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I'm stuck on a little calc question. If someone can give me a hint it will be helpful :). I am given a function $f(x)$ in the interval $[ 0,1]$ such that $f(0) = f(1) = 0$. I know that $f$ is twice-differentiable and $f''$ is continuous on $[0,1]$. I need to show that there exists $0 < x < 1$ such that $f(x) = f'(x)$. I tried to use Lagrange theorem in many ways but it didn't work out for me..any ides?

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up vote 1 down vote accepted

Hint: look at $e^{-x}f(x){}{}$.

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Thanks a lot :) – John Smith Sep 17 '12 at 10:46

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