x 1
x 2 , g x
x 4 , h x
x6
g
4
x
−2
2 −2
h
3
f
c=2
−3
(c) f x
x
−3
2 c, c 3
The graphs of f, g, and h all rise to the left and to the right. As the degree increases, the graph rises more steeply. All three graphs pass through the points 0, 0 , 1, 1 , and 1, 1 and are symmetric with
y
c=2
2
c=0
1
respect to the y-axis.
x 2
4
All of the graphs, even and odd, pass through the origin. As the powers increase, the graphs become flatter in the interval 1 x 1.
−1 −2
c = −2
(d) f x
(b) y
2, 0, 2
cx 3 , c
3 0
2, 0, 2
x 7 will look like h x
x8 will look like
even more steeply. y
y
h x
c=2
3
x5 , but rise and fall
6
x , but rise even more steeply.
2
44. (a) f x
1
c=0 −3 −2 −1
1
x
2
3
2
The leading coefficient is positive and the degree is even so the graph will rise to the left and to the right.
c = −2
42. f x
x 2 x 6
100
x3 3x 2 −4
6
10
− 25
−6
(0, 0)
(b) g x
6
(a) The graph of g is obtained from f by a vertical shift down 1 unit, followed by a reflection in the x-axis: ª¬ f x 1º¼
f x 2 1
300
x3 3x 2 1
(b) The graph of g is obtained from f by a vertical shift upwards of 1 and a horizontal shift of 2 to the right. g x
2
The leading coefficient is positive and the degree is odd so the graph will rise to the right and fall to the left.
(2, − 4) 6
g x
x3 x 6
x 2
3
−2
10
− 100
3 x 2 1 2
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Problem Solving ffo for Chapter P
(c) h x
x3 x 6
3
The leading coefficient is positive and the degree is even so the graph will rise to the left and to the right. 200
−4
10
46. For company (a) the profit rose rapidly for the first year, and then leveled off. For the second company (b), the profit dropped, and then rose again later. 47. (a) (b) (c) (d)
3 (cubic), negative leading coefficient 4 (quartic), positive leading coefficient 2 (quadratic), negative leading coefficient 5, positive leading coefficient 1.204 x 64.2667
48. (a) y
− 800
41
(b)
70
y
45. (a) x
x 0
33 0
y
2x 2 y
(c) The data point (27, 44) is probably an error. Without this point, the new model is 1.4344 x 66.4387. y
24
y
12 x
A
xy
x 12 x
49. (a) Yes, y is a function of t. At each time t, there is one and only one displacement y. (b) The amplitude is approximately 0.25 0.25 2 0.25. The period is
(b) Domain: 0 x 12 or (0, 12) 40
approximately 1.1. 0
(c) One model is y
12 0
36 in.2 . In general, the
(c) Maximum area is A
maximum area is attained when the rectangle is a square. In this case, x 6.
(d)
1 1 § 2S · cos¨ t ¸ | cos 5.7t 4 1.1 4 © ¹
0.5
(1.1, 0.25) 0
2.2
(0.5, −0.25) −0.5
The model appears to fit the data.
Problem Solving for Chapter P 1. (a)
x2 6 x y 2 8 y
x2 6 x 9 y 2 8 y 16 x
3 y 4 2
2
0 9 16 25
Center: (3, 4); Radius: 5 4 3 3 . Slope of tangent line is . So, y 0 x 0 y 3 4 4 40 4 (c) Slope of line from (6, 0) to (3, 4) is . 36 3 3 3 3 9 Slope of tangent line is . So, y 0 x Tangent line x 6 y 4 4 4 2 3 3 9 x (d) x 4 4 2 3 9 x 2 2 3 x (b) Slope of line from (0, 0) to (3, 4) is
3 x Tangent line 4
9· § Intersection: ¨ 3, ¸ 4¹ ©
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
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42
NOT FOR SALE
Chapter P 2. Let y
Preparation paration for Calculus Calc
mx 1 be a tangent line to the circle from the point (0, 1). Because the
center of the circle is at 0, 1 and the radius is 1 you have the following.
m
2
x 2 y 1
2
x 2 mx 1 1
2
1 1
1 x 4mx 3 2
0
Setting the discriminant b 2 4ac equal to zero, 16m 2 4 m 2 1 3
0
16m 2 12m 2
12
2
12
m
r
4m
3x 1 and y
Tangent lines: y 3. H x
3
3 x 1.
y
1, x t 0 ® ¯0, x 0
1, x d 0 ® ¯0, x ! 0
(d) H x
4 3 2
y
1 x −4 −3 −2 −1 −1
1
2
3
4
4
3
−2
2
−3 −4
x −4 −3 −2 −1 −1
(e)
3 2
1 ° , x t 0 ®2 ° ¯0, x 0
1 H x 2
1
y
x 2
3
4
−4
y
1
3
−3
4
−4 −3 −2 −1 −1
2
−2
1, x t 0 ® ¯2, x 0
(a) H x 2
1
4 4 3
−3
2
−4
1 x −4 −3 −2 −1 −1
1, x t 2 ® ¯0, x 2
(b) H x 2 y
1
2
3
4
−2 −3 −4
4 3
1, x t 2 ® ¯2, x 2
(f ) H x 2 2
2 1 x −4 −3 −2 −1 −1
1
2
3
4 y
−2
(c) H x
−3
4
−4
3
1, x t 0 ® ¯0, x 0
1 y
x
4
−4 −3 −2 −1 −1
3
−2
2
−3
1
2
3
4
−4
1 x −4 −3 −2 −1 −1
1
2
3
4
−2 −3 −4
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Problem Solving ffo for Chapter P
(g) f x
4. (a) f x 1
43
y
y 4
4
2
−3
x
−1
1
−4
3
x
−2
2
−2
−2
−4
−4
(b) f x 1
5. (a) x 2 y
A x
100 x 2
100 y
y 4
4
§ 100 x · x¨ ¸ 2 ¹ ©
xy
x2 50 x 2
Domain: 0 x 100 or 0, 100 x
−4
(b)
4
1600
−2 −4
(c) 2 f x
0
110 0
y
Maximum of 1250 m 2 at x
4
12 x 2 100 x 2500 1250
x
−2
2
25 m.
12 x 2 100 x
(c) A x −4
50 m, y
4
12 x 50 1250
−2
2
−4
A 50
(d) f x
x
1250 m 2 is the maximum.
50 m, y
25 m
y 4
6. (a) 4 y 3 x
300 3 x 4
300 y
2
−4
A x
x
−2
2
4
−2
§ 300 3 x · x¨ ¸ 2 © ¹
x 2 y
3 x 2 300 x 2
Domain: 0 x 100
−4
y
(b) 4000
(e) f x
3500 3000
y
2500 2000
4
1500 1000
2
500
−4
−2
2
25
4
−2
(c) A x
f x y
100
50 ft, y
37.5 ft.
x 100 x 32 x 2 100 x 2500 3750 32
2
2
A 50
2 x
−2
75
32 x 50 3750
4
−4
50
Maximum of 3750 ft 2 at x
−4
(f )
x
x
2
where x
3750 square feet is the maximum area, 50 ft and y
37.5 ft.
4
−2
INSTRUCTOR T USE ONLY −4
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
44
NOT FOR SALE
Chapter P
Preparation paration for Calculus Calc
7. The length of the trip in the water is
length of the trip over land is 4 x2 2
total time is T
22 x 2 , and the
1 3 x . So, the
y
10. 4
2
1 3 x 4
3 2
2
hours.
(4, 2)
1 x 1
8. Let d be the distance from the starting point to the beach.
Average speed
9. (a) Slope
distance time 2d d d 120 60 2 1 1 120 60 80 km h
3
(a) Slope
4
32 94
1 . Slope of tangent line is greater 5
2 1 4 1
1 . Slope of tangent line is less 3
(b) Slope than
1 . 3
2.1 2 4.41 4 10 greater than . 41
10 . Slope of tangent line is 41
(c) Slope
94 32
5. Slope of tangent line is less
4 1 2 1
3. Slope of tangent line is greater
f 4 h f 4 4 h 4
(d) Slope
than 3. 4.41 4 2.1 2 less than 4.1.
(c) Slope
(d) Slope
5
1 . 5
than
than 5. (b) Slope
2
−1
4.1. Slope of tangent line is
4 h 2 h
(e)
2 h
4 h 2 h
4 h 4
f 2 h f 2 2 h 2 2
4 h 2 h
h
4 h 2 4 h 2
4 h 2
1 ,h z 0 4 h 2
4
h
As h gets closer to 0, the slope gets closer to
4h h 2 h 4 h, h z 0
slope is
(e) Letting h get closer and closer to 0, the slope approaches 4. So, the slope at (2, 4) is 4.
1 . The 4
1 at the point (4, 2). 4
11. Using the definition of absolute value, you can rewrite the equation.
y y 2 y, ® ¯0,
x x
y ! 0 y d 0
2 x, x ! 0 . ® x d 0 ¯0,
For x ! 0 and y ! 0, you have 2 y
2x y
x.
For any x d 0, y is any y d 0. So, the graph of y y
x x is as follows.
y
(−1, 1)
5 4 3 2
(0, 12 ) (2, 14 ) x
(− 6, − 14 ) (− 4, − 12 )
(− 3, − 1)
−1
1 2 3
−2 −3 −4 −5
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
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NOT FOR SALE
Problem Solving ffo for Chapter P
I x2
12. (a)
2I
x
3
x2 6x 9
2x2
x2 6x 9
0
2
x 0
6 r
x
I x2 y 2
(b)
36 36 2
2
3
2I
x 3 y 2 2
2 x 2 y 2
x2 6x 9 y 2
2x2 2 y 2
x2 y2 6x 9
0
x
1
3 r 18 | 1.2426, 7.2426
3 y 2
x
45
2
3 y 2 2
18
18 and center 3, 0 .
Circle of radius y 8 6
2 −8
x
− 4 −2 −2
2
4
−6
13. (a)
I x2 y2
kI
x
4 y 2 2
k x2 y 2
x
4 y 2
k
1 x 2 8 x k 1 y 2
2
2 is a vertical line. Assume k z 1.
1, then x
If k
16
8x y2 k 1 8x 16 y2 x2 k 1 k 1 2
16 k 1 16 16 k 1 k 1 2
x2
2
4 · § 2 ¨x ¸ y k 1¹ ©
3, x 2 y 2 2
(b) If k
16k
k
1
2
, Circle
12
y 6 4 2 −6
−4
x
−2
2
4
−2 −4
(c) As k becomes very large,
16k 4 o 0. o 0 and 2 k 1 k 1
The center of the circle gets closer to (0, 0), and its radius approaches 0.
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
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46
NOT FOR SALE
Chapter P
Preparation paration for Calculus Calc
d1d 2
1
ª x 1 y 2 ºª x 1 y 2 º ¬ ¼¬ ¼
1
2 2 2 2 1 x 1 y 2 ª x 1 x 1 º y 4 ¬ ¼
1
14. 2
x
x 2 1
2
2
y 2 ª¬2 x 2 2º¼ y 4
x 2x 1 2x y 2 y y 4
2
x
4
2 2
2x y y 2 2
4
2
1
4
1
2x 2 y 2 x2 y 2 2
2
0 2 x 2 y 2
y 2 1
(− 2 , 0)
( 2 , 0) x
−2
2
(0, 0)
−1 −2
2x2 x
0. Then x 4
Let y
So, 0, 0 , 15. f x
0 or x 2
2.
2, 0 and
2, 0 are on the curve.
1 1 x
y
(a) Domain: all x z 1 or f, 1 1, f Range: all y z 0 or f, 0 0, f § 1 · f¨ ¸ ©1 x ¹
(b) f f x
1 § 1 · 1¨ ¸ ©1 x ¹
1 1 x 1 1 x
1 x x
x 1 x
Domain: all x z 0, 1 or f, 0 0, 1 1, f
§ x 1· f¨ ¸ © x ¹
(c) f f f x
1 § x 1· 1¨ ¸ © x ¹
1 1 x
x
Domain: all x z 0, 1 or f, 0 0, 1 1, f (d) The graph is not a line. it has holes at (0, 0) and (1, 1). y 2 1 x −2
1
2
−2
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
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NOT FOR SALE C H A P T E R 1 Limits and Their Properties Section 1.1
A Preview of Calculus..........................................................................48
Section 1.2
Finding Limits Graphically and Numerically .....................................49
Section 1.3
Evaluating Limits Analytically ............................................................61
Section 1.4
Continuity and One-Sided Limits........................................................73
Section 1.5
Infinite Limits .......................................................................................84
Review Exercises ..........................................................................................................92 Problem Solving ...........................................................................................................99
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NOT FOR SALE C H A P T E R 1 Limits and Their Properties Section 1.1 A Preview of Calculus 1. Precalculus: 20 ft/sec 15 sec
300 ft
y
(a)
2. Calculus required: Velocity is not constant.
Distance | 20 ft/sec 15 sec
6x x2
7. f ( x)
10
300 ft
8
P
6
3. Calculus required: Slope of the tangent line at x the rate of change, and equals about 0.16.
2 is x
4. Precalculus: rate of change
slope
1 bh 2
5. (a) Precalculus: Area
(b) Calculus required: Area
1 2
−2
0.08
5 4
10 sq. units
(b) slope
2
4
m
8
6 x x 2 8
bh | 2 2.5
x2 4 x , x z 2
3, m
For x
2.5, m
4 2.5
1.5
3 2
(a)
For x
1.5, m
4 1.5
2.5
5 2
1
x y
P(4, 2) 2
2 4 x x2
For x
5 sq. units
6. f ( x)
4 3
x
(c) At P 2, 8 , the slope is 2. You can improve your approximation by considering values of x close to 2. x 1
(b) slope
2
3
4
5
x 2 x 4
m
x 2 x 2
x 2
1 ,x z 4 x 2
x
1: m
x
3: m
x
5: m
1 1 3 1 2 1 | 0.2679 3 2 1 | 0.2361 5 2
(c) At P 4, 2 the slope is
1 4 2
1 4
0.25.
You can improve your approximation of the slope at x 4 by considering x-values very close to 4. 8. (a) For the figure on the left, each rectangle has width
S 4
.
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
48
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section 1.2
Area |
Finding Limits Graphically and Numerically
49
Sª
S S 3S º sin sin sin S » sin « 4¬ 4 2 4 ¼
Sª 2
2º » 2 »¼
1 « 4 «¬ 2
2 1 S | 1.8961 4 For the figure on the right, each rectangle has width
Area |
S 6
.
Sª
S S S 2S 5S º sin sin sin sin sin S » « 6¬ 6 3 2 3 6 ¼
S ª1
3 3 1º 1 » « 6 «¬ 2 2 2 2 »¼ 3 2 S | 1.9541 6
(b) You could obtain a more accurate approximation by using more rectangles. You will learn later that the exact area is 2. 9. (a) Area | 5
Area |
1 2
5 2
5 3
5 4
| 10.417
5 1.55 52 2.55 53 3.55 54 4.55 | 9.145
(b) You could improve the approximation by using more rectangles. 10. Answers will vary. Sample answer:
The instantaneous rate of change of an automobile’s position is the velocity of the automobile, and can be determined by the speedometer.
5 1 2
11. (a) D1
1
(b) D2
52
2
1 5
1
2
16 16 | 5.66
52 53
2
53 54
1
2
1
54 1
2
| 2.693 1.302 1.083 1.031 | 6.11 (c) Increase the number of line segments.
Section 1.2 Finding Limits Graphically and Numerically 1.
x
3.9
3.99
3.999
4.001
4.01
4.1
f (x)
0.2041
0.2004
0.2000
0.2000
0.1996
0.1961
lim
x o 4 x2
2.
x 4 | 0.2000 3x 4
1· § ¨ Actual limit is .¸ 5¹ ©
x
1.9
1.99
1.999
2.001
2.01
2.1
f (x)
0.2564
0.2506
0.2501
0.2499
0.2494
0.2439
lim
xo2
x 2 | 0.25 x2 4
1· § ¨ Actual limit is .¸ 4¹ ©
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
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Chapter 1
50 3.
x
–0.1
–0.01
–0.001
0.001
0.01
0.1
f (x)
0.2050
0.2042
0.2041
0.2041
0.2040
0.2033
lim
x 6 x
x
–5.1
5.01
5.001
4.999
4.99
4.9
f (x)
–0.1662
–0.1666
–0.1667
–0.1667
–0.1667
–0.1671
xo0
4.
6
1 · § .¸ ¨ Actual limit is 2 6¹ ©
| 0.2041
4 x 3 | 0.1667 x 5
lim
x o 5
5.
NOT FOR SALE
Limits its and Their Properties
1· § ¨ Actual limit is .¸ 6¹ ©
x
2.9
2.99
2.999
3.001
3.01
3.1
f (x)
–0.0641
–0.0627
–0.0625
–0.0625
–0.0623
–0.0610
ª1 x 1 º¼ 1 4 lim ¬ | –0.0625 x o3 x 3
6.
x
3.9
3.99
3.999
4.001
4.01
4.1
f (x)
0.0408
0.0401
0.0400
0.0400
0.0399
0.0392
ª x x 1 º¼ 4 5 lim ¬ | 0.04 x 4
1 · § ¨ Actual limit is .¸ 25 ¹ ©
xo4
7.
x
–0.1
–0.01
–0.001
0.001
0.01
0.1
f (x)
0.9983
0.99998
1.0000
1.0000
0.99998
0.9983
lim
xo0
8.
Actual limit is 1. Make sure you use radian mode.
–0.1
–0.01
–0.001
0.001
0.01
0.1
f (x)
0.0500
0.0050
0.0005
–0.0005
–0.0050
–0.0500
lim
cos x 1 | 0.0000 x
Actual limit is 0. Make sure you use radian mode.
x
0.9
0.99
0.999
1.001
1.01
1.1
f (x)
0.2564
0.2506
0.2501
0.2499
0.2494
0.2439
lim
x o1
10.
sin x | 1.0000 x
x
xo0
9.
1 · § ¨ Actual limit is .¸ 16 ¹ ©
x 2 | 0.2500 x2 x 6
1· § ¨ Actual limit is .¸ 4¹ ©
x
–3.1
–3.01
–3.001
–2.999
–2.99
–2.9
f (x)
1.1111
1.0101
1.0010
0.9990
0.9901
0.9091
lim
x o 3
x 3 | 1.0000 x 2 7 x 12
Actual limit is 1.
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
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NOT FOR SALE Section 1.2
11.
x f (x)
0.9
0.99
0.999
1.001
1.01
1.1
0.7340
0.6733
0.6673
0.6660
0.6600
0.6015
x 1 | 0.6666 x6 1
x o1
12.
x
–2.1
–2.01
–2.001
–1.999
–1.99
–1.9
f (x)
12.6100
12.0601
12.0060
11.9940
11.9401
11.4100
lim
x o –2
13.
Actual limit is 12.
–0.1
–0.01
–0.001
0.001
0.01
0.1
f (x)
1.9867
1.9999
2.0000
2.0000
1.9999
1.9867
lim
14.
x3 8 | 12.0000 x 2
x
xo0
sin 2 x | 2.0000 x
Actual limit is 2. Make sure you use radian mode.
x
–0.1
–0.01
–0.001
0.001
0.01
0.1
f (x)
0.4950
0.5000
0.5000
0.5000
0.5000
0.4950
lim
xo0
tan x | 0.5000 tan 2 x
15. lim 4 x 16. lim x 2 3 xo2
18. lim f x x o1
19. lim
xo2
24. lim tan x does not exist because the function increases x oS 2
without bound as x approaches
4
x o1
17. lim f x
1· § ¨ Actual limit is .¸ 2¹ ©
1
x o3
lim 4 x
xo2
lim x 3 x o1
4
from the left and
S 2
from
25. (a) f 1 exists. The black dot at (1, 2) indicates that
f 1
for values of x to the right of 2,
2.
(b) lim f x does not exist. As x approaches 1 from the x o1
x 2 For values of x to the left of 2, x 2 x 2 x 2
1, whereas 1.
2 20. lim does not exist because the function increases x o5 x 5 and decreases without bound as x approaches 5.
left, f (x) approaches 3.5, whereas as x approaches 1 from the right, f (x) approaches 1. (c) f 4 does not exist. The hollow circle at
4, 2 indicates that f is not defined at 4. (d) lim f x exists. As x approaches 4, f x approaches xo4
2: lim f x xo4
2.
0
x o1
xo0
2
decreases without bound as x approaches
2
x 2 does not exist. x 2
22. lim sec x
S
the right. 2
21. lim sin S x
51
2· § ¨ Actual limit is .¸ 3¹ ©
4
lim
Finding Limits Graphically and Numerically
1
23. lim cos 1 x does not exist because the function xo0
oscillates between –1 and 1 as x approaches 0.
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52
NOT FOR SALE
Chapter 1
Limits its and Their Properties
26. (a) f 2 does not exist. The vertical dotted line
y
30.
indicates that f is not defined at –2.
2
(b) lim f x does not exist. As x approaches –2, the
1
x o 2
values of f x do not approach a specific number. (c) f 0 exists. The black dot at 0, 4 indicates that
f 0
lim f x exists for all values of c z S .
xoc
(d) lim f x does not exist. As x approaches 0 from the xo0
31. One possible answer is y
from the right, f x approaches 4.
6 5
(e) f 2 does not exist. The hollow circle at
2,
1 2
4
indicates that f 2 is not defined.
1 x −2 − 1 −1
xo2
x
1
2
3
4
5
1 . 2
(g) f 4 exists. The black dot at 4, 2 indicates that
f 4
f
2
(f ) lim f x exists. As x approaches 2, f x approaches 1 : lim f 2 xo2
x
π
−1
4.
left, f x approaches 12 , whereas as x approaches 0
π 2
−π 2
32. One possible answer is y
2.
4
(h) lim f x does not exist. As x approaches 4, the
3
xo4
2
values of f x do not approach a specific number.
1 x
27. lim f x exists for all c z 3.
−3
−2
−1
1
2
−1
xoc
28. lim f x exists for all c z 2, 0. xoc
29.
y 6 5 4 3
f
2 1 x − 2 −1 −1
1
2
3
4
5
−2
lim f x exists for all values of c z 4.
xoc
33. C t
(a)
9.99 0.79 ª¬ª ¬ t 1 º¼º¼ 16
0
6 8
(b)
t
3
3.3
3.4
3.5
3.6
3.7
4
C
11.57
12.36
12.36
12.36
12.36
12.36
12.36
lim C t
t o 3.5
12.36
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section 1.2
(c)
Finding Limits Graphically and Numerically
t
2
2.5
2.9
3
3.1
3.5
4
C
10.78
11.57
11.57
11.57
12.36
12.36
12.36
53
The lim C t does not exist because the values of C approach different values as t approaches 3 from both sides. t o3
34. C t
5.79 0.99 ª¬ª ¬ t 1 º¼º¼ 12
(a)
0
6 4
(b)
t
3
3.3
3.4
3.5
3.6
3.7
4
C
7.77
8.76
8.76
8.76
8.76
8.76
8.76
lim C t
8.76
t o 3.5
(c)
t
2
2.5
2.9
3
3.1
3.5
4
C
6.78
7.77
7.77
7.77
8.76
8.76
8.76
The limit lim C t does not exist because the values of C approach different values as t approaches 3 from both sides. t o3
35. You need f x 3 x 2
x
1 3
36. You need f x 1
x
1 3
x 2 0.4. So, take G
0.4. If 0 x 2 0.4, then
f x 3 0.4, as desired.
1 1 x 1
2 x 0.01. Let G x 1
1 1 . If 0 x 2 , then 101 101
1 1 1 1 x 2 1 x 1 1 101 101 101 101 100 102 x 1 101 101 100 x 1 ! 101
and you have f x 1
1 1 x 1
2 x 1 101 100 101 x 1
1 100
0.01.
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54
NOT FOR SALE
Chapter 1
Limits its and Their Properties
37. You need to find such that 0 x 1 G implies
39. lim 3 x 2
1 1 0.1. That is, x
3 x
f x 1
1 1 0.1 x 1 1 0.1 1 0.1 x 9 1 11 10 10 x 10 10 x ! ! 9 11 10 10 1 ! x 1 ! 1 9 11 1 1 ! x 1 ! . 9 11
xo2
0 x 2
3 x
x 2 4 0.2. That is,
0.2 x 4 0.2
3.8
x
4.2
2
L
x· § ¨ 4 ¸ 2 0.01 2 © ¹
x 0.01 2
1 x 4 0.01 2
0 x 4 0.02
G
0.02, you have
1 x 4 0.01 2 2
x 0.01 2
x· § ¨ 4 ¸ 2 0.01 2¹ ©
4 0.2
3.8 2 x 2 So take G
x· § 40. lim ¨ 4 ¸ x o 4© 2¹
Hence, if 0 x 4 G
2
x
2 8 0.01
2
x2 1 3
3.8
you have
f x L 0.01.
38. You need to find such that 0 x 2 G implies
2
0.01 , 3
G
3 x 6 0.01
1 1 0.1. x
x2
| 0.0033
3 x 2 0.01
Using the first series of equivalent inequalities, you obtain
4 0.2
0.01 3
So, if 0 x 2 G
1 1 x 1 11 11 1 1 x 1 . 11 9
f x 3
2 8 0.01 3 x 2 0.01
1 . Then 0 x 1 G implies 11
f x 1
L
3 x 6 0.01
0.1
So take G
8
f x L 0.01.
4.2 4.2 2
4.2 2 | 0.0494.
Then 0 x 2 G implies
4.2 2 x 2
4.2 2
3.8 2 x 2
4.2 2.
Using the first series of equivalent inequalities, you obtain f x 3
x 2 4 0.2.
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
Section 1.2 41. lim x 2 3
1
xo2
x2
44. lim 2 x 5
L
3 1 0.01
Given H ! 0:
2 x
5 1 H 2x 6 H
2 x 2 0.01
2 x 3 H
x 2 x 2 0.01 x 2
x 3
0.01 x 2
If you assume 1 x 3, then G
So, if 0 x 2 G
0.01 5
0.002.
0.002, you have
x 2 0.002
So, let G
1 1 0.01 0.01 5 x 2
45. lim
xo5
x2
L
, you have
H
1 2
12 x 1 3 1 x 2
5 x 5 0.01
1 2
0.01 x 5 x 5
If you assume 4 x 6, then G
1 x 1
4
1
3
Given H ! 0:
4 29 0.01
H
2 H
x 4 H x 4 2H
0.01/11 | 0.0009.
0.01 , you have So, if 0 x 5 G 11 0.01 1 x 5 0.01 11 x 5
So, let G
2H .
So, if 0 x 4 G
2H , you have
x 4 2H 1x 2
x 5 x 5 0.01
2 H
12 x 1 3
x 2 25 0.01
x 2 4 29
2
5 1 H
x o 4 2
x 2 25 0.01
x
H
f x L H .
f x L 0.01. 29
H 2.
x 3
2 x
0.01
42. lim x 4
G
2
2 2x 6 H
x 2 4 0.01
2
H
So, if 0 x 3 G
x 2 x 2 0.01
x 2 3 1
55
1
x o 3
x 2 4 0.01
x
Finding Limits Graphically and Numerically
H
f x L H .
0.01
f x L 0.01. 43. lim x 2
6
xo4
Given H ! 0:
x
2 6 H x 4 H
So, let G
G
H . So, if 0 x 4 G
H , you have
x 4 H
x
2 6 H f x L H .
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56
NOT FOR SALE
Chapter 1
46. lim
Limits its and Their Properties
2 x 7
2 5
x o1 5
1
7
37 5
xo0
Given H ! 0:
2x 5
7
2x 5 2 5
2 5
x H x H3
5 H 2
So, let G
5 H , you 2
H
52 x 7 375
H
H 3 , you have
x H3
have 3
5 H 2
2 5
3
x H
x 0 H
f x L H .
f x L H .
x
50. lim
xo4
4
2 x 2 H
Given H ! 0:
3
x 2
Given H ! 0: 33 H 0 H So, any G ! 0 will work. So, for any G ! 0, you have
0 x 4 G
xo2
5
51. lim x 5 x o 5
H.
5
x 2
x 2 H. 10
10
x
5 0
x 5 H
So, any G ! 0 will work.
H
3H . Then,
x 5 H
0 H
1 1 f x L
x 2
x 5 10 H
Given H ! 0: 1 1 H
So, for any G ! 0, you have
x 4 H
x 5 10 H
Given H ! 0:
1
x 2
3H x 4 H
f x L H .
x 2 H
Assuming 1 x 9, you can choose G
33 H
48. lim 1
G
H 3.
So, for 0 x 0 G
x 1
xo6
x 0 H
x 1 H
So, if 0 x 1 G
47. lim 3
3
3
5 H. 2
0
H
x 1
2x 5
x
Given H ! 0:
37 5
So, let G
3
49. lim
So, let G
H.
So for x 5 G
H , you have
x 5 H x 5 10 H x 5 10 H
because x
5 0
f x L H .
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section 1.2
52. lim x 6
66
xo6
Given H ! 0:
55. lim f x
0
x oS
x 6 0 H
x oS
H.
So for x 6 G
lim f x
x 6 H
4
lim x
S
x oS
57
x 5 3 x 4 1 6
57. f x
H , you have
lim 4
x oS
56. lim f x
x 6 H So, let G
Finding Limits Graphically and Numerically
xo4
x 6 0 H
0.5
f x L H . 53. lim x 2 1 x o1
−6 − 0.1667
Given H ! 0:
x2
The domain is >5, 4 4, f . The graphing utility
1 2 H
§ does not show the hole at ¨ 4, ©
x2 1 H
x
58. f x
H
x 1
x 1
If you assume 0 x 2, then G So for 0 x 1 G x 1
H 3
H 3.
, you have
−3
The domain is all x z 1, 3. The graphing utility does not
x 1 H
§ 1· show the hole at ¨ 3, ¸. © 2¹
1 2 H f x 2 H .
59. f x
0
x o 3
5
−4
2
lim f x
x o9
Given H ! 0:
x2
4
1 1 H H 3 x 1
54. lim x 2 3 x
1· ¸. 6¹
x 3 x2 4 x 3 1 lim f x x o3 2
1 x 1 H
x2
6
2
x 9 x 3 6
10
3 x 0 H x x 3 H x 3
H
0
If you assume 4 x 2, then G So for 0 x 3 G x 3
10 0
x
H 4
H 4.
The domain is all x t 0 except x 9. The graphing utility does not show the hole at 9, 6 .
, you have
1 1 H H x 4
x x 3 H x 3x 0 H 2
f x L H .
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
58
NOT FOR SALE
Chapter 1
Limits its and Their Properties
64. (a) No. The fact that f 2
x 3 x2 9 1 lim f x x o3 6
60. f x
existence of the limit of f x as x approaches 2. (b) No. The fact that lim f x the value of f at 2.
−9
2S r
65. (a) C
3
C 2S
r −3
The domain is all x z r3. The graphing utility does not § 1· show the hole at ¨ 3, ¸. © 6¹ 25 means that the values of f approach 25 as
x o8
4 has no bearing on
xo2
3
61. lim f x
4 has no bearing on the
xoc
sides of c, but does not have to be defined at c itself. The value of f at c has no bearing on the limit as x approaches c. 63. (i) The values of f approach different numbers as x approaches c from different sides of c:
3
S
| 0.9549 cm
5.5 | 0.87535 cm 2S 6.5 When C 6.5: r | 1.03451 cm 2S So 0.87535 r 1.03451.
(b) When C
(c)
x gets closer and closer to 8. 62. In the definition of lim f x , f must be defined on both
6 2S 5.5: r
2S r
lim
x o3 S
6; H
0.5; G | 0.0796
4 3 2.48 Sr , V 3 4 3 (a) 2.48 Sr 3 1.86 r3
66. V
S
r | 0.8397 in.
y 4
2.45 d
(b)
3
1 x −4 −3 −2 −1 −1
2
1
3
d 2.51
V
4 2.45 d S r 3 d 2.51 3
2
4
0.5849 d
r 3 d 0.5992
0.8363 d r d 0.8431 (c) For H 2.51 2.48 0.03, G | 0.003
−3 −4
(ii) The values of f increase without bound as x approaches c: y
67. f x
1 x 1 x
lim 1 x
1x
e | 2.71828
xo0
6 5
y
4 7
3 2 1 x −3 −2 −1 −1
2
3
4
5
3
−2
(0, 2.7183)
2 1
(iii) The values of f oscillate between two fixed numbers as x approaches c:
x −3 −2 −1 −1
1
2
3
4
5
y 4
x
f (x)
x
f (x)
–0.1
2.867972
0.1
2.593742
–0.01
2.731999
0.01
2.704814
–0.001
2.719642
0.001
2.716942
–0.0001
2.718418
0.0001
2.718146
–0.00001
2.718295
0.00001
2.718268
3
x −4 −3 −2
2
3
4
−3 −4
INSTRUCTOR US USE SE ONLY ONLY Y –0.000001
2.718283 2.71828
0.000001
2.718280
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section 1.2
x 1 x 1
68. f x
73. False. Let
x
–1
–0.5
–0.1
0
0.1
0.5
1.0
f(x)
2
2
2
Undef.
2
2
2
2
xo0
1 x 1, x z 0, f x
x
1 x 1 x
f x
x 4, x z 2 ® 2 x ¯0,
f 2
0
lim f x
Note that for
lim x 4
xo2
2.
x 4, x z 2 ® 2 x ¯0,
f x
3
lim f x 1
2
−1
x
lim
f x
(2.001, 0.001)
1.998
Using the zoom and trace feature, G 2 G , 2 G 1.999, 2.001 . Note:
x 4 x 2
70.
1 4
x approaches
76. f x
2.002 0
2
0.5 is true.
As x approaches 0.25
(1.999, 0.001)
xo0
1 2
from either side, 0.5.
x
x
lim
0.001. So
0 z 2
x
x o 0.25 0.002
2 and f 2
xo2
75. f x
x
69.
lim x 4
xo2
1
−1
2 z 0
xo2
74. False. Let
y
−2
59
72. True
x
lim f x
Finding Limits Graphically and Numerically
0 is false.
f x
x is not defined on an open interval
containing 0 because the domain of f is x t 0.
x 2 for x z 2.
77. Using a graphing utility, you see that
0.005
sin x 1 x sin 2 x lim 2, etc. xo0 x lim
(2.999, 0.001) (3.001, 0.001)
2.998
xo0
3.002 0
So, lim
xo0
From the graph, G 0.001. So 3 G , 3 G 2.999, 3.001 . Note:
x 2 3x x 3
tan x x tan 2 x lim xo0 x
x for x z 3.
lim
L1 and lim f x xoc
1
xo0
x c has no bearing on the existence of the limit of f x as x o c. xoc
n.
78. Using a graphing utility, you see that
71. False. The existence or nonexistence of f x at
79. If lim f x
sin nx x
So, lim
xo0
2,
tan nx x
etc. n.
L2 , then for every H ! 0, there exists G1 ! 0 and G 2 ! 0 such that
x c G1 f x L1 H and x c G 2 f x L2 H . Let equal the smaller of G1 and G 2 . Then for
x c G , you have L1 L2
L1 f x f x L2 d L1 f x f x L2 H H . Therefore,
L1 L2 2H . Since H ! 0 is arbitrary, it follows that L1
L2 .
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
60
NOT FOR SALE
Chapter 1
80. f x
Limits its and Their Properties
mx b, m z 0. Let H ! 0 be given. Take
H
G
m
83. Answers will vary. x 2 x 12 xo4 x 4
.
84. lim
H
If 0 x c G
, then
7
f 4 >0.1@
n
4 >0.1@
m x c H
1
4.1
7.1
mx mc H
2
4.01
7.01
mx b mc b H
3
4.001
7.001
4
4.0001
7.0001
n
4 >0.1@
1
3.9
6.9
2
3.99
6.99
3
3.999
6.999
4
3.9999
6.9999
m
which shows that lim mx b
mc b.
xoc
81. lim ª¬ f x Lº¼ xoc
0 means that for every H ! 0 there
exists G ! 0 such that if 0 x c G, then
f x L 0
H.
n
n
f 4 >0.1@
n
n
This means the same as f x L H when 0 x c G. So, lim f x xoc
82. (a)
85. The radius OP has a length equal to the altitude z of the h h triangle plus . So, z 1 . 2 2
L.
3x 1 3x 1 x 2 0.01
1 9 x 2 1 x 2 100 9 x4 x2
1 100
1 10 x 2 1 90 x 2 1 100
Area triangle Area rectangle
1 § h· b¨1 ¸ 2 © 2¹
bh
h 2
2h
1
5 h 2
10 x 2 1 0 and 90 x 2 1 0.
Let a, b
bh
Because these are equal,
So, 3x 1 3x 1 x 2 0.01 ! 0 if 1 § , ¨ 90 ©
1 § h· b¨1 ¸ 2 © 2¹
1 · ¸. 90 ¹
h
For all x z 0 in a, b , the graph is positive. You can
1 2 . 5
P
verify this with a graphing utility.p (b) You are given lim g x xoc
L ! 0. Let
1 L. There exists G ! 0 such that 2 0 x c G implies that
H
h
O
b
L . That is, 2 L L g x L 2 2 3L L g x 2 2 g x L H
For x in the interval c G , c G , x z c, you have g x !
L ! 0, as desired. 2
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section 1.3
Evaluating Limits Analytically
86. Consider a cross section of the cone, where EF is a diagonal of the inscribed cube. AD
side of the cube. Then EF By similar triangles, EF AG BC AD
A
E
3 2x
Solving for x,
3 2 2x x
2. Let x be the length of a
x 2.
3 x 3
x 2 2
3, BC
61
6 2x
B
G
D
F
C
6 6 3 2 2
9 2 6 | 0.96. 7
Section 1.3 Evaluating Limits Analytically 5
1.
4.
−4
8
10
−5
10
−3
− 10
(a) lim h x
0
xo4
(b) lim h x
f t
5
x o 1
(a) lim f t
0
t o4
(b) lim f t
10
2.
t t 4
5
t o 1
0
10
−5
g x
x 9
(a) lim g x
2.4
(b) lim g x
4
xo4 xo0
x o 2
8
2
4
7. lim 2 x 1
16
2 0 1
xo0
8. lim 3x 2 x o 3
3
x o 3
−
10. lim x 2 1 x o1
−4
f x
(a) lim f x xo0
(b)
§ ¨ ©
S· ¸ 6¹
1 1 2
99
0 2
18 12 1
0 12. lim 3 x3 2 x 2 4
x 1
x o3
14. lim
xo4
3
3
31
x 4
15. lim x 3
3
2
x o 4
3
2
5
2
4 4
4
7
3 1 2 1 4
x o1
13. lim
0
2 3 4 3 1
x o3
lim f x | 0.524
x oS 3
7
3 3
2
11. lim 2 x 2 4 x 1
x cos x
1
3 3 2
9. lim x 2 3 x
4
3.
23
xo2
6. lim x 4
x 3
12
5. lim x3
2 2
1
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
62
NOT FOR SALE
Chapter 1
16. lim 2 x 1
Limits its and Their Properties ª¬2 0 1º¼
3
xo0
1 17. lim xo2 x
3
2 42 3 4 1
26. (a) lim f x
1
xo4
(b) lim g x
1 2
21 6
3
x o 21
(c) lim g f x
x 19. lim 2 x o1 x 4
20. lim
x o1
21. lim
xo7
22. lim
xo2
2 3 2
2
27.
1 12 4
2x 3 x 5
1 5
2 1 3
x 2 x 4 x o1
(b) lim g x xo4
21 3
2 2 2 4
2 2
51 43
24. (a) lim f x xo4
42
x o3
(c) lim g f x x o1
37. (a) lim ¬ª5 g x ¼º
2
g 3
2
xoc
g x
lim f x lim g x
xoc
xoc
(c) lim ¬ª f x g x ¼º xoc
xoc
f x
g x
cos
S
sin
S 2
xoc
§ 3· 4¨ ¸ © 2¹ xoc
ª lim f x ºª lim g x º »«x o c ¬«x o c ¼¬ ¼»
lim g x
2
0
cos 3S
1
1
lim sin x
sin
5S 6
1 2
lim cos x
cos
5S 3
1 2
§S x · 35. lim tan ¨ ¸ x o3 © 4 ¹
tan
3S 4
1
§S x · 36. lim sec¨ ¸ xo7 © 6 ¹
sec
7S 6
2 3 3
x o 5S 6
x o 5S 3
3 2
3 2
5 6
6
lim f x lim g x
xoc
lim f x
xoc
1 2
3
3 2
4 lim f x
(b) lim ª¬ f x g x º¼ xoc
(d) lim
0
10
lim f x lim g x
xoc
xoc
38. (a) lim ª¬4 f x º¼ xoc
5 2
ª lim f x º ª lim g x º ¬« x o c ¼» «¬x o c ¼»
(c) lim ª¬ f x g x º¼ xoc xoc
16
31
xoc
f x
tan S
32. lim cos 3x
34.
4
1
2
sec 0
64
3
(b) lim ¬ª f x g x º¼
(d) lim
S
3
31. lim sec 2 x
33.
g 4
5 lim g x
xoc
2
x oS
7
4 1
(b) lim g x
Sx
xo0
4
g 4
x o 3
x o1
1
16
(c) lim g f x 25. (a) lim f x
30. lim sin
7
3
sin
64
3
x o 3
Sx
xo2
g f 1
x o1
29. lim cos x o1
7 2
(c) lim g f x
(b) lim g x
x oS
3 7
3x x 2
23. (a) lim f x
28. lim tan x 1 6
15
lim sin x
x oS 2
3
g 21
xo4
2 18. lim x o 3 x 2
21
32 12
3 1 2 2
§ 3 ·§ 1 · ¨ ¸¨ ¸ © 2 ¹© 2 ¹
2
3 4
3
INSTRUCTOR U USE ONLY xoc
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section 1.3
39. (a) lim ª¬ f x º¼ xoc
ª lim f x º ¬« x o c ¼»
3
f x
(b) lim
xoc
xoc
ª lim f x º ¬«x o c ¼»
18
lim f x
lim 18
xoc
3
27
lim f x
8
27
xo0
(b) lim g x x o 1
−4
2 x2 x 3 and g x x 1 1.
at x
lim f x
729
xo0
lim g x
x o 1
4
23
27
0 1
lim f x
23
9 −8
4
−8
x3 8 and g x x 2 2.
47. f x
1
1 1
x o 1
2 x 3 agree except
5
x o 1
x2 x agree except at x
2
x 3x agree except at x 2
x 3 and h x
42. f x
x
lim f x
2
4
0.
(a) lim g x
lim g x
x o 1
−3
46. f x
2
x 1 agree except at
3
3
2
63
1.
x o 1
12
3 2
ª lim f x º ¬« x o c ¼»
x 1 and g x
x
2
4 3 2
x2 1 and g x x 1
45. f x
64
27 18
ª lim f x º ¬«x o c ¼»
23
32
xoc
xoc
(d) lim ª¬ f x º¼ xoc
at x
lim f x
lim g x
xo2
x 2 2 x 4 agree except
12
xo2 12
0.
(a) lim h x xo2
(b) lim h x xo0
43. f x
x
lim f x
2 3
1
lim f x
0 3
3
xo2 xo0
−9
x3 x agree except at x 1
x x 1 and g x 1.
(a) lim g x x o1
(b) lim g x x o 1
lim f x x o1
lim f x
x o 1
1 and f x x 1
44. g x
x
3 4
lim f x
f x
(c) lim ª¬ f x º¼ xoc
x
3 lim f x
3
2
41. f x
4
32
xoc
xoc
lim f x
xoc
40. (a) lim 3 f x
(b) lim
4 3
xoc
(c) lim ¬ª3 f x ¼º (d) lim ª¬ f x º¼ xoc
3
Evaluating Limits Analytically
9 0
x3 1 and g x x 1
48. f x
x
1.
lim f x
2
3
x o 1 7
0 x agree except at x2 x
−4
4 −1
(a) lim f x does not exist. x o1
xo0
lim g x
x o 1
0.
(b) lim f x
x 2 x 1 agree except at
1
49. lim
x x
50. lim
3x 2x
x o 0 x2
x o 0 x2
lim
xo0 x
lim
x
xo0 x
x 1 3x 2
x
lim
xo0 x
lim
1 1
xo0 x
3 2
1 3 2
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
64
NOT FOR SALE
Chapter 1
51. lim
Limits its and Their Properties
x 4 16
lim
x o 4 x2
xo4
x 4 x 4 x 4
lim
xo4 x
3 x x o3 x2 9
xo4
x 5 3 x 4
56. lim
x 1 2 x 3
57. lim
x 5 x
xo0
lim
x o3
5
lim
xo0
2 x x
2
lim
1 1 x 4 4 60. lim xo0 x
61. lim
5 2
2 x 2 2 x
lim
4 x 4 4 x 4
1 x 5 3
xo0
lim
x o3
x 5 x 5
lim
x 2 3
5 6
x 4 x 1 x 4 x 2 x 1 3 lim x o 4 x 2 6
5 6
lim
xo4
1 2
1 6
x 3 3 ¬ª x 1 2¼º
lim
x o3
1 x 1 2
1 4
x
3 x 3 x
1 5
2 5
5 10
2 2
1 2 x
lim
xo0
1 9 3
5 5
2 x 2 x
x
1 x 5
lim
xo0
2 x
lim
3 x 3 x
54. lim
lim
x 5
3 3 x
lim
xo0
1 2
1
3 x 3
2 2
2 4
1 9
xo0
x 1 lim x o 0 4 x 4
2 x 'x 2 x 'x
'x o 0
x2 5x 4 x o 4 x2 2x 8
1 6
xo4
3
5 5
lim
xo0
5
2 x x
lim
xo0
xo0
1 1 3 3 x 59. lim xo0 x
x
x 1 2 x 1 2
x 5 x
lim
x 3 x 2 x 3 x 3
x 5 3 x 5 3
x 1 2 x 3
xo0 x
58. lim
1 x o3 x 3 lim
x 5 9 x 4 x 5
xo0
lim
x o 3
x o 3 x
x 5 3 x 4
lim
xo4
lim
x2 x 6 x2 9
1 8
lim
xo4
x o3
x o 3
3 x x o 3 x 3 x 3
52. lim
55. lim
1 4
53. lim
1 16
2 x 2'x 2 x 'x lim 2 2
lim
'x o 0
'x o 0
62. lim
x
'x o 0
2
'x
'x o 0
63. lim
'x x 2
x
x 2 2 x'x 'x x 2 2
lim
'x o 0
lim
'x
'x 2 x 'x 1 x 2 2 x 1
'x o 0
2
'x
'x 2 x ' x
lim 2 x 'x
'x
'x o 0
2x
x 2 2 x'x 'x 2 x 2'x 1 x 2 2 x 1 2
lim
'x o 0
lim 2 x 'x 2
'x o 0
'x 2x 2
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
Section 1.3
x
64. lim
'x x3
2
lim
'x
'x 3 x 2 3 x'x 'x
lim
'x
ª§ sin x ·§ 1 ·º lim «¨ ¸¨ ¸» x o 0 © x ¹© 5 ¹ ¬ ¼
sin x x o 0 5x
66. lim
3 1 cos x
xo0
67. lim
sin x 1 cos x
1· ¸ © 5¹
lim
cos T tan T
lim
T
T o0
lim 3x 2 3 x'x 'x
'x o 0
I oS
73.
3 0
0 74.
lim
cos x x
3x 2
S
1
x oS 2
1 tan x x cos x
lim
x o S 4 sin
x o S 4 sin
sin x cos x
lim
x sin x cos x
1
lim
x o S 4 cos
1
T
cos x sin x x cos x cos 2 x
x o S 4 cos
0
sin T
T o0
lim sin x
x o S 2 cot
lim
2
S 1
72. lim I sec I
1 5
ª sin x 1 cos x º » x x ¼
1 0 68. lim
2
x o 0« ¬
x2
xo0
1 §¨
ª § 1 cos x ·º lim «3¨ ¸» xo0 x «¬ © ¹»¼
x
3
'x
'x o 0
'x o 0
65. lim
65
x3 3 x 2 'x 3x 'x 'x x3
3
'x o 0
Evaluating Limits Analytically
x
lim sec x
x oS 4
ª sin x º lim sin x» x o 0« ¬ x ¼
sin 2 x 69. lim xo0 x tan 2 x xo0 x
sin 2 x x o 0 x cos 2 x
70. lim
lim
1 0
1 sin 0
2
0
ª sin x sin x º lim « » xo0 cos 2 x ¼ ¬ x
75. lim
sin 3t 2t
76. lim
sin 2 x 3x
t o0
0
§ sin 3t ·§ 3 · lim¨ ¸¨ ¸ 3t ¹© 2 ¹
t o 0©
1
ho0
cos h
2
h
lim
77. f x
x 2 x
3 2
xo0
ª1 cos h 1 cos h º» h ¼
§1· 2 1 ¨ ¸ 1 © 3¹
h o 0« ¬
0 0
3· ¸ © 2¹
ª § sin 2 x ·§ 1 ·§ 3x ·º lim «2¨ ¸» ¸¨ ¸¨ ¬ © 2 x ¹© 3 ¹© sin 3 x ¹¼
x o 0 sin
71. lim
1 §¨
2 3
0
2
x
–0.1
–0.01
–0.001
0
0.001
0.01
0.1
f (x)
0.358
0.354
0.354
?
0.354
0.353
0.349
It appears that the limit is 0.354. 2
−3
3
−2
The graph has a hole at x Analytically, lim
xo0
x 2 x
0. 2
x 2 x
lim
xo0
lim
xo0 x
2
x 2 2 x 2
x 2 x 2
2
lim
xo0
2 2
1 x 2
2
1 2 2
2 | 0.354. 4
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
66
NOT FOR SALE
Chapter 1
78. f x
Limits its and Their Properties
4 x x 16
x
15.9
15.99
15.999
16
16.001
16.01
16.1
f (x )
–0.1252
–0.125
–0.125
?
–0.125
–0.125
–0.1248
It appears that the limit is –0.125. 1
0
20
−1
The graph has a hole at x
16.
4 x x o16 x 16
Analytically, lim
79. f x
lim
x o16
4 x x 4 x 4
1 x 4
lim
x o16
1 . 8
1 1 2 x 2 x
x
–0.1
–0.01
–0.001
0
0.001
0.01
0.1
f (x )
–0.263
–0.251
–0.250
?
–0.250
–0.249
–0.238
It appears that the limit is –0.250. 3
−5
1
−2
The graph has a hole at x
0.
1 1 x 2 2 Analytically, lim xo0 x
80. f x
lim
2 2 x
xo0
2 2 x
1 x
lim
x
2 x
xo0 2
1 x
lim
1
1 . 4
2 x
xo0 2
x5 32 x 2
x
1.9
1.99
1.999
1.9999
2.0
2.0001
2.001
2.01
2.1
f (x )
72.39
79.20
79.92
79.99
?
80.01
80.08
80.80
88.41
It appears that the limit is 80. 100
−4
3 −25
The graph has a hole at x x5 32 xo2 x 2
Analytically, lim
2. lim
x
2 x 4 2 x3 4 x 2 8 x 16 x 2
xo2
lim x 4 2 x3 4 x 2 8 x 16
xo2
80.
INSTRUCTOR USE ONLY Hint Use long division to fact factor x5 32. ) (Hint:
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section 1.3
81. f t
Evaluating Limits Analytically
67
sin 3t t
t
–0.1
–0.01
–0.001
0
0.001
0.01
0.1
f (t)
2.96
2.9996
3
?
3
2.9996
2.96
It appears that the limit is 3. 4
− 2
2 −1
The graph has a hole at t Analytically, lim
t o0
82. f x
sin 3t t
0. § sin 3t · lim 3¨ ¸ © 3t ¹
3 1
t o0
3.
cos x 1 2x2
x
–1
–0.1
–0.01
0.01
0.1
1
f (x)
–0.2298
–0.2498
–0.25
–0.25
–0.2498
–0.2298
It appears that the limit is –0.25. 1
−
−1
The graph has a hole at x Analytically,
0.
cos x 1 cos x 1 2x cos x 1
cos 2 x 1 2 x 2 cos x 1 sin 2 x 2 x cos x 1 2
sin 2 x 1 x2 2 cos x 1
ª sin 2 x º 1 lim « 2 » 2 cos x 1 »¼ ¬ x
x o 0«
§ 1 · 1¨ ¸ © 4¹
1 4
0.25
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
68
NOT FOR SALE
Chapter 1
Limits its and Their Properties
sin x 2 x
83. f x
x
–0.1
–0.01
–0.001
0
0.001
0.01
0.1
f (x)
–0.099998
–0.01
–0.001
?
0.001
0.01
0.099998
It appears that the limit is 0. 1
− 2
2
−1
The graph has a hole at x sin x xo0 x
0. § sin x 2 · lim x¨ ¸ xo0 © x ¹
2
Analytically, lim
0 1
0.
sin x 3 x
84. f x
x
–0.1
–0.01
–0.001
0
0.001
0.01
0.1
f (x)
0.215
0.0464
0.01
?
0.01
0.0464
0.215
It appears that the limit is 0. 2
−3
3
−2
The graph has a hole at x Analytically, lim
sin x x
xo0 3
85. lim
f x 'x f x
86. lim
f x 'x f x
'x o 0
'x o 0
'x
'x
0. lim
3
xo0
lim
§ sin x · x2 ¨ ¸ © x ¹
'x o 0
'x
lim
'x
x 'x 'x
lim
'x o 0
lim
87. lim
0.
3 x 'x 2 3 x 2
'x o 0
'x o 0 'x
f x 'x f x
0 1
'x o 0
x
x 'x
x 'x 'x
lim
'x o 0
x 'x x x
3 x 30 x 2 3x 2 'x
lim
'x o 0
x
1 x 'x
x 'x x 'x 1 x 2 x
lim
'x o 0
3'x 'x
3
x x
1 1 x 3 lim x 'x 3 'x o 0 'x x 3 x 'x 3 1 lim 'x o 0 x 'x 3 x 3 'x lim
x
'x 'x 3 x 3 'x
lim
x
1 'x 3 x 3
'x o 0
'x o 0
1
x 3
2
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section 1.3
88. lim
f x 'x f x 'x
'x o 0
lim
'x o 0
lim
x
'x 4 x 'x x 2 4 x 2
'x 'x 2 x 'x 4 'x
'x o 0
89. lim 4 x 2 d lim f x d lim 4 x 2 xo0
xo0
Evaluating Limits Analytically
xo0
4 d lim f x d 4
lim
'x o 0
lim 2 x 'x 4
xo0
95. f x
x sin
4. − 0.5
b d lim f x d b xoa
xoa
91. f x
1 x
0.5
0.5
90. lim ª¬b x a º¼ d lim f x d lim ª¬b x a º¼ xoa xoa xoa
Therefore, lim f x
x 2 2 x'x 'x 2 4 x 4'x x 2 4 x 'x
2x 4
'x o 0
xo0
Therefore, lim f x
69
b.
− 0.5
1· § lim ¨ x sin ¸ x¹
96. h x
x cos x
0
x o 0©
x cos
4
1 x
0.5
− 3 2
3 2
− 0.5
0.5
−4
lim x cos x
92. f x
− 0.5
0
xo0
1· § lim ¨ x cos ¸ x¹
x o 0©
x sin x 6
97. You say that two functions f and g agree at all but one point (on an open interval) if f x g x for all x in
− 2
2
x2 1 and g x x 1 except x 1.
0
xo0
93. f x
the interval except for x 98. f x
−2
lim x sin x
x sin x
−2
2
lim
xoc
94. f x
x 1 agree at all points
f x
g x
for which lim f x
−6
lim x sin x
c, where c is in the interval.
99. An indeterminant form is obtained when evaluating a limit using direct substitution produces a meaningless fractional expression such as 0 0. That is,
6
lim g x
xoc
0
xo0
0
xoc
0
100. If a function f is squeezed between two functions h and g, h x d f x d g x , and h and g have the same limit
x cos x
L as x o c, then lim f x exists and equals L. xoc
6
− 2
2
−6
lim x cos x
xo0
0
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
70
NOT FOR SALE
Chapter 1
101. f x
Limits its and Their Properties
x, g x
sin x, h x
103. s t
sin x x
3
lim
f g
t o2
h
−5
16t 2 500 s 2 s t 2 t
436 16t 2 500 lim t o2 2t
−3
lim
When the x-values are "close to" 0 the magnitude of f is approximately equal to the magnitude of g. So, g f | 1 when x is "close to" 0. x, g x
2t
t o2
5
102. f x
16 2 500 16t 2 500 2
lim
sin 2 x x
sin 2 x, h x
t o2
lim
16 t 2 4
2 t 16 t 2 t 2
2t lim 16 t 2
t o2 t o2
64 ft/sec
The wrench is falling at about 64 feet/second.
2
g −3
3
h f −2
When the x-values are "close to" 0 the magnitude of g is "smaller" than the magnitude of f and the magnitude of g is approaching zero "faster" than the magnitude of f. So, g f | 0 when x is "close to" 0. 104. s t
16t 2 500
§5 5 · s¨¨ ¸ s t 2 ¸¹ lim © §5 5 · 5 5 t o ¨¨ ¸¸ t © 2 ¹ 2
500 16
0 when t
lim
5 5 sec. The velocity at time a 2
5 5 is 2
0 16t 2 500
§5 5 · t o ¨¨ ¸¸ © 2 ¹
5 5 t 2 125 · § 16¨ t 2 ¸ 4 ¹ © lim §5 5 · 5 5 t o ¨¨ ¸¸ t © 2 ¹ 2 § 5 5 ·§ 5 5· 16¨¨ t ¸¨ t ¸ ¸¨ 2 2 ¸¹ © ¹© lim §5 5 · 5 5 t o ¨¨ ¸¸ t © 2 ¹ 2 ª § 5 5 ·º lim «16¨¨ t ¸» 5 5« 2 ¸¹»¼ to © ¬ 2
80 5 ft/sec | 178.9 ft/sec.
The velocity of the wrench when it hits the ground is about 178.9 ft/sec.
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section 1.3
105. s t
lim
Evaluating Limits Analytically
71
4.9t 2 200
3t
t o3
lim
4.9 3 200 4.9t 2 200
lim
4.9 t 2 9
2
s 3 s t
3t
t o3
t o3
lim
3t 4.9 t 3 t 3
t o3
3t
lim ª 4.9 t 3 º¼ t o 3¬ 29.4 m/sec The object is falling about 29.4 m/sec. 106. 4.9t 2 200
lim
s a s t
20 5 sec. The velocity at time a 7
20 5 is 7
0 ª¬4.9t 2 200º¼ t oa a t 4.9 t a t a lim t oa a t lim
a t
t oa
200 4.9
0 when t
ª § 20 5 ·º lim «4.9¨¨ t ¸» 20 5 « 7 ¸¹»¼ to © ¬ 7
28 5 m/sec | 62.6 m/sec.
The velocity of the object when it hits the ground is about 62.6 m/sec. 107. Let f x
1 x and g x
1/ x. lim f x and
111. If b 0, the property is true because both sides are equal to 0. If b z 0, let H ! 0 be given. Because
x o0
lim g x do not exist. However,
lim f x
x o0
ª 1 § 1 ·º lim ª f x g x º¼ lim « ¨ ¸» x o 0¬ xo0 x © x ¹¼ ¬ and therefore does not exist.
xoc
lim >0@
xo0
f x L H b whenever 0 x c G . So,
0
whenever 0 x c G , we have b f x L H or
108. Suppose, on the contrary, that lim g x exists. Then, xoc
because lim f x exists, so would lim ª¬ f x g x º¼ , xoc xoc which is a contradiction. So, lim g x does not exist. xoc
109. Given f x
112. Given lim f x xoc
lim x
xoc
n
Now f x 0
f x
ª lim xº ª lim x n 1 º ¬«x o c ¼» «¬x o c ¼» c ª« lim xº» ª« lim x n 2 º» ¬x o c ¼ ¬x o c ¼ cn.
M f x d f x g x d M f x
113. c ª« lim xx n 2 º» ¬x o c ¼ c c lim xx n 3 xoc
0.
xoc
x , n is a positive integer, then
lim xx n 1
f x 0 H for
x c G . Therefore, lim f x
n
xoc
"
0:
f x 0 H whenever 0 x c G .
0 H for
every H ! 0, any value of G ! 0 will work. 110. Given f x
bL.
For every H ! 0, there exists G ! 0 such that
a G ! 0 such that f x b H whenever b b
bf x bL H
which implies that lim ª¬bf x º¼ xoc
b, show that for every H ! 0 there exists
x c G . Because f x b
L, there exists G ! 0 such that
lim M f x
xoc
d
lim f x g x d lim M f x
xoc
xoc
M 0 d lim f x g x d M 0 xoc
0 d lim f x g x d 0 xoc
Therefore, lim f x g x xoc
0.
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
72
Chapter 1
NOT FOR SALE
Limits its and Their Properties
If lim f x
114. (a)
0, then lim ª¬ f x º¼ xoc
xoc
122. False. Let
0.
f x
f x d f x d f x xoc
and g x
x2.
Then f x g x for all x z 0. But
lim ª f x º¼ d lim f x d lim f x
x oc¬
1 x2 2
lim f x
xoc
0 d lim f x d 0
lim g x
xo0
0.
xo0
xoc
Therefore, lim f x
0.
xoc
(b) Given lim f x
123. lim
x o0
1 cos x x
lim
x o0
1 cos 2 x x o 0 x 1 cos x
L:
xoc
1 cos x 1 cos x x 1 cos x sin 2 x x o 0 x 1 cos x
lim
For every H ! 0, there exists G ! 0 such that f x L H whenever 0 x c G . Since
lim
x o0
f x L d f x L H for x c G , then lim f x
sin x sin x x 1 cos x
sin x º ª sin x º ª » «lim » «lim ¬x o 0 x ¼ ¬x o 01 cos x ¼
L.
xoc
lim
1 0
0
115. Let
f x
4, if x t 0 ® ¯4, if x 0
lim f x
xo0
lim 4
4.
xo0
lim f x does not exist because for
xo0
x 0, f x
116. lim f x x o 2
4 and for x t 0, f x lim f x
lim f x
x o 2
The value of f at x
xo2
124. f x
0, if x is rational ® ¯1, if x is irrational
g x
0, if x is rational ® ¯x, if x is irrational
lim f x does not exist.
xo0
4.
No matter how "close to" 0 x is, there are still an infinite number of rational and irrational numbers so that lim f x does not exist.
3
xo0
lim g x
2 is irrelevant.
0
xo0
sin x 118. False. lim x oS x
0
when x is "close to" 0, both parts of the function are "close to" 0.
2
117. The limit does not exist because the function approaches 1 from the right side of 0 and approaches 1 from the left side of 0.
−3
3
sec x 1 x2
(a) The domain of f is all x z 0, S /2 nS .
−2
0
S
125. f x
2
(b) − 3 2
119. True.
3 2
120. False. Let
f x
x x z 1 , ® ¯3 x 1
Then lim f x x o1
−2
c
The domain is not obvious. The hole at x apparent. 1 (c) lim f x xo0 2
1.
1 but f 1 z 1.
121. False. The limit does not exist because f x
approaches 3 from the left side of 2 and approaches 0 from the right side of 2.
(d)
4
−3
sec x 1 x2
tan 2 x x 2 sec x 1
6
−2
sec x 1 sec x 1 x2 sec x 1
sec x 1 x o0 x2
So, lim
0 is not
sec 2 x 1 x sec x 1 2
1 § sin 2 x · 1 ¨ ¸ cos 2 x © x 2 ¹ sec x 1
1 § sin 2 x · 1 ¨ 2 ¸ x o 0 cos 2 x x sec x 1 © ¹
lim
§1· 1 1 ¨ ¸ © 2¹
1 . 2
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section 1.4
126. (a) lim
xo0
1 cos x x2
lim
xo0
lim
Continuity and OneOne One-Sided Limits
73
1 cos x 1 cos x 1 cos x x2 1 cos 2 x 1 cos x
x o 0 x2
sin 2 x 1 x o 0 x2 1 cos x lim
· ¸ © 2¹
1 §¨
1 2
(b) From part (a), 1 cos x 1 1 1 | 1 cos x | x 2 cos x | 1 x 2 for x | 0. x2 2 2 2
(c) cos 0.1 | 1
1 2 0.1 2
0.995
(d) cos 0.1 | 0.9950, which agrees with part (c). 127. The graphing utility was set in degree mode, instead of radian mode.
Section 1.4 Continuity and One-Sided Limits lim f x
3
(b) lim f x
3
1. (a)
x o 4 x o 4
(c) lim f x xo4
(b)
xo2
lim f x
lim f x
(b) lim f x
0
x o 3
(c) lim f x x o3
(b)
3
lim f x
3
x o 3
(c) lim f x x o 3
2
x o 1
The function is NOT continuous at x 2. x o 8
1 x 8
8. lim x o 5
0
lim f x
x o 3
lim f x
2.
(c) lim f x does not exist.
The function is NOT continuous at x 4. (a)
0
x o 1
7. lim 0
lim f x
x o 1
(b)
2
lim f x
x o 3
6. (a)
2
The function is continuous at x 3. (a)
The function is NOT continuous at x
4 and is continuous
2
x o 2 x o 2
3
x o 2
(c) lim f x does not exist
x o 2
(c) lim f x
3
x o 2
(b) lim f x
3
The function is continuous at x on f, f . 2. (a)
lim f x
5. (a)
x 5 x 2 25
10. lim
2 x x2 4
x o 5
x o 2
11.
3
The function is NOT continuous at x f 3 4 z lim f x .
3 because
3 x 5
9. lim
3.
1 88
x
lim
x o 3
x 9
x x 9 2
2
1 16
3 55
3 10
lim
1 x 5
1 10
lim
1 x 2
x o 5
x o 2
1.
1 4
does not exist because
decreases without bound as x o 3.
x o 3
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
74
NOT FOR SALE
Chapter 1
12. lim
x o 9
Limits its and Their Properties
x 3 x 9
lim
x o 9
lim
x o 9
x 3 x 9 x 9
x
lim
x o 9
1 1 ' x x x 15. lim 'x 'x o 0
x 3 x 3 x 3
1 x 3
1 6
x x 'x x x 'x
'x o 0
x o 0
9
lim
13. lim
1 'x
'x 1 x x 'x 'x
lim
1 x x 'x
1 x x 0
16.
lim
'x x 'x x 2 x 2
'x
'x o 0
x o10
lim
'x o 0
'x o 0
x
lim
14.
x
x x
lim
x o 0
x
x 10
1
lim
x 10
x o10
x 10 x 10
1
1 x2
lim
x 2 2 x 'x 'x x 'x x 2 x
lim
2 x 'x 'x 'x
2
'x
'x o 0 2
'x lim 2 x 'x 1
'x o 0 'x o 0
2x 0 1
17. lim f x
lim
x o 3
x o 3
x o 2
lim x 2 4 x 6
lim f x
x o 2
lim f x x o1
20. lim f x x o1
2
2 2 2
2
25. lim 2 a xb does not exist because x o3
lim 2 a xb
x o 3
2 3
5
2 4
6.
and lim x 1
x o1 x o1
2
2
19. lim f x lim f x
x o 2
x o 2
x o 2
lim f x
24. lim 2 x a xb
5 2
lim x 2 4 x 2
18. lim f x
xo2
x 2 2
2x 1
x o1
lim x 3 1
x o1
lim 2 a xb
x o 3
2
c x f· § 26. lim¨1 dd gg¸ x o1 e 2 h¹ ©
2
1 1
2
2 lim 1 x
x o1
27. f x 0
1 x2 4
has discontinuities at x 2 and x f 2 and f 2 are not defined.
21. lim cot x does not exist because
2 because
x oS
lim cot x and lim cot x do not exist.
x oS
22.
x oS
lim sec x does not exist because
x o S 2
23. lim 5a xb 7 x o 4
axb
lim sec x do not exist.
3 for 3 d x 4
8
1 because f 1 is not
defined.
x o S 2
5 3 7
x2 1 x 1
has a discontinuity at x
x oS 2
lim sec x and
28. f x
29. f x
axb 2
x
has discontinuities at each integer k because lim f x z lim f x . x ok
x ok
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section 1.4
x 1 x, ° x 1 has a discontinuity at 30. f x ®2, °2 x 1, x ! 1 ¯ x 1 because f 1 2 z lim f x 1.
has a nonremovable discontinuity at x 6 because lim f x does not exist, and has a removable x o 6
discontinuity at x
49 x is continuous on >7, 7@. 2
32. f t
33. lim f x
3
x o 0
lim f x
lim
x o6
9 t 2 is continuous on >3, 3@.
3
lim f x . f is continuous on >1, 4@.
x o 0
47. f x
xo6 x
x
6 because 1 6
x 2 2 x 5
x o5
2 because
discontinuity at x 35. f x 36. f x
3 has a nonremovable discontinuity at x 2
x 37. f x 38. f x
0.
2. x 9 is continuous for all real x. 2
x 2 2 x 1 is continuous for all real x.
1 4 x2
39. f x
1
discontinuities at x
lim f x
48. f x
x
x o 2 x
1 5
1 . 7
x 1 2 x 1
has a nonremovable discontinuity at x 2 because lim f x does not exist, and has a removable x o 2
discontinuity at x lim f x
1 because 1 2
lim
x o1 x
1 . 3
has nonremovable
r2 because lim f x and
x 7
49. f x
x 7
xo2
lim f x do not exist.
lim
x o 2
x o1
2 x 2 x
1 . 12
has a nonremovable discontinuity at x 5 because lim f x does not exist, and has a removable
34. g 2 is not defined. g is continuous on >1, 2 .
6 has a nonremovable discontinuity at x x
75
x 6 x 2 36
46. f x
x o1
31. g x
Continuity and OneOne One-Sided Limits
has a nonremovable discontinuity at x lim f x does not exist.
x o 2
7 because
x o 7
1 is continuous for all real x. 2 x 1
40. f x 41. f x 42. f x
3x cos x is continuous for all real x. cos
Sx 2
is continuous for all real x.
x is not continuous at x 0, 1. Because x2 x x 1 for x z 0, x 0 is a removable x2 x x 1 discontinuity, whereas x 1 is a nonremovable discontinuity.
43. f x
44. f x
x
x has nonremovable discontinuities at x2 1 1 and x 1 because lim f x and lim f x do x o1
x o 1
not exist. 45. f x
50. f x
x
x 8
has a nonremovable discontinuity at x 8 8 because lim f x does not exist. x o8
51. f x
x, x d 1 ® 2 ¯x , x ! 1
has a possible discontinuity at x 1.
f 1
1
lim f x
2.
lim x
x o1
x o1
lim f x
lim x 2
x o1
3.
f 1
1.
x o1
1½ ° ¾lim f x 1° x o1 ¿
1
lim f x x o1
f is continuous at x real x.
1, therefore, f is continuous for all
x is continuous for all real x. x2 1
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
76
Chapter 1
Limits its and Their Properties
2 x 3, x 1 ® 2 x t1 ¯x ,
52. f x
has a possible discontinuity at x 1.
f 1
2
1
1
lim f x
2.
lim 2 x 3
x o1
lim f x
f 1
lim x 2
x o1
1
1
lim f x x o1
f is continuous at x real x. 53. f x
1½ ° f x ¾lim ° x o1 ¿
x o1
x o1
3.
1.
1, therefore, f is continuous for all
x ° 1, x d 2 ®2 °3 x, x ! 2 ¯
has a possible discontinuity at x 1.
f 2
2.
x o 2
2 1 2
2
½ 2° ° f x does not exist. ¾ xlim o2 1° °¿
§x · lim ¨ 1¸ ¹ lim 3 x
lim f x
x o 2 © 2
lim f x
x o 2
2.
x o 2
Therefore, f has a nonremovable discontinuity at x 54. f x
x d 2 2 x, ® 2 x 4 x 1, x ! 2 ¯
has a possible discontinuity at x 1.
f 2
2 2
lim f x
2.
2.
4 lim 2 x
x o 2
x o 2
4
lim x 2 4 x 1
lim f x
x o 2
x o 2
½ ° ¾ lim f x does not exist. 3° x o 2 ¿
Therefore, f has a nonremovable discontinuity at x 55. f x
2.
Sx °tan , 4 ® ° x, ¯
2.
x 1 x t1
Sx °tan , 1 x 1 4 ® ° x, x d 1 or x t 1 ¯ has possible discontinuities at x 1. 2. 3.
f 1 lim f x
x o 1
f 1
1, x
f 1
1
1
x o1
lim f x
f is continuous at x
1
lim f x
1 x o1
1.
f 1
lim f x x o1
r1, therefore, f is continuous for all real x.
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section 1.4
Sx °csc , 6 ® °2, ¯
56. f x
Continuity and OneOne One-Sided Limits
77
x 3 d 2 x 3 ! 2
Sx °csc , 1 d x d 5 6 ® °2, x 1 or x ! 5 ¯
has possible discontinuities at x
1, x
5.
S
f 5
csc
1.
f 1
2.
lim f x
3.
f 1
csc
6
2
lim f x
2
x o1
lim f x
f 5
x o1
1 and x
2
2
x o5
f is continuous at x 57. f x
5S 6
lim f x
x o5
5, therefore, f is continuous for all real x.
csc 2 x has nonremovable discontinuities at
integer multiples of S 2.
63. f 1
3
Find a so that lim ax 4
3
a 1 4
3
a
7.
Find a so that lim ax 5
3
a 1 5
3
x o1
58. f x
Sx
has nonremovable discontinuities at each 2 2k 1, k is an integer.
59. f x
tan
ax 8b has nonremovable discontinuities at
each integer k. 60. f x
64. f 1
3 x o1
5 a xb has nonremovable discontinuities at
a
each integer k. 61. lim f x
0
lim f x
0
x o 0 x o 0
65. f 2
8
Find a so that lim ax 2 x o 2
f is not continuous at x
2.
66. lim g x x o 0
50
lim g x
x o 0 −8
Let a
8
2.
8 a
8 22
2.
4 sin x 4 x a lim a 2 x
lim
x o 0
x o 0
4.
− 10
62. lim f x
0
lim f x
0
x o 0 x o 0
f is not continuous at x
4.
20
−8
8
−10
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
78
NOT FOR SALE
Chapter 1
Limits its and Their Properties
67. Find a and b such that lim ax b
a b
x o 1
a b
3a
x o 3
b
f x
1
a
2 1
b
x2 a2 xoa x a lim x a
1
xoa
8 a
Find a such 2a
x
1
°x 2 3 x, x ! 4 ® °¯2 x 5, x d 4
75. g x
lim
69. f g x
2.
x d 1 2, ° x 1, 1 x 3 ® °2, x t 3 ¯
2 4
xoa
3a b
2
4a
68. lim g x
2 and lim ax b
2a
There is a nonremovable discontinuity at x
4.
4.
10
2
Continuous for all real x.
−2
8 −2
1 x 1
70. f g x
Nonremovable discontinuity at x all x ! 1. 1 x 5 6
71. f g x
2
1. Continuous for
72. f g x
sin x
f 0
1 x 1 2
Nonremovable discontinuities at x
axb
cos x 1 , x 0 ° x ® °5 x, x t 0 ¯
5 0
lim f x
r1
cos x
x o 0
lim f x
1
x
lim 5 x
x o 0
2
0 lim
x o 0
0
0
x o 0
Therefore, lim f x
0
xo0
Continuous for all real x 73. y
76. f x
f 0 and f is continuous on
the entire real line. x 0 was the only possible discontinuity.
x
3
Nonremovable discontinuity at each integer 0.5 −7
−3
2
3
−3
− 1.5
74. h x
77. f x
1
x
Continuous on f, f
1 x 2
Nonremovable discontinuities at x 2
x x2 x 2
1 and x
2.
78. f x
x
x 3
Continuous on >3, f −3
4
79. f x −2
sec
Sx 4
Continuous on:
!, 6, 2 , 2, 2 , 2, 6 , 6, 10 , ! 80. f x
x 1 x
INSTRUCTOR USE ONLY Continuous on 0, f
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section 1.4
81. f x
87. f x
sin x x
f 0 −4
x3 x 1
1 and f 1
1
By the Intermediate Value Theorem, f c
4
−2
The graph appears to be continuous on the interval >4, 4@. Because f 0 is not defined, you know that f has a discontinuity at x 0. This discontinuity is removable so it does not show up on the graph. x3 8 x 2
least one value of c between 0 and 1. Using a graphing utility to zoom in on the graph of f x , you find that
x | 0.68. Using the root feature, you find that x | 0.6823. 88. f x
x3 5 x 3
f 0
3 and f 1
3
By the Intermediate Value Theorem, f c
−4
4
The graph appears to be continuous on the interval >4, 4@. Because f 2 is not defined, you know that f has a discontinuity at x 2. This discontinuity is removable so it does not show up on the graph.
>1, 2@. f 1
x3 4 is continuous on the interval 37 12
and f 2
83 . By the Intermediate
Value Theorem, there exists a number c in >1, 2@ such that f c 84. f x
f 0
0.
x 5 x 3 is continuous on the interval >0, 1@. 3
3 and f 1
3. By the Intermediate Value
Theorem, there exists a number c in >0, 1@ such that
f c 85. f x
f 0
0. x 2 2 cos x is continuous on >0, S @.
3 and f S
S 2 1 | 8.87 ! 0. By the
Intermediate Value Theorem, f c
0 for at least one
value of c between 0 and S . 86. f x
f 1 f 4
0 for at
least one value of c between 0 and 1. Using a graphing utility to zoom in on the graph of f x , you find that
0
1 x4 12
0 for at
f x is continuous on >0, 1@.
14
83. f x
79
f x is continuous on >0, 1@.
3
82. f x
Continuity and OneOne One-Sided Limits
5 §S x · tan ¨ ¸ is continuous on the interval >1, 4@. x © 10 ¹
x | 0.56. Using the root feature, you find that x | 0.5641. 89. g t
2 cos t 3t
g is continuous on >0, 1@. g 0
2 ! 0 and g 1 | 1.9 0.
By the Intermediate Value Theorem, g c
0 for at
least one value of c between 0 and 1. Using a graphing utility to zoom in on the graph of g t , you find that
t | 0.56. Using the root feature, you find that t | 0.5636. 90. h T
1 T 3 tan T
h is continuous on >0, 1@. h 0
1 ! 0 and h 1 | 2.67 0.
By the Intermediate Value Theorem, h c
0 for at
least one value of c between 0 and 1. Using a graphing utility to zoom in on the graph of h T , you find that
T | 0.45. Using the root feature, you find that T | 0.4503.
§S · 5 tan ¨ ¸ | 4.7 and © 10 ¹ 5 § 2S · tan ¨ ¸ | 1.8. By the Intermediate 4 © 5 ¹
Value Theorem, there exists a number c in >1, 4@ such that f c
0.
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
80
NOT FOR SALE
Chapter 1
91. f x
Limits its and Their Properties
x2 x 1
f is continuous on >0, 5@. f 0
1 and f 5
ª5 º f is continuous on « , 4». The nonremovable ¬2 ¼ discontinuity, x 1, lies outside the interval.
29
1 11 29 The Intermediate Value Theorem applies. x x 12
0
4 x 3
0
35 §5· and f 4 f¨ ¸ 6 © 2¹ 35 20 6 6 3
4 or x
3
The Intermediate Value Theorem applies.
x x 1 2
2
x
x 3 x
c
So, f 3 92. f x
11
4 is not in the interval.
x2 x x 1
11.
x2 x
x2 6x 8
f 0
8 and f 3
1
The Intermediate Value Theorem applies. x2 6x 8
0
2 x 4
0
x
2 or x
2 x
c
4
4 is not in the interval.
So, f 2
0.
3
2 is not in the interval.
So, f 3
6.
95. (a) The limit does not exist at x c. (b) The function is not defined at x c. c, but it is not equal to the (c) The limit exists at x value of the function at x c. (d) The limit does not exist at x c. 96. Answers will vary. Sample answer:
x3 x 2 x 2
5 4 3 2 1
2 and f 3
19
−2 −1
2 4 19 The Intermediate Value Theorem applies.
x2
2 or x
3 x
f is continuous on >0, 3@.
x
0
y
93. f x
f 0
2 x 3
c
1 0 8
6x 6 0
x
20 3
6
x2 5x 6
x
f is continuous on >0, 3@.
x
x2 x x 1
94. f x
x3 x 2 x 2
4
x3 x 2 x 6
0
2 x x 3
0
x
2
2
x 3 has no real solution. c
So, f 2
4.
2
x 1
3 4 5 6 7
−2 −3
The function is not continuous at x lim f x 1 z 0 lim f x . x o 3
3 because
x o 3
97. If f and g are continuous for all real x, then so is f g (Theorem 1.11, part 2). However, f g might not be
continuous if g x g x
0. For example, let f x
x and
x 1. Then f and g are continuous for all real 2
x, but f g is not continuous at x
r1.
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section 1.4
98. A discontinuity at c is removable if the function f can be made continuous at c by appropriately defining (or redefining) f c . Otherwise, the discontinuity is
nonremovable. x 4 x 4
(c) f x
x
3 a xb
g x
3 x
3 a xb
f x
3 x½° ¾ for x an integer °¿
3 x
3 a xb
f
12
3 0
4 x 4 4
x
4 is removable
4 is nonremovable, x
3, g
12
3 1
4.
0.40, 0 t d 10 ° ®0.40 0.05at 9b, t ! 10, t not an integer °0.40 0.05 t 10 , t ! 10, t an integer ¯
105. C t
x 4
2 a xb.
g x 1
For example,
x t 4
1, ° °0, ® °1, °0, ¯
104. The functions agree for integer values of x:
f x
sin x 4
(b) f x
81
However, for non-integer values of x, the functions differ by 1.
x 4
(a) f x
Continuity and OneOne One-Sided Limits
C
y
0.7
4
0.6
3
0.5
2
0.4 0.3
1
0.2 x −6 −4 −2
−1
2
4
0.1
6
t 2
−2
99. True
f c
1. 2.
lim f x
3.
f c
0 t d 10 °0.40, ® 0.40 0.05 10 t , t a b ! 10 °¯
C t
lim f x
xoc
All of the conditions for continuity are met.
lim f x
xoc
8 10 12 14
Note: You could also express C as
L exists.
100. True. If f x
6
There is a nonremovable discontinuity at each integer greater than or equal to 10.
L is defined.
xoc
4
g x , x z c, then
lim g x (if they exist) and at least one of
xoc
these limits then does not equal the corresponding function value at x c.
and n, respectively. It can have, at most, n discontinuities. 102. False. f 1 is not defined and lim f x does not exist. x o1
103. lim f t | 28
t
0
1
1.8
2
3
3.8
N t
50
25
5
50
25
5
Discontinuous at every positive even integer. The company replenishes its inventory every two months. N 50
Number of units
101. False. A rational function can be written as P x Q x where P and Q are polynomials of degree m
§ c t 2 fg · 25¨ 2dd g t¸ 2 e h © ¹
106. N t
40 30 20 10
t
t o 4
2
4
6
8
10 12
Time (in months)
lim f t | 56
t o 4
At the end of day 3, the amount of chlorine in the pool has decreased to about 28 oz. At the beginning of day 4, more chlorine was added, and the amount is now about 56 oz.
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
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82
Chapter 1
NOT FOR SALE
Limits its and Their Properties
107. Let s t be the position function for the run up to the
campsite. s 0 s 20
0 t
0 corresponds to 8:00 A.M.,
112. sgn x
k (distance to campsite)). Let r t be the
position function for the run back down the mountain: r 0 k , r 10 0. Let f t s t r t . When t 0 (8:00 A.M.), f 0 s 0 r 0 0 k 0. 10 (8:00 A.M.), f 10
When t
1, if x 0 ° 0 ®0, if x °1, if x ! 0 ¯
(a)
lim sgn x
1
x o 0
(b) lim sgn x
1
x o 0
s 10 r 10 ! 0.
(c) limsgn x does not exist. xo0
y
Because f 0 0 and f 10 ! 0, then there must be a
4
value t in the interval >0, 10@ such that f t
2
f t
0, then s t r t
s t
r t . Therefore, at some time t, where
3
0. If
1
0, which gives us
4 3 S r be the volume of a sphere with radius r. 3 500S V is continuous on >5, 8@. V 5 | 523.6 and 3 2048S V 8 | 2144.7. Because 3 523.6 1500 2144.7, the Intermediate Value Theorem guarantees that there is at least one value r between 5 and 8 such that V r 1500. (In fact,
3
4
−4
113. (a)
S 60 50 40 30 20 10 t 5
10
15 20
25 30
(b) There appears to be a limiting speed and a possible cause is air resistance. 114. (a) f x
f x1 ! 0 and there exists x2 in >a, b@ such that f x2 0. Then by the Intermediate Value Theorem,
2
−3
r | 7.1012.) 109. Suppose there exists x1 in >a, b@ such that
1
−2
0 d t d 10, the position functions for the run up and the run down are equal. 108. Let V
x
−4 −3 −2 −1
0, 0 d x b ® ¯b, b x d 2b
y
2b
f x must equal zero for some value of x in
> x1, x2 @ or > x2 , x1@ if x2 x1 . So, f would have a zero in >a, b@, which is a contradiction. Therefore, f x ! 0 for all x in >a, b@ or f x 0 for all x in >a, b@. 110. Let c be any real number. Then lim f x does not exist
b
x b
NOT continuous at x
xoc
because there are both rational and irrational numbers arbitrarily close to c. Therefore, f is not continuous at c. 111. If x
0, then f 0
continuous at x
0 and lim f x xo0
If x z 0, then lim f t lim f t
tox
tox
lim kt
tox
(b) g x
0. So, f is
0.
2b
b.
x 0 d x d b °° 2 , ® °b x , b x d 2b °¯ 2
y
2b
0 for x rational, whereas
kx z 0 for x irrational. So, f is not
b
continuous for all x z 0.
x b
2b
Continuous on >0, 2b@.
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section 1.4
2 °1 x , x d c ® x, x ! c °¯
115. f x
116. Let y be a real number. If y 0, then x y ! 0, then let 0 x0 S 2 such that
f is continuous for x c and for x ! c. At x 1 r
c
1 4
1 r 2
2
5
83
0. If
tan x0 ! y this is possible since the tangent
M
c, you
function increases without bound on >0, S 2 . By the
c. Solving c 2 c 1, you obtain
need 1 c 2
Continuity and OneOne One-Sided Limits
Intermediate Value Theorem, f x
.
tan x is
continuous on >0, x0 @ and 0 y M , which implies that there exists x between 0 and x0 such that tan x y. The argument is similar if y 0.
x c2 c ,c ! 0 x
117. f x
Domain: x c 2 t 0 x t c 2 and x z 0, ª¬c 2 , 0 0, f x c2 c x
lim
x o0
Define f 0
x c2 c x c c 2
1 2c to make f continuous at x
lim f x
xoc
f c exists.
'x o 0
Therefore, f is continuous at x
x ª x c cº ¬ ¼
1 2c
x c c 2
If y t 0 and y d 1, then y y 1 d 0 d x 2 , as desired. So assume y ! 1. There are now two cases.
y y 1
c.
y y 1 2 y d x 1 2 y 2
xa xb
119. h x
1
lim
xo0
2
Case l: If x d y 12 , then 2 x 1 d 2 y and
f c .
xoc
x o0
x c2 c2
0.
c 'x. As x o c, 'x o 0]
lim f x
lim
121. The statement is true.
lim f c 'x
[Let x 3.
x o0
x c2 c x
f c is defined.
118. 1. 2.
lim
x2 2x 1 2 y d x2 2 y 2 y
15
x2 −3
Case 2: If x t y
3
−3
x2 t y
h has nonremovable discontinuities at x
f 2 x f1 x . Because f1 and
f 2 are continuous on >a, b@, so is f. f a
f 2 a f1 a ! 0 and
f b
f 2 b f1 b 0
f 2 c f1 c
(b) Let f1 x
x and f 2 x
>0, S 2@, f1 0
0 f1 c
f 2 c
cos x, continuous on
f 2 0 and f1 S 2 ! f 2 S 2 .
So by part (a), there exists c in >0, S 2@ such that c
2
1 4
! y2 y y y 1
In both cases, y y 1 d x 2 .
By the Intermediate Value Theorem, there exists c in >a, b@ such that f c 0. f c
y2 y
r1, r 2, r 3, !.
120. (a) Define f x
1 2
1 2
122. P 1
P 02 1
P 0 1
1
P 2
P 12 1
P 1 1
2
P 5
P 22 1
P 2 1
2
2
2
5
Continuing this pattern, you see that P x
x for
infinitely many values of x. So, the finite degree polynomial must be constant: P x x for all x.
cos c .
INSTRUCTOR USE ONLY Using a graphing utility, c | 0.739.
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
84
NOT FOR SALE
Chapter 1
Limits its and Their Properties
Section 1.5 Infinite Limits 1. f x
1 x 4
4. f x
As x approaches 4 from the left, x 4 is a small negative number. So, lim f x
2. f x
lim f x
5.
6.
f.
As x approaches 4 from the right, x 4 is a small positive number. So, lim f x
f.
x o 4
3. f x
7.
1
x
4
As x approaches 4 from the left or right, x 4 is a 2
small positive number. So, lim f x
lim f x
x o 4
9. f x
x x2 4
f
lim 2
x x2 4
f
1 x 2 1 lim x o 2 x 2 lim
x o 2
lim tan
x o 4
f
Sx
f
4 Sx lim tan 4 x o 2 lim sec
Sx
x o 2
4 Sx lim sec 4 x o 2
f.
f
x o 2
2
8.
x o 4
lim 2
x o 2
x o 2
As x approaches 4 from the left, x 4 is a small negative number. So, lim f x
lim f x
x o 4
1 x 4
x o 4
2
small positive number. So,
f
x o 4
4
2
As x approaches 4 from the right, x 4 is a small positive number. So, lim f x
x
As x approaches 4 from the left or right, x 4 is a
f
x o 4
1
f
f f
f.
1 x 9 2
x f x
–3.5
–3.1
–3.01
–3.001
2.999
–2.99
–2.9
–2.5
0.308
1.639
16.64
166.6
166.7
16.69
1.695
0.364
lim f x
f
lim f x
f
x o 3 x o 3
2
−6
6
−2
10. f x
x x2 9
x f x
–3.5
–3.1
–3.01
–3.001
2.999
–2.99
–2.9
–2.5
1.077
5.082
50.08
500.1
499.9
49.92
4.915
0.9091
lim f x
f
lim f x
f
x o 3 x o 3
2
−6
6
INSTRUCTOR USE ONLY −2 −2
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section 1.5
IIn Infinite Limits
85
x2 x 9
11. f x
2
x f x
–3.5
–3.1
–3.01
–3.001
2.999
–2.99
–2.9
–2.5
3.769
15.75
150.8
1501
1499
149.3
14.25
2.273
lim f x
f
lim f x
f
x o 3 x o 3
4
−6
6
−4
12. f x
sec
x f x
Sx 6
–3.5
–3.1
–3.01
–3.001
2.999
–2.99
–2.9
–2.5
3.864
19.11
191.0
1910
1910
191.0
19.11
3.864
lim f x
f
lim f x
f
x o 3 x o 3
4
−6
6
−4
13. lim
x o 0
1 x2
f
Therefore, x 14. lim
x o 2
lim
x o 2
3
f
3
f
2 4
x
2
Therefore, x 2
15.
lim
x o 2
x x2 4
Therefore, x
1 x2
18.
0 is a vertical asymptote.
4
x
lim
x o 0
lim h s
s o 5
Therefore, s lim h s
s o 5
Therefore, s 19. lim
x o 2
2 is a vertical asymptote. f and lim
x o 2
2
x x2 4
f
2 is a vertical asymptote.
x2 lim 2 x o 2 x 4
x2 f and lim 2 x o 2 x 4
Therefore, x
2 is a vertical asymptote.
f
16. No vertical asymptote because the denominator is never zero.
f and lim h s s o 5
5 is a vertical asymptote. f and lim h s s o 5
x2 2 x o 2 x 2 x 1 lim
lim
x o 1
f f
2 is a vertical asymptote.
x2 2 x 2 x 1
x2 2 x o 1 x 2 x 1 lim
Therefore, x
f.
5 is a vertical asymptote.
x2 2 x 2 x 1
Therefore, x
f.
f f
1 is a vertical asymptote.
17. No vertical asymptote because the denominator is never zero.
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
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86
NOT FOR SALE
Chapter 1 Limitss and Their Properties Prope
20. lim
x o 0
2 x x 1 x
lim
2
Therefore, x
x o 0
f
2 x 2 x o1 x 1 x
f
lim
Therefore, x
f
Therefore, t
28. h t
2 1 x x 2 x 8 6 x2 2x 8
x2 4x 2 3 x 6 x 24
1 x, 6
2
3 2 x 1
x
2 and x
Vertical asymptotes at x
1.
x x3 2 x 2 9 x 18
25. f x
3. The graph has
1 x x 1 2
x 1
lim x 2 x 1
lim f x
x o 1
x2 4 x 2x2 x 2 3
1.
t
tan T
2n
1 S
S
2
2
lim
T o0
tan T
lim x 1
x o 1
0 because
2
2
−3
x 2 x 2 x 2 x 2 1
nS , n any interger.
1.
T
x2 1 x o 1 x 1
1.
sin T has vertical asymptotes at T cos T
T
There is no vertical asymptote at T
3.
x o 1
The graph has a hole at x 26. h x
t
33. lim
has no vertical asymptote because
nS , n a
nonzero integer. There is no vertical asymptote at t 0 since
T
Vertical asymptotes at x 0 and x holes at x 3 and x 2.
x
t has vertical asymptotes at t sin t
32. g T
4 , x z 3, 2 x x 3
1 has vertical asymptotes at cos S x
2n 1 , n any integer. 2
x
lim
4 x 3 x 2
2. The graph has a hole at
sin S x has vertical asymptotes at cos S x
sec S x
t o 0 sin
x x 2 x 2 9
x 1 x 1
t ,t z 2 2 t 2 4
tan S x
31. s t
4 x 2 x 6
3
t
30. f x
No vertical asymptote. The graph has holes at x and x 4.
24. f x
2 t 2 t 2 4
2n 1 , n any integer. 2
x
x z 2, 4
x2 x 2
t
29. f x
0 is a vertical asymptote.
5.
t t 2
Vertical asymptote at t t 2.
4· § lim ¨1 2 ¸ t ¹ t o 0 ©
1 2 x3
x 3 ,x z 5 x2 1
No vertical asymptote. The graph has a hole at x
1 is a vertical asymptote.
4· § 21. lim ¨1 2 ¸ t ¹ t o 0 ©
23. f x
x 5 x 3 x 5 x 2 1
27. f x
f
0 is a vertical asymptote.
2 x lim 2 x o1 x 1 x
22. g x
2 x x 1 x 2
3
−5
Removable discontinuity at x
1
has no vertical asymptote because x 2 1
4 . 5
The graph has a hole at x
2.
lim h x
x o 2
lim
x o 2 x 2
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section 1.5
34. lim
x o 1
x2 6 x 7 x 1
lim x 7
8
x o 1
43.
lim
x o 3
x 3 x x 6
IIn Infinite Limits
lim
x 3 x 3 x 2
lim
1 x 2
x o 3
2
2 −3
3
x o 3
44. −12
Removable discontinuity at x 35.
x2 1 x o 1 x 1
f
lim
lim
x o 1
6x2 x 1 2 x o 1 2 4 x 4 x 3 lim
x o1
x2 1 x 1
f
Vertical asymptote at x
1
3x 1 2 x 1 2 x 3 2 x 1
lim
3x 1 2x 3
x o 1 2
45. lim
x 1 1 x 1
x2
46. lim
x o3
x 2 x2
1 5
lim
x o 1 2
1
87
lim
x o1 x 2
1 1
5 8
1 2
1 9
8
−3
3
sin x 1
x o 0 sin
Removable discontinuity at x
50.
1
51. lim
x oS
f
x
f
x csc x
lim
x oS
x sin x
0
3
52. lim
lim
x o 1
38. lim
x o1
39. lim
x o 2
1 x 1 1
x
f
lim x sec S x
f and
x o 1 2
0
lim x sec S x
x o 1 2
2
x2 42. lim 2 x o 4 x 16
lim x 2 tan S x
f and
lim x 2 tan S x
f. Therefore,
x o 1 2 x o 1 2
f
2
1
lim ª x 2 tan xº¼
x o 0¬
f.
x o 1 2
54.
x x 2
x
cot x
f
2
2 x x o1 1 x
x o1
2
Therefore, lim x sec S x does not exist.
40. lim
41. lim
53.
f
1
x
x
xo0
−2
37.
f
2 x o S 2 cos x lim
2
−3
2
49. lim
1
x 1
x o 1
f
1· § 48. lim ¨ x 2 ¸ x¹ x o 0 ©
−8
36. lim
1· § 47. lim ¨1 ¸ x¹ x o 0 ©
f
lim x 2 tan S x does not exist.
x o 1 2
55. f x
x2 x 1 x3 1
lim f x
1 2
lim
x o1
x o1
x2 x 1 x 1 x 2 x 1
1 x 1
f
3
−4
5
INSTRUCTOR USE ONLY 3 −3
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
88
NOT FOR SALE
Chapter 1 Limitss and Their Properties Prope
x 1 x 2 x 1 x3 1 x2 x 1 x2 x 1 lim f x lim x 1 0
56. f x x o1
x o1
64. No, it is not true. Consider p x
−8
m0
65. m
8
−4
1 v 2 c 2
lim m
1 x 2 25 lim f x f
57. f x x o 5
k
lim
k f
V o 0 V −8
− 0.3
x o 4
Sx 8 f
50S sec 2
(b) r
50S sec2
f
lim
x o 25
2x
69. (a) Average speed
50 50 50 y 50 x
61. One answer is
62. No. For example, f x
asymptote.
x 3 . x 2 4 x 12
f Total distance Total time 2d d x d y 2 xy y x 2 xy
50 x
2 xy 50 y
50 x
2 y x 25
25 x x 25
1 has no vertical x2 1
3 ft sec 2
625 x 2
60. The line x c is a vertical asymptote if the graph of f approaches r f as x approaches c.
x 3 x 6 x 2
7 ft sec 12
625 225
says how the limit fails to exist.
f x
f
2 15
−6
lim f x
200S ft sec
3
2 7
(b) r
bound as x approaches c is called an infinite limit. f is not a number. Rather, the symbol
S
625 49
9
(c)
200S ft sec 3
6
lim ª¬50S sec 2 T º¼
68. (a) r
59. A limit in which f x increases or decreases without
S
T o S 2
6
xoc
f
67. (a) r
(c)
−9
f
(In this case you know that k ! 0. )
8
lim f x
1 v 2 c 2
v o c
k V
66. P
0.3
m0
lim
v o c
sec
x 1
has a hole at 1, 2 , not a vertical asymptote.
4
58. f x
p x
x2 1 x 1
f x
x 2 1. The function
y
Domain: x ! 25 (b)
x
30
40
50
60
y
150
66.667
50
42.857
y
63. 3 2
(c)
1 x
lim
x o 25
25 x x 25
f
INSTRUCTOR ST USE ONLY −2
−1
1
−1
3
As x gets close to 25 mi/h, m y becomes larger and larger.
−2
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE Section 1.5
70. (a)
x
1
0.5
0.2
0.1
0.01
0.001
0.0001
f (x)
0.1585
0.0411
0.0067
0.0017
0
0
0
In Infinite Limits
89
0.5
− 1.5
1.5
− 0.25
(b)
x o 0
lim
x sin x x
x
1
0.5
0.2
0.1
0.01
0.001
0.0001
f (x)
0.1585
0.0823
0.0333
0.0167
0.0017
0
0
0
0.25
− 1.5
1.5
− 0.25
(c)
x o 0
lim
x sin x x2
x
1
0.5
0.2
0.1
0.01
0.001
0.0001
f (x)
0.1585
0.1646
0.1663
0.1666
0.1667
0.1667
0.1667
0
0.25
− 1.5
1.5
− 0.25
(d)
x o 0
lim
x sin x x3
x
1
0.5
0.2
0.1
0.01
0.001
0.0001
f (x)
0.1585
0.3292
0.8317
1.6658
16.67
166.7
1667.0
0.1667 1 6
1.5
− 1.5
1.5
−1.5
lim
x o 0
x sin x x4
For n ! 3, lim
x o 0
f x sin x xn
f.
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
90
NOT FOR SALE
Chapter 1 Limitss and Their Properties Prope 1 1 bh r 2T 2 2
71. (a) A
1 1 2 10 10 tan T 10 T 2 2
50 tan T 50 T
§ S· Domain: ¨ 0, ¸ © 2¹ (b)
T
0.3
0.6
0.9
1.2
1.5
f T
0.47
4.21
18.0
68.6
630.1
100
0
1.5 0
(c)
f
lim A
T o S 2
72. (a) Because the circumference of the motor is half that of the saw arbor, the saw makes 1700 2
850 revolutions per minute.
(b) The direction of rotation is reversed. § §S ·· (c) 2 20 cot I 2 10 cot I : straight sections. The angle subtended in each circle is 2S ¨ 2¨ I ¸ ¸ 2 ¹¹ © ©
So, the length of the belt around the pulleys is 20 S 2I 10 S 2I
S 2I .
30 S 2I .
60 cot I 30 S 2I
Total length
§ S· Domain: ¨ 0, ¸ © 2¹ (d)
(e)
I
0.3
0.6
0.9
1.2
1.5
L
306.2
217.9
195.9
189.6
188.5
450
2
0 0
(f)
60S | 188.5
lim L
I o S 2
(All the belts are around pulleys.) (g) lim L I o 0
f
73. False. For instance, let f x
x 1 or x 1
g x
x . x2 1
76. False. Let
2
f x
The graph of f has a vertical asymptote at x f 0
74. True 75. False. The graphs of y tan x, y cot x, y
1 ° , x z 0 ®x °3, x 0. ¯
sec x and y
0, but
3.
csc x have
INSTRUCTOR USE ONLY vertical asymptotes.
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
Section 1.5 1 and g x x2
77. Let f x 1 lim x o 0 x2
1 , and c x4
1 f and lim 4 xo0 x
78. Given lim f x
f z 0.
L:
xoc
L 2 ! 0 there exists
G1 ! 0 such that g x L L 2 whenever 0 x c G1. So, L 2 g x 3L 2. Because lim f x
xoc
f then for M ! 0, there exists
G 2 ! 0 such that f x ! M 2 L whenever x c G 2 . Let G be the smaller of G1 and
G 2 . Then for 0 x c G , you have f x g x ! M 2 L L 2 lim f x g x
M . Therefore
f. The proof is similar for L 0.
xoc
g x
f x
1 f x
Then, lim
1 f x
x oc
(3) Quotient: Let H ! 0 be given. There exists G1 ! 0 such that f x ! 3L 2H whenever 0 x c G1 and
f, let g x
91
1. Then
0 by Theorem 1.15.
80. Given lim
xoc
(2) Product: If L ! 0, then for H
xoc
lim
f and lim g x
xoc
79. Given lim f x xoc
f, but
§ x2 1· lim ¨ ¸ 4 xo0 © x ¹
1· §1 lim ¨ 4¸ x o 0© x 2 x ¹
0.
In Infinite Limits
0. Suppose lim f x exists and equals L. xoc
lim 1
xoc
lim f x
xoc
1 L
0.
This is not possible. So, lim f x does not exist. xoc
1 is defined for all x ! 3. Let M ! 0 be x 3 given. You need G ! 0 such that 1 f x ! M whenever 3 x 3 G . x 3
81. f x
Equivalently, x 3
1 whenever M
x 3 G , x ! 3. 1 . Then for x ! 3 and M 1 1 x 3 G, ! M and so f x ! M . 8 x 3
So take G
there exists G 2 ! 0 such that g x L L 2 whenever 0 x c G 2 . This
inequality gives us L 2 g x 3L 2. Let G be the smaller of G1 and G 2 . Then for 0 x c G , you have g x
f x
3L 2 3L 2H
Therefore, lim
xoc
H.
g x
f x
0.
1 1 is defined for all x 5. Let N 0 be given. You need G ! 0 such that f x N whenever x 5 x 5 1 1 1 5 G x 5. Equivalently, x 5 ! whenever x 5 G , x 5. Equivalently, whenever N x 5 N
82. f x
1 . Note that G ! 0 because N 0. For x 5 G and N 1 1 N , and N. x 5 x 5
x 5 G , x 5. So take G x 5,
1 1 ! x 5 G
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Chapter 1 Limitss and Their Properties Prope
92
Review Exercises for Chapter 1 1. Calculus required. Using a graphing utility, you can estimate the length to be 8.3. Or, the length is slightly longer than the distance between the two points, approximately 8.25. 11
−9
9 −1
9
2. Precalculus. L
3. f x
1 3 1 2
2
| 8.25
4 2 x 2 x
x
–0.1
–0.01
–0.001
0.001
0.01
0.1
f (x)
–1.0526
–1.0050
–1.0005
–0.9995
–0.9950
–0.9524
lim f x | 1.0
xo0
1 −3
3
−5
4.
x
–0.1
–0.01
–0.001
0.001
0.01
0.1
f (x)
1.432
1.416
1.414
1.414
1.413
1.397
lim f x | 1.414
xo0
3
−3
3
−3
5. lim x 4 x o1
1 4
5
Let H ! 0 be given. Choose G
H . Then for 0 x 1 G
H , you have
x 1 H
x
4 5 H f x L H .
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
Review Exercises for f Chapter 1
x
6. lim
x o9
9
93
3
Let H ! 0 be given. You need x 3 H
x 3 H
x 3
x 3 x 9 H
Assuming 4 x 16, you can choose G So, for 0 x 9 G x 9 5H
x 3.
5H .
5H , you have
x 3H
x 3 H f x L H .
7. lim 1 x 2 xo2
1 22
3
Let H ! 0 be given. You need 1 x 2 3 H x 2 4
x 2 x 2 H x 2
Assuming 1 x 3, you can choose G
H
So, for 0 x 2 G x 2
H 5
5
H 5
1 H x 2
.
, you have
H
x 2
x 2 x 2 H x2 4 H 4 x2 H
1 x 2 3
H
f x L H . 9. Let H ! 0 be given. G can be any positive
8. lim 9 x o5
number. So, for 0 x 5 G , you have
11. lim x 2
12. lim 10 x
99 H
6
2
xo6
10
4
xo7
f x L H . 4x x2 x
(a) lim h x xo0
(b) lim h x x o 1
10. g x
x 4 x x
40
4 x, x z 0
4 2
14. lim 3 y 1
34 1
yo4
4
4 1
5
15. lim
t 2 4
t o 2 t 2
t2 9 t o3 t 3
2 x x 3
16. lim
lim
t o 2 t
2
7
13. lim t 2 t o4
9. h x
2
1 2
lim t 3 t o3
16 4
34
6
81
2.45
9
1 4
6
(a) lim g x does not exist x o3
(b) lim g x xo0
2 0 0 3
0
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
94
NOT FOR SALE
Chapter 1 Limitss and Their Properties Prope x 3 1 x 4
17. lim
xo4
x 3 1 x 4
lim
xo4
x
lim
xo4
1 x 3 1
lim
4 x 2 x
lim
1 4 x 2
4 x 2 x
xo0
xo0
xo0
20. lim
1
ª1 lim « s o 0« ¬
1 s 1
s o0
s
x3 125 x o 5 x 5
21. lim
lim
1 s 1
1 1
s
x
lim
x o 2 x 2
23. lim
xo0
24.
lim
25. lim
'x o 0
x
1
x
x 2 2x 4
4 12
'x
1 3
1 0
0
S
lim
sin S 6 cos 'x cos S 6 sin 'x 1 2
'x o 0
lim
'x o 0
1 s 1º ¼
1 2
2 x 2 2 x 4
sin ª¬ S 6 'xº¼ 1 2 'x
cos S 'x 1
2 x 2
1
'x o 0 2
26. lim
1 s ª¬ 1
1
75
xo0
4 S 4
4x
x o S 4 tan
s o0
x 5
§ x ·§ 1 cos x · lim ¨ ¸¨ ¸ x ¹ © sin x ¹©
1 cos x sin x
1 s 1º » 1 s 1» ¼ lim
lim x 5 x 25
x o 2
1
5 x 2 5 x 25
x
x o 5
lim
1 1
4 x 2 4 x 2 1 4
x o 5
x2 4 x o 2 x 3 8
1 x 1 x x 1
1 2
2
22. lim
xo0
xo0 x
ª1 1 s º¼ 1 lim ¬ s ª 1 1 s 1º ¬ ¼
s o0
lim
lim
x 3 1
lim
xo4
18. lim
3 1
x 4
ª1 x 1 º¼ 1 19. lim ¬ x o0 x
x 3 1 x 3 1
lim
'x o 0
cos 'x 'x
1
'x lim
'x o 0
3 sin 'x 2 'x
0
3 1 2
3 2
cos S cos 'x sin S sin 'x 1 'x
ª cos 'x 1 º sin 'x º ª lim « sin S » 'lim « x o 0 'x 'x »¼ ¬ ¬ ¼
'x o 0
0 0 1 27. lim ª¬ f x g x º¼ xoc
f x x o c g x
28. lim
lim f x x oc
lim g x
34 23 3 4 23
29. lim ª¬ f x 2 g x º¼ xoc
12
0
9 8
30. lim ª¬ f x º¼ x oc
2
23
34 2
ªlim f x º «¬x o c »¼
2
§ 3· ¨ ¸ © 4¹
7 12 2
9 16
INSTRUCTOR USE ONLY x oc
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Review Exercises for f Chapter 1
2x 1 x 1
31. f x
(a)
3
x
1.1
1.01
1.001
1.0001
f(x)
0.5680
0.5764
0.5773
0.5773
2x 1 x 1
lim
x o1
3
Actual limit is
| 0.577
2
(b)
95
The graph has a hole at x
3 3.
1.
lim f x | 0.5774.
x o1
−1
2 0
2x 1 x 1
(c) lim
x o1
3
2x 1 x 1
lim
x o1
lim
x o1
3
2x 1 2x 1
2 x 1 3 x 1 2 x 1 2 2x 1
lim
x o1
2
3
1 3
2 3
3
3 3
3 3
1 3 x x 1
32. f x
(a)
x
1.1
1.01
1.001
1.0001
f(x)
–0.3228
–0.3322
–0.3332
–0.3333
1 3 x | 0.333 x o1 x 1
1· § ¨ Actual limit is .¸ 3¹ ©
lim
2
(b)
The graph has a hole at x
−3
1.
lim f x | 0.333.
x o1
3
−3
1 3 x (c) lim x o1 x 1
1 3 x 1 lim x o1 x 1 1 lim
x o1
lim
x o1
3 3
x x x x
3
1 x ª x 1 «1 3 x ¬ 1 1
3
x
x 3
3
2
2 2
x º»¼ 2
3
1 3
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
96
NOT FOR SALE
Chapter 1 Limitss and Their Properties Prope
33. v
lim
s 4 s t 4t
t o4
2 ª ¬ 4.9 16 250º¼ ª¬4.9t 250º¼ lim t o4 4t
lim
t o4
lim
4.9 t 16
4t 4.9 t 4 t 4 4t
t o4
lim ª4.9 t 4 º¼
39.2 m/sec
t o 4¬
The object is falling at about 39.2 m/sec. 34. 4.9t 2 250
0 t
50 sec 7
50 , the velocity is 7
When a
s a s t t oa a t
ª4.9a 2 250º¼ ª¬4.9t 2 250º¼ lim ¬ t oa a t
lim
lim
4.9 t 2 a 2
a t 4.9 t a t a lim t oa a t t oa
lim ª 4.9 t a º¼ t oa¬ § ¨a ©
4.9 2a
50 · ¸ 7¹
70 m/sec.
The velocity of the object when it hits the ground is about 70 m/sec. 35. lim
x o 3
x 3 x 3
lim
x o 3
x 3 x3
42. f x
1
36. lim a x 1b does not exist. There is a break in the graph xo4
at x
0
38. lim g x
11
x o1
lim a x 3b
x o k
2
3 where k is an integer.
k 2 where k is an integer.
Nonremovable discontinuity at each integer k
39. lim h t does not exist because lim h t t o1
t o1
and lim h t
1 2
t o1
40. lim f s 41. f x
Continuous on f, 0 0, f
ax 3b lim a x 3b k x o k
37. lim f x
s o 2
2 x
43. f x
4.
xo2
x2
1 1
2
3 x 2 7
Continuous on f, f
1.
11
2
Continuous on k , k 1 for all integers k 44. f x
3x 2 x 2 x 1
3 x
2 x 1 x 1
Continuous on f, 1 1, f
3x 2 x 1 3x 2 x 2 x 1 x 1 lim f x lim 3x 2 5
45. f x x o1
x o1
Removable discontinuity at x
1
INSTRUCTOR USE ONLY f, 1 1, f Continuous on f
© 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Review Exercises for f Chapter 1
x 1 2x 2 x 1 1 lim x o 1 2 x 1 2
5 x, x d 2 ® ¯2 x 3, x ! 2
46. f x
lim 5 x
50. f x
3
x o 2
lim 2 x 3
1
x o 2
Continuous on f, 1 1, f
2
Continuous on f, 2 2, f
lim
xo2
2
1
x
51. f x
1
x 2
2
2k , 2k
x 1 x
x o 0
1
1 x
1
52. f x
f
4
§ 2n 1 S 2n 1 S · , ¨ ¸ 4 4 © ¹
0
for all integers n. 53. f 2
3 x 1 lim f x f
49. f x
x o 2
c 2 6
f 1
Nonremovable discontinuity at x Continuous on f, 1 1, f 2
lim x 1
4
x o 3
Find b and c so that lim x 2 bx c x o1
Consequently you get
1 b c
Solving simultaneously, 55. f is continuous on >1, 2@. f 1
f 2
b
2 and lim x 2 bx c 2
and 9 3b c
3 and
1 0 and
c
5.
5
2c
1
c
1 2
4.
x o 3
4. 4. 56. C x
13 ! 0. Therefore by the Intermediate Value
Theorem, there is at least one value c in 1, 2 such that 2c 3 3
5
Find c so that lim cx 6
x o1
x o1
1 S
Continuous on
Continuous on f, 1@ 0, f
54. lim x 1
tan 2 x
2n
x
Nonremovable discontinuity at x
x o1
2
Nonremovable discontinuities when
1 x
Domain: f, 1@, 0, f
lim f x
2
for all integers k.
2
Continuous on f, 2 2, f
lim
Sx
Continuous on
Nonremovable discontinuity at x
48. f x
csc
Nonremovable discontinuities at each even integer.
f
2
1
Removable discontinuity at x
Nonremovable discontinuity at x
47. f x
97
12.80 2.50 ¬ªa xb 1¼º,
x ! 0
12.80 2.50 ª¬a xb 1º¼ ,
x ! 0
25
0.
0
5 10
C has a nonremovable discontinuity at each integer 1,, 2, 3,!.
INSTRUCTOR USE ONLY © 2010 Brooks/Cole, Cengage Learning
© Cengage Learning. All Rights Reserved.
98
Chapter 1 Limitss and Their Properties Prope x2 4 x 2
57. f x
(a)
x
lim f x
ª x 2º 2 « » ¬« x 2 »¼
66.
4
x o 2
4
x o 2
(c) lim f x does not exist.
68.
xo2
x
1 x
(b) lim f x
0
(c) lim f x
0
x o 0
0
Vertical asymptotes at x
sec x x
73. lim
csc 2 x x
x o 0
2
2 and x
10
2
10
75. C
csc S x
lim
2x x 1 x 2
lim
x o 1
x 1 x3 1
f
f
80,000 p , 0 d p 100 100 p
f
$80.000
(c) C 90
$720,000
f
lim
x o 1
1 x2 x 1
lim
80,000 p p
p o100 100
f
1 3
tan 2x x
76. f x
(a)
1 x sin 2 x
(b) C 50
(d)
x 64. lim x o 1 2 2 x 1 65.
lim
x o 0
(a) C 15 | $14,117.65
2
x o 2
4 5
f
cos 2 x x x o 0
Vertical asymptote at every integer k 63.
ª 4 § sin 4 x ·º lim « ¨ ¸ 4 x ¹»¼
x o 0 ¬ 5 ©
74. lim
Vertical asymptote at x 62. f x
72. lim
1 4
f
x2 4
sin 4 x 5x
f
1
x o 0
8
x
f
71. lim 2 x
1 x 1 x 1 2
x2 2x 1 x 1 x o 1 lim
x o 0
4x 4 x2
61. f x
x o 1
f
70. lim
Vertical asymptote at x 60. h x
lim
x2 2 x 1 x 1 x o1
x o 2 3
x o1
1
x 1 x4 1
1· § 69. lim ¨ x 3 ¸ x ¹ x o 0 ©
(a) Domain: f, 0@ >1, f
59. g x
x o 1
67. lim
(b) lim f x
58. f x
lim
x
–0.1
–0.01
–0.001
0.001
0.01
0.1
f(x)
2.0271
2.0003
2.0000
2.0000
2.0003
2.0271
lim
xo0
tan 2 x x
2
(b) Yes, define f x
tan 2 x , x z 0 ° . ® x °2, 0 x ¯
Now f x is continuous at x
INSTRUCTOR USE ONLY 0.
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Problem Solving ffo for Chapter 1
99
Problem Solving for Chapter 1 x 2 y 1 2
1. (a) Perimeter 'PAO
x2 y2 1
x 2 x 2 1 2
Perimeter 'PBO
x
1 y 2
x2 y2 1
x
1 x 4
x2 x4 1
2
2
x 2 x 2 1
x2 x4 1
x
x2 x4 1
2
(b) r x
(c)
1 x 4 2
x
4
2
1
0.1
0.01
Perimeter 'PAO
33.02
9.08
3.41
2.10
2.01
Perimeter 'PBO
33.77
9.60
3.41
2.00
2.00
r x
0.98
0.95
1
1.05
1.005
lim r x
x o 0
1 01 1 01
2 2
1 bh 2 1 bh 2
1 1 x 2 1 1 y 2
2. (a) Area 'PAO
Area 'PBO
Area 'PBO Area 'PAO
(b) a x
(c)
x2 x4 1
1
x2 2 x 2
x 2 y 2
x2 2
x
x
4
2
1
0.1
0.01
Area 'PAO
2
1
12
1 20
1 200
Area 'PBO
8
2
12
1 200
1 20,000
a x
4
2
1
1 10
1 100
lim a x
x o 0
lim x
x o 0
0
3. (a) There are 6 triangles, each with a central angle of 60q
Area hexagon
ª1 º 6« bh» ¬2 ¼
Sº ª1 6« 1 sin » 3¼ ¬2
S 3. So,
3 3 | 2.598. 2
h = sin 60° 1 60° h = sin θ 1
θ
Area ((Circle)) Area ((Hexagon) g )
S
3 3 | 0.5435 2
INSTRUCTOR USE ONLY Error
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100
NOT FOR SALE
Chapter 1 Limitss and Their Properties Prope
(b) There are n triangles, each with central angle of T
(c)
n sin 2S n
2S º ª1 n « 1 sin » n¼ ¬2
ª1 º n « bh» ¬2 ¼
An
2
2S n. So,
.
n
6
12
24
48
96
An
2.598
3
3.106
3.133
3.139
(d) As n gets larger and larger, 2S n approaches 0. Letting x which approaches 1 S 4. (a) Slope
(b) Slope
40 30
4 3
mx (d) lim mx x o3
x,
3 x 3 4 3 25 x 4 4
25 x 2
y 12
y
25 x 2 4
mx
25 x 2 4
25 x 16
x o3
x
3
25 x 2 4
3 x 3 x x o3 x 3 25 x 2 lim
lim
x o3
3 x
25 x 2 4
4
x o5
lim
12
xo5
xo5
169 x 2
3 4
lim
xo5
This is the slope of the tangent line at P.
lim
169 x 2 12 x 5 12
144 169 x
lim
6 4 4
x,
169 x 2 12 x 5
(d) lim mx
5 . 12
5 x 5 12 5 169 x Tangent line 12 12
2
lim
sin x S x
S
12 5
x, y
(c) Q
25 x 2 4 x 3
x o3
(b) Slope of tangent line is
25 x 2 4 x 3 lim
2S n
2n
5. (a) Slope
3 Tangent line: y 4 4
x, y
sin 2S n
S.
y
(c) Let Q
sin 2S n
2S n, An
x x
5 12
x 12
169 x 2
169 x 2
x 2 25
5 12
xo5
2
169 x 2
169 x 2
5
169 x
2
10 12 12
5 12
a bx a bx
3 3
This is the same slope as part (b). a bx x
6.
a bx x
3
x Letting a
3
a bx 3 a bx
3
3 simplifies the numerator.
So, lim
xo0
3 bx x
3
lim
xo0 x
lim
xo0
b 3 3 3 and b 6.
Setting
bx 3 bx b 3 bx
3, you obtain b
3
3 6. So,
INSTRUCTOR USE ONLY a
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NOT FOR SALE
Problem Solving ffo for Chapter 1
101
7. (a) 3 x1 3 t 0 x1 3 t 3 x t 27
Domain: x t 27, x z 1 or >27, 1 1, f (b)
0.5
− 30
12 −0.1
3 27
13
(c)
lim f x
(d) lim f x
3 x1 3 2 x 1
lim
x o1
2
x o1
3 x
13
lim
x o1
x
1
3 x1 3 2
4
x1 3 1
lim
x
x
13
1
1
lim f x
x o 0
lim a 2 2
x o 0
ax tan x
1 12 y
10. 3
tan x § a¨ because lim x 0 o x ©
· 1¸ ¹
2 1 x
−1
Thus, a2 2
0
a 2 a 1
0
a
9. (a) lim f x xo2
lim f x
−2
(a) f
14 a4b
4 0
1, 2
f 3
c1f ed 3 hg
3: g1 , g 4
f 1
a1b
(b) f continuous at 2: g1 x o 2
1
−1
a
a2 a 2
(c)
2
3 x1 3 2
a2 2
x o 0
lim
3 x1 3 2
1 23
1 1 1 2 x o 0
3 x1 3 2
x1 3 1 x 2 3 x1 3 1
x o1
1 | 0.0714 14
3 x1 3 2
lim
x o1
8. lim f x
2 28
27 1
x o27
3: g1 , g3 , g 4
(b)
1
lim f x
1
lim f x
0
lim f x
f
lim f x
f
x o1 x o1
x o 0 x o 0
(c) f is continuous for all real numbers except x
0, r1, r 12 , r 13 , !
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Calculus AP Edition 9th Edition Larson Solutions Manual Full Download: http://alibabadownload.com/product/calculus-ap-edition-9th-edition-larson-solutions-manual/ 102
NOT FOR SALE
Chapter 1 Limitss and Their Properties Prope
y
13. (a)
y
11. 4
2
3 2 1
1
x
−4 −3 −2 −1
1
2
3
4
−2
x
−3
a
b
−4
a1b a1b
f 1
(a)
f 0 f
0 1 3 2
lim f x
1
lim f x
1
lim f x
1
x o1 x o1
x o1 2
(b) (i) lim Pa , b x
1
v2
(b)
48
r
Let v0 r
(c)
lim r
vo0
Let v0
0 x 0 G x
G1 a . Then for
G1 a , you have
G1 a
ax G1 f ax L H .
As a counterexample, let 1, x z 0 a 0 and f x . ® 0 ¯2, x
1920 v0 2 2.17 r
1920 2 v v0 2 2.17
vo0
f x L H . Let G
4 3 mi sec.
v 2 v0 2 2.17
a, b.
(d) The area under the graph of U, and above the x-axis, is 1.
192,000 48 v0 2
1920 r
lim r
1
14. Let a z 0 and let H ! 0 be given. There exists G1 ! 0 such that if 0 x 0 G1 then
192,000 v0 2 48 r
192,000 v 2 v0 2 48
v2
(iv) lim Pa , b x
except x
r
Let v0
0
(c) Pa , b is continuous for all positive real numbers
v 2 v0 2 48
vo0
(iii) lim Pa , b x x o b
192,000 r
lim r
0
x o b
1
0, r1, r 2, r 3, !
12. (a)
(ii) lim Pa , b x x o a
(c) f is continuous for all real numbers except x
1
x o a
0
0
1 2
f 2.7 (b)
1 1
Then lim f x xo0
lim f ax
xo0
L, but
1
lim f 0
xo0
lim 2
xo0
2.
1920 2.17 v0 2
| 1.47 mi/sec .
2.17 mi sec
10,600 v 2 v0 2 6.99 10,600 6.99 v0 2 6.99 | 2.64 mi sec.
Because this is smaller than the escape velocity for Earth, the mass is less.
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