# 29 171 x 4 y y y y y 03 02 1 014 171 010 x 3 2 1 0

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Unformatted text preview: ce the roots are x = −2.45, 0.65, 2.75. f (x) = 2 2x5 + 5x3 + 14x2 + 30x − 7 . Points of inﬂection will occur when the numerator changes sign, (x2 + 1)5/2 since the denominator is always positive. A plot of y = 2x5 + 5x3 + 14x2 + 30x − 7 suggests that there is only one root and it lies in [0, 1]. Subdivide into ten subintervals and determine that the root lies between x = 0.2 and x = 0.3. Thus to one decimal place the point of inﬂection is located at x = 0.25. f (x) = f (x1 ) − f (x2 ) = x2 − x2 = (x1 + x2 )(x1 − x2 ) < 0 if x1 < x2 for x1 , x2 in [0, +∞), so f (x1 ) < f (x2 ) 1 2 and f is thus increasing. 46. f (x1 ) − f (x2 ) = is decreasing. 1 1 x2 − x1 − = > 0 if x1 < x2 for x1 , x2 in (0, +∞), so f (x1 ) > f (x2 ) and thus f x1 x2 x1 x2 145 47. Chapter 5 (a) If x1 < x2 where x1 and x2 are in I , then f (x1 ) < f (x2 ) and g (x1 ) < g (x2 ), so f (x1 ) + g (x1 ) < f (x2 ) + g (x2 ), (f + g )(x1 ) < (f + g )(x2 ). Thus f + g is increasing on I . (b) Case I: If f and g are ≥ 0 on I , and if x1 < x2 where x1 and x2 are in I , then 0 < f (x1 ) < f (x2 ) and 0 < g (x1 ) < g (x2 ), so f (x1 )g (x1 ) < f (x2 )g (x2 ), (f · g )(x1 ) < (f · g )(x2 ). Thus f · g is increasing on I . Case II: If f and g are not necessarily positive on I then no conclusion can be drawn: for example, f (x) = g (x) = x are both increasing on (−∞, 0), but (f · g )(x) = x2 is decreasing there. 48. (a) f (x) = x, g (x) = 2x 49. (a) (b) f (x) = x, g (x) = x + 6 (c) f (x) = 2x, g (x) = x b b f (x) = 6ax + 2b = 6a(x + ), f (x) = 0 when x = − . f changes its direction of concavity 3a 3a b b so − is an inﬂection point. at x = − 3a 3a (b) If f (x) = ax3 + bx2 + cx + d has three x-intercepts, then it has three roots, say x1 , x2 and x3 , so we can write f (x) = a(x − x1 )(x − x2 )(x − x3 ) = ax3 + bx2 + cx + d, from which it follows that 1 b = (x1 + x2 + x3 ), which is the average. b = −a(x1 + x2 + x3 ). Thus − 3a 3 (c) f (x) = x(x2 − 3x2 + 2) = x(x − 1)(x − 2) so the intercepts are 0, 1, and 2 and the average is 1. f (x) = 6x − 6 = 6(x − 1) changes sign at x = 1. 50. b f (x) = 6x + 2b, so the point of inﬂection is at x = − . Thus an increase in b moves the point of 3 inﬂection to the left. 51. (a) Let x1 < x2 belong to (a, b). If both belong to (a, c] or both belong to [c, b) then we have f (x1 ) < f (x2 ) by hypothesis. So assume x1 < c < x2 . We know by hypothesis that f (x1 ) < f (c), and f (c) < f (x2 ). We conclude that f (x1 ) < f (x2 ). (b) Use the same argument as in part (a), but with inequalities reversed. 52. By Theorem 5.1.2, f is increasing on any interval ((2n − 1)π, 2(n + 1)π ) (n = 0, ±1, ±2, · · ·), because f (x) = 1 + cos x > 0 on [(2n − 1)π, (2n + 1)π ]. By Exercise 51 (a) we can piece these intervals together to show that f (x) is increasing on (−∞, +∞). 53. t = 7.67 1000 0 15 0 54. By zooming on the graph of y (t), maximum increase is at x = −0.577 and maximum decrease is at x = 0.577. 55. 56. y y 4 2 infl pt 3 1 t infl pts 2 1 x Exercise Set 5.2 146 EXERCISE SET 5.2 1. (a) (b) y y f (x) f (x) x x (c) (d) y y f (x) f (x) x 2. (a) x (b) y y x x (c) (d) y y x x 4. (a) f (x) = 6x − 6 and f (x) = 6, with f (1) = 0. For the ﬁrst derivative test, f < 0 for x < 1 and f > 0 for x > 1. For the second derivative test, f (1) > 0. (b) 3. f (x) = 3x2 − 3 and f (x) = 6x. f (x) = 0 at x = ±1. First derivative test: f > 0 for x < −1 and x > 1, and f < 0 for −1 < x < 1, so there is a relative maximum at x = −1, and a relative minimum at x = 1. Second derivative test: f < 0 at x = −1, a relative maximum; and f > 0 at x = 1, a relative minimum. (a) f (x) = 2 sin x cos x = sin 2x (so f (0) = 0) and f (x) = 2 cos 2x. First derivative test: if x is near 0 then f < 0 for x < 0 and f > 0 for x > 0, so a relative minimum at x = 0. Second derivative test: f (0) = 2 > 0, so relative minimum at x = 0. (b) g (x) = 2 tan x sec2 x (so g (0) = 0) and g (x) = 2 sec2 x(sec2 x + 2 tan2 x). First derivative test: g < 0 for x < 0 and g > 0 for x > 0, so a relative minimum at x = 0. Second derivative test: g (0) = 2 > 0, relative minimum at x = 0. (c) 5. Both functions are squares, and so are positive for values of x near zero; both functions are zero at x = 0, so that must be a relative minimum. (a) f (x) = 4(x − 1)3 , g (x) = 3x2 − 6x + 3 so f (1) = g (1) = 0. (b) f (x) = 12(x − 1)2 , g (x) = 6x − 6, so f (1) = g (1) = 0, which yields no information. (c) f < 0 for x < 1 and f > 0 for x > 1, so there is a relative minimum at x = 1; g (x) = 3(x − 1)2 > 0 on both sides of x = 1, so there is no relative extremum at x = 1. 147 Chapter 5 (a) f (x) = −5x4 , g (x) = 12x3 − 24x2 so f (0) = g (0) = 0. (b) f (x) = −20x3 , g (x) = 36x2 − 48x, so f (0) = g (0) = 0, which yields no information. (c) 6. f < 0 on bot...
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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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