Let $f$ be a continuous and differentiable function such that $f(a)=f(b)=0$ , show that $f'(c)=\pi f(c)$ for some $c \in (a,b)$ Let $f:[a,b] \rightarrow \mathbb{R} $  be a continuous function in $[a,b]$ and differentiable in $(a,b)$   such that $f(a)=f(b)=0$ . Show that $f'(c)=\pi f(c)$ for some $c \in (a,b)$
The initial hypothesis of that $f(a)=f(b)$ makes me think that I need to use the Rolle's Theorem of some way but I don't see how to use it.
Any hint will be appreciated.
 A: Here's the crux of the matter (I think). Suppose $f>0$ on $(a,b).$ Then $\ln f$ is well defined and differentiable on $(a,b).$ We have $\ln f(x) \to -\infty$ at $a^+,b^-.$ (Good to draw a picture.) It follows from the MVT that $(\ln f)'$ takes on arbitrarily large positive and negative values. By Darboux, $(\ln f)'$ takes on all values in $\mathbb {R},$ in particular the value $\pi.$ Since $(\ln f)' = f'/f,$ we're done.
Now we don't have $f>0$ on $(a,b)$ probably. So the argument is not complete, but I think this is the idea.
A: Define $g(x)=f(x)\cdot e^{-\pi x}$. Notice that $g$ is, like $f$, continuous on $[a,b]$, differentiable on $(a,b)$, and zero at the endpoints $a$ and $b$.  Also notice that $$g'(x)=f'(x)\cdot e^{-\pi x}+f(x)\cdot(-\pi)e^{-\pi x}= (f'(x)-\pi f(x))e^{-\pi x}.$$  Apply Rolle's theorem.
A: Consider the function g(x) = f(x)exp(-πx) .
Then g(a) = f(a)exp(-πa) =   0.exp(-πa) = 0
     g(b) = f(b)exp(-πn) =   0.exp(-πb) = 0
and note that the derivative g'(x) = f'(x)exp(-πx) - πf(x)exp(-πx)
so we can apply Rolle's theorem, so there exists c such that
 g'(c) = 0, f'(c).exp(-πc) - πf(c).exp(-πc) = 0 

we can divide by exp(-πc) to give the result 
   f'(c)  - πf(c)  = 0

