Q4Solutions - Now, since f 00 < 0 for all x 6 = 0, x = 0...

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MAC 2233 Quiz 4 Solutions July 1, 2009 You must show all work to receive credit 1. Let f ( x ) = 3 x 4 - 6 x 3 + x - 8. Find the open intervals where f is concave down. Solution It suffices to find all open intervals where f 00 ( x ) < 0. Begin by computing f 0 and f 00 . We have f 0 ( x ) = 12 x 3 - 18 x 2 + 1 f 00 ( x ) = 36 x 2 - x = 36 x ( x - 1) . The zeros of f 00 are x = 0 and x = 1. By computing f 00 ( - 1) ,f 00 ( 1 2 ) and f 00 (2), we see that f 00 ( x ) < 0 on (0 , 1) and f 00 ( x ) > 0 on ( -∞ , 0) (1 , ). It follows that f is concave down on (0 , 1). 2. Find all inflection points of the function f ( x ) = x 4 7 . If there are none, explain why . Solution A routine computation shows that f 00 ( x ) = - 12 49 x 10 / 7 . By inspecting f 00 , we see that f 00 6 = 0 for any x in the domain of f . Also, we see that f 00 does not exist at x = 0, so x = 0 is the only candidate to be an inflection point.
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Unformatted text preview: Now, since f 00 < 0 for all x 6 = 0, x = 0 is not an inection point of f . 3. Consider the function f ( x ) = 2 x 3 + 3 x 2-12 x-4. The critical points of f are x 1 = 1 and x 2 =-2. Moreover, f (1) = 0 and f (-2) = 0. If f 00 ( x ) = 6(2 x + 1), classify each of x 1 and x 2 as local maximizers or local minimizers. Solution Since each of x 1 = 1 and x 2 =-2 are zeros of f , we need only com-pute the signs of f 00 ( x 1 ) and f 00 ( x 2 ). A routine calculation shows that f 00 (1) > 0 and f 00 (-2) < 0. Therefore, x 1 = 1 is a local minimizer of f and x 2 =-2 is a local maximizer of f ....
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This note was uploaded on 12/27/2011 for the course MAC 2233 taught by Professor Smith during the Spring '08 term at University of Florida.

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