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Unformatted text preview: MAA 4200: Homework 7 Solutions July 25, 2009 [4.18] Proof: Let f ( x ) = x 3 3 x + b = 0 . Then f ( x ) = 3 x 2 3 < 0 for x ( 1 , 1) . Suppose f has two roots r 1 ,r 2 [ 1 , 1] . Then f satisfies the hypotheses of the Mean Value Theorem on [ r 1 ,r 2 ] and hence there exists c ( r 1 ,r 2 ) ( 1 , 1) such that f ( c ) = 0 , a contradiction. Therefore f has at most one root in [ 1 , 1] . [4.19] Proof: Let f ( x ) = cos( x ) x 3 x 2 4 x, then f (0) = 1 > 0 and f ( / 2) = 3 / 8 2 / 4 2 < . Hence, by the Intermediate Value Theorem, f has at least one root in [0 ,/ 2] . Moreover, f ( x ) = sin( x ) 3 x 2 2 x 4  3 x 2 2 x 3 < 3 for x (0 ,/ 2). Suppose f has two roots r 1 ,r 2 [0 ,/ 2] . Then f satisfies the hypotheses of the Mean Value Theorem on [ r 1 ,r 2 ] and hence there exists c ( r 1 ,r 2 ) (0 ,/ 2) such that f ( c ) = 0 , a contradiction. Therefore f has at most one root in [0 ,/ 2] . So f has exactly one root in [0...
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This note was uploaded on 08/13/2009 for the course MAT 371 taught by Professor Thieme during the Spring '07 term at ASU.
 Spring '07
 thieme
 Mean Value Theorem

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