# 10 x 2 2 0 24 x 12 x 22 can never be less than

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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 x-axis at x = 0, 1, −1 4 2 x -2 -1 1 -2 -4 -6 (b) lim y = +∞; y x→±∞ curve never crosses x-axis 0.2 0.1 -1 (c) 1 lim y = −∞, lim y = +∞; x→−∞ x y x→+∞ curve crosses x-axis at x = −1 0.4 0.2 x -1 1 -0.2 163 Chapter 5 (d) lim y = +∞; y x→±∞ curve crosses x-axis 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=x-4 ≈(-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=x-6 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 right-hand side of (7...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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