lab6_solns_f11

lab6_solns_f11 - Math 116 - Lab 6 - Solutions - Fall 2011....

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Math 116 - Lab 6 - Solutions - Fall 2011. 1. a) f 00 ( π 4 ) = - 4 < 0 concave down. 1. b) f 00 (1) = 25 e - 5 > 0 concave up. 1. c) f 00 (12) = - 1 144 < 0 concave down. 2. a) f 00 ( x ) = 1 x . Since the domain of f ( x ) = x ln( x ) is x (0 , ), we have f 00 ( x ) > 0 for every x in the domain. Therefore, there are no critical points. 2. b) f 00 ( x ) = cosh( x ) > 0 for every x . So there are no second order critical points. 2. c) f 00 ( x ) = sinh( x ). The second order critical point is the solution of f 00 ( x ) = 0, which is x = 0; it is an inflection point as f 00 ( x ) changes sign at x = 0. 2. d) f 0 ( x ) = e - x (1+ e - x ) 2 = 1 ( e x/ 2 + e - x/ 2 ) 2 = 1 4 [cosh( x/ 2)] - 2 . f 00 ( x ) = - 1 4 1 cosh 3 ( x 2 ) sinh( x 2 ). Hence the solution of f 00 ( x ) = 0 is x = 0; it’s easy to verify that it is an inflection point. 3. f 00 ( x ) = b 2 e - bx > 0 for every x . So the answer is for any a and b . 4. a) f 0 ( x ) = x 3 - 4 x = x ( x - 2)( x + 2). So the solutions of f 0 ( x ) = 0 are x = 0, x = 2 and x = - 2. Since f 00 ( x ) = 3 x 2 - 4, it is easy to verify that f 00 (0) < 0, whereas f 00 (2) > 0 and f 00 ( - 2) > 0. Hence the second derivative test tells us that x = 0 is a local maximum, while x = 2 and x = - 2 are local minima. 4. b)
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This note was uploaded on 12/28/2011 for the course MATH 116 taught by Professor Robertandre during the Spring '11 term at Waterloo.

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lab6_solns_f11 - Math 116 - Lab 6 - Solutions - Fall 2011....

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