140Bquiz13asols - 1 MATH 140B 1. QUIZ#13 Solutions Dec 14,...

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1 MATH 140B QUIZ#13 Solutions Dec 14, 2007 1. i) True. Since C (0) = 0 , it is enough to show that C ( t ) 0 for all t 0 . Note that C ( t ) 0 for all t 0 if e t/ 4 e t/ 3 for all t 0 . . Multiplying by e t/ 3 means C ( t ) 0 if e t/ 12 1 for all t 0 . Finally, raising each side to the power 12, means C ( t ) 0 if e t 1 for all t 0 . But this is true by looking at the graph of e t . ii) True C ( t ) = - 1 4 e t/ 4 + 1 3 e t/ 3 . Therefore, C ( t ) = 0 when 1 4 e t/ 4 = 1 3 e t/ 3 . Multiplying by e t/ 3 gives 1 4 e t/ 12 = 1 3 . Therefore, t = 12 ln 4 3 . This critical point is a local maximum since, by the same reasoning, C ( t ) > 0 when t < 12 ln 4 3 and C ( t ) < 0 when t > 12 ln 4 3 . Since this critical point is unique , it is actually a global maximum. iii) False lim t →∞ C ( t ) = lim t →∞ p 1 e t/ 4 - 1 e t/ 3 P = 0 - 0 = 0 . Therefore, C) is the answer.
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This note was uploaded on 03/30/2008 for the course MATH 140B taught by Professor Fabbri,marc during the Fall '07 term at Pennsylvania State University, University Park.

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