Memorandum: MAT1613 May2016.
1. (a)The y intercept is where x = 0 : 1 − 2x2 x2 − 1 then 1 = −1 f (0) = −1
f (x) =
(b) For the horizontal asymptote we divide by the hghest power of x : f (x) =
1 − 2x2 1/x2 − 2 = x2 − 1 1 − 1/x2
hence −2 1/x2 − 2 = = −2 x→±∞ 1 − 1/x2 1
lim f (x) = lim
x→±∞
and so the horizontal asymptote is y = −2. Fot the vertical asymptote: The function is not defined for x = ±1.We check whether or not there are vertical asymptotes at x = ±1 by calculating the following limits: lim+
x→1
1 − 2x2 = −∞; x2 − 1
x→1
1 − 2x2 =∞ x2 − 1
lim +
1 − 2x2 = ∞; x2 − 1
lim −
1 − 2x2 = −∞ x2 − 1
x→−1
lim−
x→−1
Hence the verical asymptotes are x = 1 and x = −1. (c) 0
f (x) =
(x2 − 1)(−4x) − (1 − 2x2 )2x −4x3 + 4x − 2x + 4x3 2x = = (x2 − 1)2 (x2 − 1)2 (x2 − 1)2
and 2(x2 − 1)2 − 2x2(x2 − 1)2x (x2 − 1)4 2 2(x − 1)(x2 − 1 − 4x2 ) = (x2 − 1)4 2(−3x2 − 1) −2(3x2 + 1) = = (x2 − 1)3 (x2 − 1)3 −2(3x2 + 1) = (x2 − 1)2 (x − 1)(x + 1)
f 00 (x) =
1
Since (x2 − 1)2 ≥ 0 and (3x2 + 1) > 0 and f 00 (x) is undefined at x = ±1 we consider y = −2 , the undefined values and y = (x − 1) and y = (x + 1) in the sign pattern. Sign pattern for f 00 (x) : y =x−1 y =x+1 y = −2 f 00 (x)
− − − −
− + − −1 +
1
+ + − −
i) f is concave up on (−1, 1) and concave down on (∞, −1) ∪ (1, ∞) −2(3x2 +1) 2 ii) Since f 00 (x) = 0 where (x2 −1) 2 (x−1)(x+1) = 0 .which is never possible since −2(3x + 1) < 0 for all while x, there is no reflection point. 2. Find the exact value of 1 sin[tan−1 (−1) + 3 cos−1 (− )] 2 Let tan−1 (−1) = θ then tan θ = −1 and so θ = − π4 Let cos−1 (− 21 ) = φ then cos φ = − 12 and so φ = 2π 3 Then sin[− 3.
π 2π π −1 + 3( )] = sin[− ] = √ 4 3 4 2
√ 1 f (x) = 2 x = 2x 2 1 f 0 (x) = x− 2 3 f 00 (x) = − 12 x− 2 5 f 000 (x) = 34 x− 2 7 f iv (x) = − 15 x− 2 8
f (1) = 2 f 0 (1) = 1 f 00 (1) = − 12 000 f (1) = 43 f iv (1) = − 15 8
(x − 1) f 0 (1) (x − 1)2 f 00 (1) (x − 1)3 f 000 (1) (x − 1)4 f 00v (1) + + + 1! 2! 3! 4! 3 15 1 = 2 + (x − 1) − (x − 1)2 + (x − 1)3 − (x − 1)4 2.2! 4.3! 8.4! 1 1 5 = 2 + (x − 1) − (x − 1)2 + (x − 1)3 − (x − 1)4 4 8 64
P4,1 f (x) = f (1) +
4.a) lim+ x2 ln x x→0
ln x −∞ which is of the form 2 x→0 1/x ∞ 1/x = lim+ x→0 −2x−3 x2 = lim+ x→0 −2 = 0
lim+ x2 ln x =
x→0
lim+
2
b) lim+ (
x→1
1 x − ) = x ln x x − 1 = = = = =
c) lim (1 + tan x)1/x x→0
lim+
x→1
lim
x→1+
lim
x→1+
lim
x→1+
lim
x→1+
(x − 1) − x2 ln x 0 which is of the form 2 (x − x) ln x 0 2 0 1 − [x /x + 2x ln x] which is of the form 2 (x − x)/x + ln x(2x − 1) 0 0 1 − x − 2x ln x which is of the form x − 1 + (2x − 1) ln x 0 −1 − [2x/x + 2 ln x] 1 + (2x − 1)/x + 2 ln x −3 − 2 ln x 1 + 2 − 1/x + 2 ln x
−3 2
We have a form 1∞ .
ln lim (1 + tan x)1/x = lim ln(1 + tan x)1/x x→0
x→0
ln(1 + tan x) 0 which is of the form x→0 x 0 1 2 sec x (1+ tan x) = lim x→0 1 1 sec2 0 = 1 + tan 0 = 1
=
lim
Hencelim (1 + tan x)1/x = e1 = e x→0 5. Rotation around the y q − axis so all notaion in terms of y. y−1 2 y−1 Inner radius: r(y) = 2 q Intersection points: ⇔ y−1 2
Outer radius : R(y) =
=
y−1 2
so 2(y −1) = (y −1)2 i.e. y 2 −4y +3 = (y −3)(y −1) = 0
then y = 1 or y = 3.
3
Z V
(
= π Z1 3 = π
r
3
y−1 2 y−1 2 ) −( ) ]dy 2 2
[(y − 1)/2 − (y 2 − 2y + 1)/4]dx
Z1 3
π [−y 2 + 4y − 3]dy 4 1 π −1 3 [ y + 2y 2 − 3y] 31 = 4 3 π π 4 π = [−9 + 18 − 9 + 1/3 − 2 + 3] = . = 4 4 3 3
=
6. Z a)
√ sec2 (3+ln x) dx x
Let u = 3 + ln
√
,x > 0
= x , du dx Z
1 2x
i.e. 2du =
dx x
√ sec2 (3+ ln x) dx x
Z = 2 = =
Z b)
2 tan u + c √ 2 tan(3+ ln x) + c
2
ln x4 dx, x > 1 2
du dx
Z
Z c)
sec2 udu
u = ln x4 = x42 . 2x = 4
2 x
dv = 1 v=x
Z x2 x2 ln dx = x ln − 2 1dx 4 4 x2 = x ln − 2x + c 4
1
dx 3 (3−2x) −∞
We first look at the indefinite integral: Let u = 3 − 2x then du = −2 dx So we have Z Z 1 dx = u−3 du (3 − 2x)3 −2 1 −2 = u +c 4 1 = +c 4(3 − 2x)2 4
Z
1
−∞
dx = (3 − 2x)3
1 |1 t→−∞ 4(3 − 2x)2 t 1 1 − lim = 4 t→−∞ 4(3 − 2t)2 1 = 4
7.
Z
lim
x−1 dx (x + 1)(x2 + x + 1)
We see that for y = x2 + x + 1, b2 − 4ac = −3 < 0, so x−1 A Bx + C = + (x + 1)(x2 + x + 1) x + 1 x2 + x + 1 A(x2 + x + 1) + (Bx + C)(x + 1) = (x + 1)(x2 + x + 1) Then x − 1 = A(x2 + x + 1) + (Bx + C)(x + 1) = (A + B)x2 + (A + B + C)x + (A + C) Then A + B = 0, A + B + C = 1 and A + C = −1 We have So C = 1, A = −2 and B = 2 Z
x−1 dx = (x + 1)(x2 + x + 1) = = = =
Z −2 2x + 1 dx + dx x+1 x2 + x + 1 Z Z 1 1 −2 du + dv where u = x + 1, du/dx = 1, v = x2 + x + 1, dv/dx = u v −2 ln |u| + ln |v| + c −2 ln |x + 1| + ln x2 + x + 1 + c (x2 + x + 1) ln +c (x + 1)2 Z
8.a) g(x) = x2 + 2x + 5 = (x + 1)2 + 4 5
b) Use (a) and trigonometric substitution to determine the integral. Z Z dx dx √ p . dx. = x2 + 2x + 5 (x + 1)2 + 4 Z 1 dx q = 2 ( x+1 )2 + 1 2
Let x+1 dx = tan θ i.e. x = 2 tan θ − 1 then = 2 sec2 θ 2 dθ Then 1 2
Z
Z 2 sec2 θdθ 1 q √ = 2 tan2 θ + 1 ( x+1 )2 + 1 2 Z = sec θdθ dx
ln |sec θ + tan θ| + c p (x + 1)2 + 4 x + 1 = ln + +c 2 2 √ x2 + 2x + 5 + x + 1 = ln +c 2 =
6
(2)
9. Z
dx = 0 1 + sin x + cos x
Z
Z =
2 1+z 2 2z 1−z 2 1 + 1+z 2 + 1+z 2 2 1+z 2 dz 1+z 2 +2z+1−z 2 2 1+z
dz
Z
2 dz 2(z + 1) Z 1 = dz (z + 1) = ln |z + 1| + c x = ln tan + 1 + c 2 =
Z 0
π 2
π dx = ln tan + 1 − ln |tan 0 + 1| = ln 2 − ln 1 = ln 2 1 + sin x + cos x 4
7
