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One of my friend gave the following problem to me but I am unable to solve it even after trying it for nearly two days.

Problem. Let $f:[0,1]\to\mathbb{R}$ be a continuous function such that $f(0)=f(1)=0$. If $f$ is differentiable on $(0,1)$ then prove that there exists $c\in(0,1)$ such that $f'(c)=f(c)$.

Can anyone help me solving this problem?

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  • $\begingroup$ It's a result of the Mean Value Theorem of the differential calculus. $\endgroup$
    – user90369
    Jul 13, 2016 at 14:20
  • $\begingroup$ @user90369: How so? $\endgroup$
    – user170039
    Jul 13, 2016 at 14:20
  • $\begingroup$ Ugh you are right. I delete my comment. $\endgroup$
    – Vincent
    Jul 13, 2016 at 15:20

2 Answers 2

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Let $g(x)=f(x)e^{-x}$. We have that $g(0)=g(1)=0$, so there (by Rolle's theorem) exists $c\in (0,1)$ such that $g'(c)=0$, which is equivalent with $f'(c)-f(c)=0$ or $f'(c)=f(c)$.

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    $\begingroup$ Cute. So for any real $k$ there is $c_k$ such that $f'(c_k)=kf(c_k)$. $\endgroup$ Jul 13, 2016 at 15:02
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Hint: Let $g(x)=\frac{f(x)}{e^x}$. Then by the Mean Value Theorem there is a $c$ between $0$ and $1$ such that $g'(x)=0$. Now calculate $g'(x)$.

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