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 March 23rd, 2009, 10:49 PM #1 Newbie   Joined: Sep 2007 Posts: 17 Thanks: 0 Give this one a shot?? Suppose f is an odd function and is differentiable everywhere. Prove that for every positive number b, there exists a number c in (-b,b) such that f '(c) = f(b)/b Where do you even begin?
March 24th, 2009, 06:08 AM   #2
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 Originally Posted by titans4ever0927 Suppose f is an odd function and is differentiable everywhere. Prove that for every positive number b, there exists a number c in (-b,b) such that f '(c) = f(b)/b
By definition of an odd function, f(-b) = -f(b).

The Mean Value Theorem tells you that, for this sort of function, there must be some value c on the interval (-b, b) such that:

[color=white]. . . . .[/color]$f'(c)\, =\, \frac{f(b)\, -\, f(-b)}{b\, -\, (-b)}$

Plug the result of f being odd into the above, and simplify.

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