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March 23rd, 2009, 10:49 PM   #1
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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?
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March 24th, 2009, 06:08 AM   #2
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Quote:
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]

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