# Prove that there exists $c\in\mathbb{R}$ such that $f'(c)=f(c)$

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?

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

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 $f'(c)=f(c)$.
• Cute. So for any real $k$ there is $c_k$ such that $f'(c_k)=kf(c_k)$. Jul 13, 2016 at 15:02
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 $g'(x)$.