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Unformatted text preview: tion to each other. The only place that can go wrong the assumption that f ( b ) ≥ 0 at some point b > 0. Therefore, f ( x ) < 0 for all x > 0. (This argument is called “proof by contradiction”.) Solution. (a) Since we assume f ( b ) ≥ , f ( b )f (0) b= f ( b ) b ≥ . 1 (b) Since the function f ( x ) is continuous on [0 ,b ] and diﬀerentiable on (0 ,b ), we can apply the MVT. From MVT, we conclude that there exists c in (0 ,b ) such that f ( c ) = f ( b )f (0) b= f ( b ) b ≥ (c) f ( x ) = 1 2 √ x +11 2 = 2(1√ x +1) 2 √ x +1 < , for x > 0. But f ( c ) ≥ 0 from (b), where c in (0 ,b ), so we have a contradiction. 2...
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This note was uploaded on 12/03/2010 for the course M 408c taught by Professor Mcadam during the Fall '06 term at University of Texas at Austin.
 Fall '06
 McAdam
 Calculus

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