# Rolle's theorem $\beta \cdot f(x)+f'(x)=0$

Prove that if $f$ is differentiable on $[a,b]$ and if $f(a)=f(b)=0$ then for any real $\beta$ there is an $x \in (a,b)$ such that $\beta \cdot f(x)+f'(x)=0$. (Using rolle's theorem)

My attempt:

Using Rolle's theorem we can say that there exists some $c \in (a,b)$, where $f'(c)=0$ . Therefore one factor in the expression $\beta \cdot f(x)+f'(x)=0$ is $0$ at $c$ but I am unable to prove that the other factor will simultaneously be $0$ at $c$.

Hint: Apply Rolle's Theorem to $g(x) = e^{\beta x} f(x)$.
$g(a) = g(b) = 0$, therefore there is a $c \in (a, b)$ such that $g'(c) = 0$. The conclusion follows with $g'(x) = e^{\beta x}(\beta f(x) + f'(x))$ since $e^{\beta x}$ is never zero.