False example in r 1 n 9 t f if f 0 1 r is a

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False Example: In R , { ( - 1) n } . 9. T F If f : [0 , 1] R is a continuous function and R 1 0 f ( x ) dx = 0, then f ( x ) is positive somewhere and negative somewhere in this interval (unless it is identically zero). True If f is not identically zero, then it is either positive somewhere or negative somewhere (or both). Say it is positive at x 0 . Then by continuity, it is positive in a neignborhood of x 0 . If f ( x ) 0, everywhere, then R 1 0 f ( x ) dx > 0, a contradiction. Thus f must be negative at some x 1 – and hence also in a neighborhood of x 1 . 10. T F f ( x ) := X 1 sin(3 n πx ) 2 n is a continuous function on R . True Since sin(3 n πx ) 2 n 1 2 n , by the Weierstrass M-Test the series converges uniformly and absolutely – and hence to a continuous function. Part B: Traditional Problems (5 problems, 14 points each) B–1. Let f : [ - 2 , 2] be a smooth function with the property that f ( - 1) = 1 , f (0) = 0 , f (1) = 2 . Show that at some point c ( - 1 , 1) we have f 00 ( c ) > 0. In fact, find an explicit constant m > 0 so that f 00 ( c ) m . Solution By the mean value theorem applied to the intervals [ - 1 , 0] and [0 , 1] there are points a ( - 1 , 0) and b (0 , 1) so that f 0 ( a ) = - 1 and f 0 ( b ) = 2. Applying the mean value theorem to f 0 , we conclude there is a point c ( a, b ) such that f 00 ( c ) = 3 / ( b - a ) > 3 / 2. B–2. Let A ( t ) and B ( t ) be n × n matrices that are differentiable for t [ a, b ] and let t 0 ( a, b ). Directly from the definition of the derivative, show that the product M ( t ) := A ( t ) B ( t ) is differentiable at t = t 0 and obtain the usual formula for M 0 ( t 0 ). Solution [Caution: Usually A ( t ) and B ( t ) will not commute.] M ( t 0 + h ) - M ( t 0 ) h = A ( t 0 + h ) B ( t 0 + h ) - A ( t 0 ) B ( t 0 ) h = [ A ( t 0 + h ) - A ( t 0 )] h B ( t 0 + h ) + A ( t 0 ) [ B ( t 0 + h ) - B ( t 0 )] h .
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