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

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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 = 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 everywhere in IR, so f is constant. 2. Define a function p as follows: p ( x ) := n 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 ( 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) = ( x - 0) f ( c ), hence f ( x ) = xf ( c ), i.e., f ( x ) /x = f ( c ). Since f is monotonically increasing, the condition c < x implies f ( c ) f ( x ), so f ( x ) /x f ( x ), hence 0 xf ( x ) - f ( x ) x 2 = g ( x ) where g ( x ) := f ( x ) x . Thus g ( x ) 0 for all x > 0, which means g is monotonically increasing. 4. Since f ( x ) = | x | 3 = x 3 , x 0 , - x 3 x < 0 , we get f ( x ) 3 x 2 , x > 0 , - 3 x 2 x < 0 , and f (0) = lim x 0+ f ( x ) = lim x 0 - f ( x ) = 0 . Next, f ( x ) = 6 x, x > 0 , - 6 x x < 0 , and f (0) = lim x 0+ f ( x ) = lim
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