If h 3r then b 2 3r the triangle is equilateral r the

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Unformatted text preview: be less than zero because it is the product of two squares; the minimum value is 0 for x = 1 or −2, no maximum because lim f (x) = +∞. 15 x→+∞ -3 2 0 179 25. Chapter 6 5(8 − x) , f (x) = 0 when x = 8 and f (x) does not exist 3x1/3 when x = 0; f (−1) = 21, f (0) = 0, f (8) = 48, f (20) = 0 so the maximum value is 48 at x = 8 and the minimum value is 0 at x = 0, 20. 50 f (x) = -1 20 0 26. f (x) = (2 √ x2 )/(x2 + 2)2 , f (x) √ 0 for√ in the interval (−1, 4) − = x when x = 2; f (−1) = −1/3, f ( 2) = 2/4, f (4) = 2/9 so the √ √ maximum value is 2/4 at x = 2 and the minimum value is −1/3 at x = −1. 0.4 -1 4 -0.4 27. f (x) = −1/x2 ; no maximum or minimum because there are no critical points in (0, +∞). 25 0 10 0 28. f (x) = (1 − x2 )/(x2 + 1)2 ; critical point x = 1. Maximum value f (1) = 1/2, minimum value 0 because f (x) is never less than zero on [0, +∞) and f (0) = 0. 0.5 0 20 0 29. f (x) = 2 sec x tan x − sec2 x = (2 sin x − 1)/√ 2 x, f (x) = √ for x in cos 0 (0, π/4) when x = π/6; f (0) = 2, f (π/6) = 3, f (π/4) = 2 √− 1 so 2 the maximum value is 2 at x = 0 and the minimum value is 3 at x = π/6. 2 3 0 1.5 30. f (x) = 2 sin x cos x − sin x = sin x(2 cos x − 1), f (x) = 0 for x in (−π, π ) when x = 0, ±π/3; f (−π ) = −1, f (−π/3) = 5/4, f (0) = 1, f (π/3) = 5/4, f (π ) = −1 so the maximum value is 5/4 at x = ±π/3 and the minimum value is −1 at x = ±π . 1.5 C c -1.5 Exercise Set 6.1 31. 180 f (x) = x2 (2x − 3)e−2x , f (x) = 0 for x in [1, 4] when x = 3/2; 27 if x = 1, 3/2, 4, then f (x) = e−2 , e−3 , 64e−8 ; 8 27 −3 e at x = 3/2, critical point at x = 3/2; absolute maximum of 8 absolute minimum of 64e−8 at x = 4 0.2 1 4 0 32. 1 − ln x f (x) = , f (x) = 0 when x = e; x2 absolute minimum of 0 at x = 1; absolute maximum of 1/e at x = e 0.4 1 e 0 33. f (x) = −[cos(cos x)] sin x; f (x) = 0 if sin x = 0 or if cos(cos x) = 0. If sin x = 0, then x = π is the critical point in (0, 2π ); cos(cos x) = 0 has no solutions because −1 ≤ cos x ≤ 1. Thus f (0) = sin(1), f (π ) = sin(−1) = − sin(1), and f (2π ) = sin(1) so the maximum value is sin(1) ≈ 0.84147 and the minimum value is − sin(1) ≈ −0.84147. 1 o 0 -1 34. f (x) = −[sin(sin x)] cos x; f (x) = 0 if cos x = 0 or if sin(sin x) = 0. If cos x = 0, then x = π/2 is the critical point in (0, π ); sin(sin x) = 0 if sin x = 0, which gives no critical points in (0, π ). Thus f (0) = 1, f (π/2) = cos(1), and f (π ) = 1 so the maximum value is 1 and the minimum value is cos(1) ≈ 0.54030. 1.5 0 0 35. c 4, x < 1 so f (x) = 0 when x = 5/2, and f (x) does not exist when x = 1 because 2x − 5, x > 1 lim− f (x) = lim+ f (x) (see Theorem preceding Exercise 75, Section 3.3); f (1/2) = 0, f (1) = 2, f (x) = x→1 x→1 f (5/2) = −1/4, f (7/2) = 3/4 so the maximum value is 2 and the minimum value is −1/4. 36. f (x) = 2x + p which exists throughout the interval (0, 2) for all values of p so f (1) = 0 because f (1) is an extreme value, thus 2 + p = 0, p = −2. f (1) = 3 so 12 + (−2)(1) + q = 3, q = 4 thus f (x) = x2 − 2x + 4 and f (0) = 4, f (2) = 4 so f (1) is the minimum value. 37. sin 2x has a period of π , and sin 4x a period of π/2 so f (x) is periodic with period π . Consider the interval [0, π ]. f (x) = 4 cos 2x + 4 cos 4x, f (x) = 0 when cos 2x + cos 4x = 0, but cos 4x = 2 cos2 2x − 1 (trig identity) so 2 cos2 2x + cos 2x − 1 = 0 (2 cos 2x − 1)(cos 2x + 1) = 0 cos 2x = 1/2 or cos 2x = −1. From cos 2x = 1/2, 2x = π/3 or 5π/3 so x = π/6 or √π/6. From cos 2x = −1, 2x = π so x =√ 2. 5 π/ √ f (0) = 0, f (π/6) = 3 3/2, f (π/2) = 0, f (5π/6) = −3 3/2, f (π ) = 0. The maximum value is 3 3/2 √ at x = π/6 + nπ and the minimum value is −3 3/2 at x = 5π/6 + nπ , n = 0, ±1, ±2, · · ·. 181 38. Chapter 6 x x has a period of 6π , and cos a period of 4π , so f (x) has a period of 12π . Consider the interval 3 2 x x x x [0, 12π ]. f (x) = − sin − sin , f (x) = 0 when sin + sin = 0 thus, by use of the trig identity 3 2 3 2 a−b 5x x 5x x a+b cos , 2 sin cos − = 0 so sin = 0 or cos = 0. Solve sin a + sin b = 2 sin 2 2 12 12 12 12 5x x = 0 to get x = 12π/5, 24π/5, 36π/5, 48π/5 and then solve cos = 0 to get x = 6π . The sin 12 12 corresponding values of f (x) are −4.0450, 1.5450, 1.5450, −4.0450, 1, 5, 5 so the maximum value is 5 and the minimum value is −4.0450 (approximately). cos 39. Let f (x) = x − sin x, then f (x) = 1 − cos x and so f (x) = 0 when cos x = 1 which has no solution for 0 < x < 2π thus the minimum value of f must occur at 0 or 2π . f (0) = 0, f (2π ) = 2π so 0 is the minimum value on [0, 2π ] thus x − sin x ≥ 0, sin x ≤ x for all x in [0, 2π ]. 40. Let f (x) = ln x − x + 1, then f (x) = 1/x − 1 and so f (x) = 0 at x = 1. Since lim+ f (x) = −∞ and x→0 lim f (x) = −∞, f (x) has a maximum of f (1) = 0 at x = 1 and so f (x) ≤ 0 for 0 < x < +∞, so x...
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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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