Copy of s10 - Solutions to homework # 10. 1. If |f (x) −...

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Solutions to homework # 10. 1. If | f ( x ) - f ( y ) | ≤ ( x - y ) 2 , then 0 ± ± ± ± ± f ( x ) - f ( y ) x - y ± ± ± ± ± ≤ | x - y | for all x 6 = y IR . Note that the leftmost and the rightmost expressions tend to 0 as x y . Therefore by the Sandwich theorem lim x y ± ± ± ± ± f ( x ) - f ( y ) x - y ± ± ± ± ± = 0 for all y IR . Hence f 0 = 0 everywhere in IR, so f is constant. 2. Define a function p as follows: p ( x ) := n X j =0 c j x j +1 j + 1 = c 0 x + c 1 x 2 2 + ··· + c n x n +1 n + 1 . Then p (0) = 0 and p (1) = c 0 + c 1 2 + ··· + c n n + 1 = 0 by the assumption of the problem. The function p is a polynomial, so in particular it is continuously differentiable everywhere in [0 , 1]. Therefore, by Theorem 5.10, there exists x (0 , 1) such that p 0 ( x ) = 0, i.e., c 0 + c 1 x + ··· + c n x n = 0 . 3. Fix x > 0. By the assumption of the problem, f is continuous on [0 , x ] and differentiable on (0 , x ). Then the mean value theorem implies that there exists c (0 , x ) such that f ( x ) - f (0) = (
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Copy of s10 - Solutions to homework # 10. 1. If |f (x) −...

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