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Unformatted text preview: MA 115 Homework Solutions for Week 5 4.2 #6 Absolute maximum value is f (8) = 5; absolute minimum value is f (2) = 0; local maximum values are f (1) = 2, f (4) = 4, and f (6) = 3; local minimum values are f (2) = 0, f (5) = 2, and f (7) = 1. 4.2 #24 f ( x ) = x 3 + x 2 x f ( x ) = 3 x 2 + 2 x 1 f ( x ) = 0 ( x + 1)(3 x 1) = 0 x = 1 , 1 3 . These are the only critical numbers. 4.2 #38 f ( x ) = x 3 3 x + 1 , [0 , 3]. f ( x ) = 3 x 2 3 = 0 x = 1, but 1 / [0 , 3]. So by testing the values x = 0 , 1 , 3 we see f (0) = 1, f (1) = 1 and f (3) = 19. So f (3) = 19 is the absolute maximum value and f (1) = 1 is the absolute minimum value. 4.2 #46 f ( x ) = x 2 cos x, [ , ]. f ( x ) = 1+2 sin x = 0 sin x = 1 2 x = 5 6 , 6 . By comparing the function value at both endpoints and the critical points we see f ( ) = 2 1 . 14, f ( 5 6 ) = 3 5 6 . 886, f ( 6 ) = 6 3 2 . 26, f ( ) = + 2 5 . 14. So f ( ) = + 2 is the absolute maximum value and f ( 6 ) = 6 3 is the absolute minimum value. 4.3 #6 (a) f is increasing on the intervals where f ( x ) > 0, namely, (2 , 4) and (6 , 9). (b) f has a local maximum where it changes from increasing to decreasing, that is, where f changes from positive to negative (at x = 4). Similarly, where f changes from negative to positive, f has a local minimum (at x = 2 and at x = 6). (c) When f is increasing, its derivative f is positive and hence, f is concave upward. this happens on (1 , 3) , (5 , 7) and (8 , 9). Similarly, f is concave downward when f...
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This note was uploaded on 04/07/2008 for the course MA 115 taught by Professor Mahalanobis during the Fall '08 term at Stevens.
 Fall '08
 Mahalanobis

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