TITLE: If $f(x)=f'(x)+f''(x)$ then show that $f(x)=0$
QUESTION [18 upvotes]: A real-valued function $f$ which is infinitely differentiable on $[a.b]$ has the following properties:

$f(a)=f(b)=0$
$f(x)=f'(x)+f''(x)$ $\forall x \in [a,b]$

Show that $f(x)=0$ $\forall x\in [a.b]$

I tried using the Rolle's Theorem, but it only tells me that there exists a $c \in [a.b]$ for which $f'(c)=0$. 
All I get is:

$f'(a)=-f''(a)$
$f'(b)=-f''(b)$
$f(c)=f''(c)$

Somehow none of these direct me to the solution.

REPLY [20 votes]: Hint $f$ can't have a positive maximum at $c$ since then $f(c)>0, f'(c)=0, f''(c) \le 0$ implies that $f''(c)+f'(c)-f(c) < 0$. Similarly $f$ can't have a negative minimum. Hence $f = 0$.