If $f(x)=f'(x)+f''(x)$ then show that $f(x)=0$ 
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.

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 $f = 0$.
