Chapter 5 let y be the height of the box then s x2 4xy

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Unformatted text preview: h sides of x = 0, so there is no relative extremum there; g (x) = 12x2 (x − 2) < 0 on both sides of x = 0 (for x near 0), so again there is no relative extremum there. 7. f (x) = 3x2 + 6x − 9 = 3(x + 3)(x − 1), f (x) = 0 when x = −3, 1 (stationary points). √ (b) f (x) = 4x(x2 − 3), f (x) = 0 when x = 0, ± 3 (stationary points). 8. (a) f (x) = 6(x2 − 1), f (x) = 0 when x = ±1 (stationary points). (b) f (x) = 12x3 − 12x2 = 12x2 (x − 1), f (x) = 0 when x = 0, 1 (stationary points). (a) √ f (x) = (2 − x2 )/(x2 + 2)2 , f (x) = 0 when x = ± 2 (stationary points). (b) f (x) = 2 x−1/3 = 2/(3x1/3 ), f (x) does not exist when x = 0. 3 (a) f (x) = 8x/(x2 + 1)2 , f (x) = 0 when x = 0 (stationary point). 9. 10. (a) (b) f (x) = 1 (x + 2)−2/3 , f (x) does not exist when x = −2. 3 11. (a) (b) 12. (a) (b) 13. (a) 4(x + 1) , f (x) = 0 when x = −1 (stationary point), f (x) does not exist when x = 0. 3x2/3 f (x) = −3 sin 3x, f (x) = 0 when sin 3x = 0, 3x = nπ, n = 0, ±1, ±2, · · · x = nπ/3, n = 0, ±1, ±2, · · · (stationary points) f (x) = f (x) = 4(x − 3/2) , f (x) = 0 when x = 3/2 (stationary point), f (x) does not exist when x = 0. 3x2/3 cos x, sin x > 0 sin x, sin x ≥ 0 and f (x) does not so f (x) = − cos x, sin x < 0 − sin x, sin x < 0 exist when x = nπ , n = 0, ±1, ±2, · · · (sin x = 0) because limx→nπ− f (x) = limx→nπ+ f (x) (see Theorem preceding Exercise 75, Section 3.3). Now f (x) = 0 when ± cos x = 0 provided sin x = 0 so x = π/2 + nπ , n = 0, ±1, ±2, · · · are stationary points. f (x) = | sin x| = x = 2 because f (x) changes sign from − to + there. (b) x = 0 because f (x) changes sign from + to − there. (c) x = 1, 3 because f (x) (the slope of the graph of f (x)) changes sign at these points. 14. (a) x=1 15. (a) √ critical points x = 0, ± 5; f : (b) x=5 (c) x = −1, 0, 3 − − 0+ + 0− − 0 + + −5 0 5 √ x = 0: relative maximum; x = ± 5: relative minimum (b) critical point x = 0; f : −−− 0 +++ 0 x = 0: relative minimum 16. (a) critical points x = 0, −1/2, 1; f : +++0− 0− − −0 + −1 2 0 1 x = 0: neither; x = −1/2: relative maximum; x = 1: relative minimum Exercise Set 5.2 (b) critical points: x = ±3/2, −1; f : 148 + +0−−?+ +0− − −3 2 −1 3 2 x = ±3/2: relative maximum; x = −1: relative minimum 17. f (x) = −2(x + 2); critical point x = −2; f (x): +++0−−− -2 f (x) = −2; f (−2) < 0, f (−2) = 5; relative max of 5 at x = −2 18. f (x) = 6(x − 2)(x − 1); critical points x = 1, 2; f (x) : +++0−−−0+++ 1 2 f (x) = 12x − 18; f (1) < 0, f (2) > 0, f (1) = 5, f (2) = 4; relative min of 4 at x = 2, relative max of 5 at x = 1 19. f (x) = 2 sin x cos x = sin 2x; critical points x = π/2, π , 3π/2; f (x) : + +0− −0+ +0 − − π 2 π f (x) = 2 cos 2x; f (π/2) < 0, f (π ) > 0, f (3π/2) < 0, f (π/2) = f (3π/2) = 1, f (π ) = 0; relative min of 0 at x = π , relative max of 1 at x = π/2, 3π/2 20. f (x) = 1/2 − cos x; critical points x = π/3, 5π/3; f (x): − −0+++++0− − π 3 5π 3 f (x) = − sin x; f (π/3) < 0, f (5π/3) > 0 √ √ f (π/3) = π/6 − 3/2, f (5π/3) = 5π/6 + 3/2; √ √ relative min of π/6 − 3/2 at x = π/3, relative max of 5π/6 + 3/2 at x = 5π/3 21. f (x) = 3x2 + 5; no relative extrema because there are no critical points. 22. f (x) = 4x(x2 − 1); critical points x = 0, 1, −1 f (x) = 12x2 − 4; f (0) < 0, f (1) > 0, f (−1) > 0 relative min of 6 at x = 1, −1, relative max of 7 at x = 0 23. f (x) = (x − 1)(3x − 1); critical points x = 1, 1/3 f (x) = 6x − 4; f (1) > 0, f (1/3) < 0 relative min of 0 at x = 1, relative max of 4/27 at x = 1/3 24. f (x) = 2x2 (2x + 3); critical points x = 0, −3/2 relative min of −27/16 at x = −3/2 (first derivative test) 25. f (x) = 4x(1 − x2 ); critical points x = 0, 1, −1 f (x) = 4 − 12x2 ; f (0) > 0, f (1) < 0, f (−1) < 0 relative min of 0 at x = 0, relative max of 1 at x = 1, −1 26. f (x) = 10(2x − 1)4 ; critical point x = 1/2; no relative extrema (first derivative test) 27. f (x) = 4 x−1/5 ; critical point x = 0; relative min of 0 at x = 0 (first derivative test) 5 28. f (x) = 2 + 2 x−1/3 ; critical points x = 0, −1/27 3 relative min of 0 at x = 0, relative max of 1/27 at x = −1/27 29. f (x) = 2x/(x2 + 1)2 ; critical point x = 0; relative min of 0 at x = 0 3π 2 149 30. Chapter 5 f (x) = 2/(x + 2)2 ; no critical points (x = −2 is not in the domain of f ) no relative extrema 31. f (x) = 2x/(1 + x2 ); critical point at x = 0; relative min of 0 at x = 0 (first derivative test) 32. f (x) = x(2 + x)ex ; critical points at x = 0, −2; relative min of 0 at x = 0 and relative max of 4/e2 at x = −2 (first derivative test) 33. f (x) = 2x if |x| > 2, f (x) = −2x if |x| < 2, f (x) does not exist when x = ±2; critical points x = 0, 2, −2 relative min of 0 at x = 2, −2, relative max of 4 at x = 0 34. f (x) = −1 if x < 3, f (x) = 2x if...
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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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