Use the Mean Value Theorem to prove that there exists a $c\in(0,1)$ such that $f(c) = f'(c)$ 
Suppose that $f:[0,1]\to\Bbb{R}$,  $f$ is continuous on $[0,1]$ and is differentiable on $(0,1)$. Furthermore, $f(0)=f(1)=0$.
  Prove that there exists a $c\in(0,1)$ such that $f(c)=f'(c)$. 

The only hint I am given is to use the Mean Value Theorem on $g:[0,1]\to\Bbb{R}$ where it is defined $\forall{x}\in[0,1]:g(x)=e^{-x}f(x)$
I am genuinely confused as to what the question is trying to ask me. Would trying to solve this lead me to a proof similar to Rolle's Theorem? Any help is appreciated!
 A: You can use eithen MVT or Rolle's theorem for $g$:


*

*$g$ is continous in $[0,1]$

*$g$ is differentiable in $(0,1)$

*$g(0)=g(1)=0$
So from Rolle's theorem $\exists c\in (0,1)$ s.t. $g'(c)=0$
But $g'(x)=e^{-x}f'(x)-e^{-x}f(x)=e^{-x}(f'(x)-f(x))$. So $g'(c)=0\Rightarrow f'(c)-f(c)=0\Rightarrow f(c)=f'(c)$
I don't know wether this answer is what you were looking for
A: Note that $0$ and $1$ are the roots of the equation $f(x)=0$. Also $f$ is                                 continuous everywhere and it has a tangent at every point on it with abscissa between $0$ and $1$. Then $\exists$ a point $c\in(0,1)$ such that the tangent to the curve at $(c,f(c))$ and is parallel to the  line segment joining the points $(0,f(0))$ and $(1,f(1))$.  
Rolle's theorem is a particular case of MVT. If $f(a)=f(b)$ holds in addition to the two conditions of MVT, then $f(b)-f(a)=0$ and consequently, $f'(c)=0$. Geometrically we can say that there is a point $c\in(a,b)$ such that the tangent of                  $f(x)$ at $c$ is parallel to the $x$-axis.
