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Unformatted text preview: 0.3
0.4 √
√
lim (ln x)/ x = −∞ by inspection, lim (ln x)/ x = lim x→0+ (e1/2, 1 e 1) (e 5/6, 5 e 5/3)
2
6 x→+∞ 2 3 1/x
2
√ = lim √ = 0, L’Hˆpital’s
o
1/2 x x→+∞ x Rule.
2 − ln x
ln x
(b) y = √ , y =
x
2x3/2
−8 + 3 ln x
y=
4x5/2
y = 0 if x = e2 ,
y = 0 if x = e8/3 y 8
( e 8/3 , 3 e 4/3 ) ( e 2 , 2/ e ) 0.5
x
2 6 10 14 1 49. (a) lim y = −∞, lim y = +∞; x→−∞ y x→+∞ curve crosses xaxis at x = 0, 1, −1 4
2
x
2 1 1
2
4
6 (b) lim y = +∞; y x→±∞ curve never crosses xaxis
0.2 0.1 1 (c) 1 lim y = −∞, lim y = +∞; x→−∞ x y x→+∞ curve crosses xaxis at x = −1 0.4
0.2
x
1 1
0.2 163 Chapter 5 (d) lim y = +∞; y x→±∞ curve crosses xaxis at x = 0, 1
0.4
0.2
x
1 50. (a) y y y a a b y a (c) (a) b x x b y a 51. x x b y a b x
a (b) 1 b x horizontal asymptote y = 3 as x → ±∞, vertical asymptotes of
x = ±2 y
10 5 x
5 5
5 Exercise Set 5.3 (b) 164 horizontal asymptote of y = 1 as x → ±∞, vertical asymptotes
at x = ±1 y
10 x
5 5 10 (c) horizontal asymptote of y = −1 as x → ±∞, vertical
asymptotes at x = −2, 1 y
10 x
5 5 10 (d) horizontal asymptote of y = 1 as x → ±∞,
vertical asymptote at x = −1, 2 y
10 x
5 5 10 52. y a 53. b x (b) 0.5 y = (1 − bx)e−bx , y = b2 (x − 2/b)e−bx ;
relative max at x = 1/b, y = 1/be; point of
inﬂection at x = 2/b, y = 2/be2 . Increasing
b moves the relative max and the point of
inﬂection to the left and down, i.e. towards
the origin. (b) 0.4 (a) y = −2bxe−bx , y = 2b(−1 + 2bx2 )e−bx ;
relative max at x = 0, y = 1; points
√
of inﬂection at x = ± 1/2b, y = 1/ e.
Increasing b moves the points of inﬂection
towards the y axis; the relative max doesn’t
move. 3 0.2 54. (a) 1 2 2
0 2 2 165 55. Chapter 5 (a) The oscillations of ex cos x about zero
increase as x → ±∞ so the limit does not
exist. (b) y 6
4 (0,1)
(1.52, 0.22)
x 2 1 1 2 2 (c) The curve y = eax cos bx oscillates between y = eax and y = −eax . The frequency of oscillation
increases when b increases.
y y
5 1 10 b=3 1 5 x
2
b=1 a=2
a=1
x 1
5 a=3 b=2 0.5 1 b=1 a=1 56. 57. 58. P (x)
− (ax + b)
Q(x)
the degree of Q(x).
lim x→±∞ = lim x→±∞ R(x)
Q(x) = 0 because the degree of R(x) is less than 2
x2 − 2
= x − so
x
x
y = x is an oblique asymptote;
x2 + 2
,
y=
x2
4
y =− 3
x y y= 5
x2 − 2x − 3
=x−4+
so
x+2
x+2
y = x − 4 is an oblique asymptote;
x2 + 4x − 1
10
,y =
y=
2
(x + 2)
(x + 2)3 y=x
4 4 x y y= x = 2 10 10 x ≈(0.24, 1.52) y=x4 ≈(4.24, 10.48) 59. (x − 2)3
12x − 8
=x−6+
so
x2
x2
y = x − 6 is an oblique asymptote;
(x − 2)2 (x + 4)
,
y=
x3
24(x − 2)
y=
x4 y y= 10 10
(4, 13.5) (2, 0)
x
y=x6 Exercise Set 5.3 60. 166 4 − x3
,
x2
x3 + 8
y =−
,
x3
24
y= 4
x y y= (2, 3) y = x
x 61. 1
1
(x − 1)(x + 1)2
− 2=
,
xx
x2
y = x + 1 is an oblique asymptote;
(x + 1)(x2 − x + 2)
,
y=
x3
2(x + 3)
y =−
x4
y =x+1− y
y = x+1 (1, 0) x (3,  16 )
9 62.
63. The oblique asymptote is y = 2x so (2x3 − 3x + 4)/x2 = 2x, −3x + 4 = 0, x = 4/3.
lim [f (x) − x2 ] = lim (1/x) = 0 x→±∞ y x→±∞ x3 + 1
1
2x3 − 1
1
=
, y = 2x − 2 =
,
x
x
x
x2
√
2
2(x3 + 1)
y =2+ 3 =
, y = 0 when x = 1/ 3 2 ≈ 0.8,
3
x
√x
y = 3 3 2/2 ≈ 1.9; y = 0 when x = −1, y = 0 3 y = x2 + 64. 3 lim [f (x) − (3 − x2 )] = lim (2/x) = 0 x→±∞ x→±∞
3 2 + 3x − x
2
2
2(x3 + 1)
=
, y = −2x − 2 = −
,
x
x
x
x2
3
4
2(x − 2)
y = −2 + 3 = −
, y = 0 when x = −1, y = 0;
3
x
√x
y = 0 when x = 3 2 ≈ 1.3, y = 3
y = 3 − x2 + 65. y = x2
x y
y = 3  x2
1
3 x L Let y be the length of the other side of the rectangle, then
L = 2x + 2y and xy = 400 so y = 400/x and hence L = 2x + 800/x.
L = 2x is an oblique asymptote (see Exercise 48)
800
2(x2 + 400)
800
2(x2 − 400)
L = 2x +
=
, L =2− 2 =
,
x
x
x
x2
1600
L = 3 , L = 0 when x = 20, L = 80
x 100 20 x 167 66. Chapter 5 Let y be the height of the box, then S = x2 + 4xy and x2 y = 500 so S y = 500/x2 and hence S = x2 + 2000/x.
The graph approaches the curve S = x2 asymptotically 1000 (see Exercise 63)
x3 + 2000
2000
2000
2(x3 − 1000)
S = x2 +
=
, S = 2x − 2 =
,
x
x
x
x2
3
4000
2(x + 2000)
, S = 0 when x = 10, S = 300
S =2+ 3 =
x
x3 67. 30 y = 0.1x4 (6x − 5);
critical points: x = 0, x = 5/6;
relative minimum at x = 5/6,
y ≈ −6.7 × 10−3 x y 0.01
1 68. 1 y = 0.1x4 (x + 1)(7x + 5);
critical points: x = 0, x = −1, x = −5/7,
relative maximum at x = −1, y = 0;
relative minimum at x = −5/7, y ≈ −1.5 × 10−3 x y 0.001
1 69. x kL2 Ae−kLt
kL2 A
S , so P (0) =
(1 + Ae−kLt )2
(1 + A)2 (a) P (t) = (b) The rate of growth increases to its maximum, which occurs when P is halfway between 0 and
1
ln A; it then decreases back towards zero.
L, or when t =
Lk
dP
is maximized when P lies half way between 0 and L, i.e. P = L/2.
From (6) one sees that
dt
This follows since the right side of (6) is a parabola (with P as independent variable) with
1
ln A, from (8).
P intercepts P = 0, L. The value P = L/2 corresponds to t =
Lk (c) 70. Since 0 < P < L the righthand side of (7...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.
 Spring '14
 The Land

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