# Finding a point with $f(x)=f'(x)$

Let $f :[0,1] \rightarrow \mathbb{R}$ be a fixed continuous function that is differentiable on $(0,1)$ and such that $f(0)=f(1)=0$. Does there exist a $x_0 \in (0,1)$ such that $f(x_0)=f'(x_0)$?

By the mean value theorem there is a point that has derivative $0$, but this doesn't help me.

Consider $g(x) = f(x)e^{-x}$. Since $g(0), g(1) = 0$, the derivative $g'(x) = \left(f'(x) - f(x)\right)e^{-x}$ must vanish somewhere on $(0, 1)$. Thus there exists some $x_0\in (0, 1)$ with $f(x_0) = f'(x_0)$.