# Right for a combined area of 1 k 1 k 1 2k 1 so

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Unformatted text preview: if x = y the inequality also holds. (b) f (x) = sin x, f (x) = cos x, |f (x)| ≤ 1 = M , so |f (x) − f (y )| ≤ |x − y | or | sin x − sin y | ≤ |x − y |. (a) If x, y belong to I and x < y then for some c in I , (a) If x, y belong to I and x < y then for some c in I , f (y ) − f (x) = f (c), y−x so |f (x) − f (y )| = |f (c)||x − y | ≥ M |x − y |; if x > y exchange x and y ; if x = y the inequality also holds. (b) If x and y belong to (−π/2, π/2) and f (x) = tan x, then |f (x)| = sec2 x ≥ 1 and | tan x − tan y | ≥ |x − y | Exercise Set 6.5 (c) 31. (a) (b) 204 y lies in (−π/2, π/2) if and only if −y does; use part (b) and replace y with −y √ Let f (x) = x. By the Mean-Value Theorem there is a number c between x and y such that √ √ √ y− x 1 y−x 1 √ = √ < √ for c in (x, y ), thus y − x < √ y−x 2c 2x 2x multiply through and rearrange to get √ xy < 1 (x + y ). 2 32. Suppose that f (x) has at least two distinct real solutions r1 and r2 in I . Then f (r1 ) = f (r2 ) = 0 so by Rolle’s Theorem there is at least one number between r1 and r2 where f (x) = 0, but this contradicts the assumption that f (x) = 0, so f (x) = 0 must have fewer than two distinct solutions in I . 33. (a) If f (x) = x3 + 4x − 1 then f (x) = 3x2 + 4 is never zero, so by Exercise 32 f has at most one real root; since f is a cubic polynomial it has at least one real root, so it has exactly one real root. (b) Let f (x) = ax3 + bx2 + cx + d. If f (x) = 0 has at least two distinct real solutions r1 and r2 , then f (r1 ) = f (r2 ) = 0 and by Rolle’s Theorem there is at least one number between r1 and r2 where f (x) = 0. But f (x) = 3ax2 + 2bx + c = 0 for √ √ x = (−2b ± 4b2 − 12ac)/(6a) = (−b ± b2 − 3ac)/(3a), which are not real if b2 − 3ac < 0 so f (x) = 0 must have fewer than two distinct real solutions. 34. 35. √ √ √ 1 1 1 1 1 4− 3 = 2 − 3. But < √ < √ for c in (3, 4) so f (x) = √ , √ = 4−3 4 2x2c 2c 23 √ √ √ √ 1 1 < 2 − 3 < √ , 0.25 < 2 − 3 < 0.29, −1.75 < − 3 < −1.71, 1.71 < 3 < 1.75. 4 23 (a) d2 [f (x) + g 2 (x)] = 2f (x)f (x) + 2g (x)g (x) = 2f (x)g (x) + 2g (x)[−f (x)] = 0, dx so f 2 (x) + g 2 (x) is constant. (b) f (x) = sin x and g (x) = cos x 37. (a) d2 [f (x) − g 2 (x)] = 2f (x)f (x) − 2g (x)g (x) = 2f (x)g (x) − 2g (x)f (x) = 0 so f 2 (x) − g 2 (x) dx is constant. (b) 36. f (x) = 1 (ex − e−x ) = g (x) and g (x) = 1 (ex + e−x ) = f (x) 2 2 If f (x) = g (x), then f (x) = g (x) + k . Let x = 1, f (1) = g (1) + k = (1)3 − 4(1) + 6 + k = 3 + k = 2, so k = −1. f (x) = x3 − 4x + 5. 38. 39. Let h = f − g , then h is continuous on [a, b], diﬀerentiable on (a, b), and h(a) = f (a) − g (a) = 0, h(b) = f (b) − g (b) = 0. By Rolle’s Theorem there is some c in (a, b) where h (c) = 0. But h (c) = f (c) − g (c) so f (c) − g (c) = 0, f (c) = g (c). y c x 205 40. Chapter 6 (a) Suppose f (x) = 0 more than once in (a, b), say at c1 and c2 . Then f (c1 ) = f (c2 ) = 0 and by using Rolle’s Theorem on f , there is some c between c1 and c2 where f (c) = 0, which contradicts the fact that f (x) > 0 so f (x) = 0 at most once in (a, b). (b) If f (x) > 0 for all x in (a, b), then f is concave up on (a, b) and has at most one relative extremum, which would be a relative minimum, on (a, b). 41. similar to the proof of part (a) with f (c) < 0 42. similar to the proof of part (a) with f (c) = 0 CHAPTER 6 SUPPLEMENTARY EXERCISES 3. (a) If f has an absolute extremum at a point of (a, b) then it must, by Theorem 6.1.4, be at a critical point of f ; since f is diﬀerentiable on (a, b) the critical point is a stationary point. (b) It could occur at a critical point which is not a stationary point: for example, f (x) = |x| on [−1, 1] has an absolute minimum at x = 0 but is not diﬀerentiable there. 4. No; speeding up means the velocity and acceleration have the same sign, i.e. av > 0; the velocity is increasing when the acceleration is positive, i.e. a > 0. These are not the same thing. An example is s = t − t2 at t = 1, where v = −1 and a = −2, so av > 0 but a < 0. 5. Yes; by the Mean Value Theorem there is a point c in (a, b) such that f (c) = 7. (a) f (b) − f (a) = 0. b−a f (x) = −1/x2 = 0, no critical points; by inspection M = −1/2 at x = −2; m = −1 at x = −1 (b) f (x) = 3x2 − 4x3 = 0 at x = 0, 3/4; f (−1) = −2, f (0) = 0, f (3/4) = 27/256, f (3/2) = −27/16, so m = −2 at x = −1, M = 27/256 at x = 3/4 1/3 (c) (d) ex (x − 2) , stationary point at x = 2; by Theorem x→+∞ x→0 x3 6.1.5 f (x) has an absolute minimum at x = 2, and m = e2 /4. (a) √ 3 − x2 , critical point at x = 3. Since lim+ f (x) = 0, f (x) has no minimum, and f (x) = 2 2 2 x→0 √ (x + 3) √ M = 3/3 at x = 3. (b) f (x) = 10x3 (x − 2), critical points at x = 0, 2; lim− f (x) = 88, so f (x) has no maximum; m = −9 (c) at x = 2 crit...
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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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