Let $f:[0,1] \rightarrow \Bbb{R}$ be a fixed continuous function such that $f$ is differentiable on $(0,1)$ and $f(0)=f(1)=0$. Then the equation $$f(x_0)=f'(x_0)$$ admits
$1$) no solution for any $x_0 \in (0,1)$
$2$) more than one solution for $x_0 \in (0,1)$
$3$) exactly one solution for $x_0 \in (0,1)$
$4$) at least one solution for $x_0 \in (0,1)$
Actually $4$) is true! By Rolle's theorem, $f'(c)=0$ for some $c \in (0,1)$
Is $f(c)=0$ ? If so, how to prove? or any other way?
The example $f(x)\equiv0$ shows that 1) and 3) are not true. The example $f(x)=x(1-x)$ shows that 2) is not true. To show that 4) holds, consider $g(x)=e^{-x}\,f(x)$.