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Unformatted text preview: Math 508 Exam 2 Jerry L. Kazdan December 8, 2006 12:00 1:20 Directions This exam has two parts, Part A has 3 shorter problems (8 points each, so 24 points), Part B has 5 traditional problems (15 points each, so 75 points). Closed book, no calculators but you may use one 3 00 5 00 card with notes. Part A: Short Problems (3 problems, 8 points each). A1. A continuous function f : R R has the property that Z x f ( t ) dt = cos( x ) e- x + C, where C is some constant. Find both f ( x ) and the constant C . Solution: Letting x = 0, we find that 0 = 1 + C , so C =- 1. To compute f , use the fundamental theorem of calculus. Thus take the derivative of both sides f ( x ) = d dx [cos( x ) e- x + C ] =- sin( x ) e- x- cos( x ) e- x . A2. A function h : R R with two continuous derivatives has the property that h (0) = 2, h (1) = 0, and h(3)=1. Prove there is at least one point c in the interval 0 < x < 3 where h 00 ( c ) > 0 by finding some explicit m > 0 (such as m = 3 / 2) with h 00 ( c ) m . Solution: By the mean value theorem applied twice, there is some a (0 , 1) and some b (1 , 3) so that h ( a ) = h (1)- h (0) 1- =- 2 , h ( b ) = h (3)- h (1) 3- 1 = 1 2 . Thus, by the mean value theorem again there is some c ( a, b ) so that h 00 ( c ) = h ( b )- h ( a ) b- a = 1 2 + 2 b- a > 5 / 2 3 = 5 6 . A3. Say a smooth function u ( x ) satisfies u 00- c ( x ) u = 0 for 0 x 1 (here c ( x ) is some given contunuous function). If c ( x ) > 0 everywhere, show that there is no point where u ( x ) is both positive and has a local maximum. If we also knew that u (0) = 0 and u (1) = 0, why can we conclude that u ( x ) = 0 for all x 1?...
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This note was uploaded on 03/06/2012 for the course MATH 508 taught by Professor Staff during the Fall '10 term at UPenn.
- Fall '10