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Unformatted text preview: MATH 137 Assignment #7 Solutions 1. (a) Since f is continuous over the closed interval [1 , 4], the Extreme Value Theorem applies and tells us that f must attain a maximum value over the interval. (b) We need to find the critical number of f . f ( x ) = 1 ln x x 2 , which is defined on [1 , 4], and so f is differentiable on (1 , 4). So there are no critical numbers where f ( x ) does not exist. We solve f ( x ) = 0 1 ln x x 2 = 0 1 ln x = 0 since x 6 = 0 x = e Since f (1) = 0, f ( e ) = 1 e . 368, and f (4) = ln4 4 . 347, the maximum value of f on [1 , 4] is 1 e and it occurs at x = e . (c) Since 1 e is the maximum value of f ( x ) = ln x x on [1 , 4], then 1 e ln x x for all x [1 , 4]. In fact, since the location of the maximum is unique, 1 e > ln x x for all x [1 , 4] , x 6 = e . 1 e > ln (since [1 , 4]) > e ln > ln e e > e ln e (since e x is a strictly increasing function) e > e (d) The minimum value is 0 and occurs at the endpoint x = 1. (e) Since f ( x ) = ln x x = ln x 1 x , then e f ( x ) = e ln x 1 /x = x 1 x = g ( x ). Since e x is a strictly increasing function, the maximum value of g ( x ) = e f ( x ) on [1 , 4] occurs at the maximum of f ( x ) on [1 , 4], or at x = e . Similarly, the minimum of g ( x ) occurs at the minimum of f ( x ), or at x = 1. Therefore, the maximum value of g ( x ) = x 1 x on [1 , 4] is e 1 /e and the minimum value is 1....
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This note was uploaded on 12/31/2011 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.
 Fall '08
 ANDREWCHILDS
 Math, Algebra

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