B by counterexample both hx x2 and g x 2x2 have

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Unformatted text preview: ) [0, +∞) (c) none (e) none (b) (−∞, 0] (d) (−∞, 0), (0, +∞) 4(x + 1) 3x2/3 4(x − 2) f (x) = 9x5/3 (a) [−1, +∞) (c) (−∞, 0), (2, +∞) (e) 0, 2 (b) (−∞, −1] (d) (0, 2) 4(x − 1/4) 3x2/3 4(x + 1/2) f (x) = 9x5/3 (a) [1/4, +∞) (c) (−∞, −1/2), (0, +∞) (e) −1/2, 0 (b) (d) (a) (−∞, 0] (c) (−∞, −1), (1, +∞) (e) −1, 1 (b) [0, +∞) (d) (−1, 1) 19. f (x) = 20. f (x) = 21. f (x) = −xe−x /2 2 f (x) = (−1 + x2 )e−x /2 22. f (x) = (2x2 + 1)ex 2 f (x) = 2x(2x2 + 3)ex (a) (−∞, +∞) (c) (0, +∞) (e) 0 (b) (d) 23. f (x) = 2x 1 + x2 1 − x2 f (x) = 2 (1 + x2 )2 (a) [0, +∞) (c) (−1, 1) (e) −1, 1 (b) (−∞, 0] (d) (−∞, −1), (1, +∞) 24. f (x) = x(2 ln x + 1) f (x) = 2 ln x + 3 (a) [e−1/2 , +∞) (c) (e−3/2 , +∞) (e) e−3/2 (b) (0, e−1/2 ] (d) (0, e−3/2 ) 25. f (x) = − sin x f (x) = − cos x (a) [π, 2π ] (c) (π/2, 3π/2) (e) π/2, 3π/2 2 2 (−∞, 1/4] (−1/2, 0) none (−∞, 0) 1 (b) (d) [0, π ] (0, π/2), (3π/2, 2π ) 0 2p -1 26. f (x) = 2 sin 4x f (x) = 8 cos 4x (a) (0, π/4], [π/2, 3π/4] (b) [π/4, π/2], [3π/4, π ] (c) (0, π/8), (3π/8, 5π/8), (7π/8, π ) (d) (π/8, 3π/8), (5π/8, 7π/8) (e) π/8, 3π/8, 5π/8, 7π/8 1 0 p 0 Exercise Set 5.1 27. 142 f (x) = sec2 x f (x) = 2 sec2 x tan x (a) (−π/2, π/2) (c) (0, π/2) (e) 0 10 (b) none (d) (−π/2, 0) ^ 6 -10 28. f (x) = 2 − csc2 x 8 cos x f (x) = 2 csc x cot x = 2 3 sin x (a) [π/4, 3π/4] (c) (0, π/2) (e) π/2 2 (b) (0, π/4], [3π/4, π ) (d) (π/2, π ) 0 p -2 29. f (x) = cos 2x f (x) = −2 sin 2x (a) [0, π/4], [3π/4, π ] (c) (π/2, π ) (e) π/2 0.5 (b) [π/4, 3π/4] (d) (0, π/2) 0 p -0.5 30. f (x) = −2 cos x sin x − 2 cos x = −2 cos x(1 + sin x) f (x) = 2 sin x (sin x + 1) − 2 cos2 x = 2 sin x(sin x + 1) − 2 + 2 sin2 x = 4(1 + sin x)(sin x − 1/2) Note: 1 + sin x ≥ 0 2 (a) [π/2, 3π/2] (b) [0, π/2], [3π/2, 2π ] (c) (π/6, 5π/6) (d) (0, π/6), (5π/6, 2π ) (e) π/6, 5π/6 0 o -2 31. (a) (b) y (c) y 4 4 x 2 y 4 x x 2 2 143 32. Chapter 5 (a) (b) y 4 (c) y 4 4 x 2 x 34. x 2 2 (a) f (x) = 3(x − a)2 , f (x) = 6(x − a); inflection point is (a, 0) (b) 33. y f (x) = 4(x − a)3 , f (x) = 12(x − a)2 ; no inflection points For n ≥ 2, f (x) = n(n − 1)(x − a)n−2 ; there is a sign change of f (point of inflection) at (a, 0) if and only if n is odd. For n = 1, y = x − a, so there is no point of inflection. 35. f (x) = 1/3 − 1/[3(1 + x)2/3 ] so f is increasing on [0, +∞) thus if √ 3 x √ > 0, then f (x) > f (0) = 0, 1 + x/3 − 1 + x > 0, 3 1 + x < 1 + x/3. 2.5 0 10 0 36. f (x) = sec2 x − 1 so f is increasing on [0, π/2) thus if 0 < x < π/2, then f (x) > f (0) = 0, tan x − x > 0, x < tan x. 10 0 0 37. x ≥ sin x on [0, +∞): let f (x) = x − sin x. Then f (0) = 0 and f (x) = 1 − cos x ≥ 0, so f (x) is increasing on [0, +∞). 6 4 0 4 -1 38. (a) Let h(x) = ex − 1 − x for x ≥ 0. Then h(0) = 0 and h (x) = ex − 1 ≥ 0 for x ≥ 0, so h(x) is increasing. (b) Let h(x) = ex − 1 − x − 1 x2 . Then h(0) = 0 and h (x) = ex − 1 − x. By part (a), ex − 1 − x ≥ 0 2 for x ≥ 0, so h(x) is increasing. Exercise Set 5.1 (c) 6 0 6 2 0 39. 144 0 2 0 Points of inflection at x = −2, +2. Concave up on (−5, −2) and (2, 5); concave down on (−2, 2). Increasing on [−3.5829, 0.2513] and [3.3316, 5] , and decreasing on [−5, −3.5829] and [0.2513, 3.3316]. 250 -5 5 -250 40. √ √ Points of inflection at x = ±1/ 3. Concave √ on (−5, −1/ 3) and up √ √ (1/ 3, 5), and concave down on (−1/ 3, 1/ 3). Increasing on [−5, 0] and decreasing on [0, 5]. 1 -5 5 -2 41. Break the interval [−5, 5] into ten subintervals and check f (x) at each endpoint. We find f (−1) > 0 and f (0) < 0. Refine [−1, 0] into ten subintervals; f (−0.2) > 0, f (−0.1) < 0; repeat, f (−0.18) > 0, f (−0.17) < 0, so x = −0.175 is correct to two decimal places. Note also that f (1) = 0 so there are two inflection points. 42. Break the interval [−5, 5] into ten subintervals and check f (x) at each endpoint. We discover f (−1) > 0, f (0) < 0 and f (1) > 0. Refine [−1, 0] into ten subintervals and we see that f (−0.6) > 0, f (−0.5) < 0. Subdivide [−0.6, −0.5] into 10 subintervals and we see that f (−0.58) > 0 and f (−0.57) < 0. Thus x = −0.575 is within 0.005 of the true root and is thus correct to two decimal places. For the other root we could proceed in a similar manner, but it easier to note that f (x) is an even function and thus the other root is x = 0.575 to two decimal places. 43. 44. 45. 90x3 − 81x2 − 585x + 397 . The denominator has complex roots, so is always positive; hence (3x2 − 5x + 8)3 the x-coordinates of the points of inflection of f (x) are the roots of the numerator (if it changes sign). A plot of the numerator over [−5, 5] shows roots lying in [−3, −2], [0, 1], and [2, 3]. Breaking each of these intervals into ten subintervals locates the roots in [−2.5, −2.4], [0.6, 0.7] and [2.7, 2.8]. Thus to one decimal pla...
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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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