How prove this $f'(c)=2c(f(c)-f(0))$ let $f(x)$ is  continuous on $[0,1/2]$, and derivative on $(0,1/2)$,such
$$f'(1/2)=0$$
show that
there exsit $c\in(0,1/2)$, such
$$f'(c)=2c(f(c)-f(0))$$
My try: let
$$F(x)=e^{-x^2}[f(x)-f(0)]\Longrightarrow F'(x)=e^{-x^2}[f'(x)-2x(f(x)-f(0))]$$
then I can't
 A: I dont think the above answer is not accurate.  If one wants to use that method, a little more work should be done with the sign of $g(x)$ by stating this:
If $h$ is a differential function on [a,b] and $h'(a)\le h'(a)( a<b)$ hence: $\forall \lambda \in [h'(a);h'(b)]: \exists c \in [a,b]: h'(c)=\lambda$ 

Another approach:
Denote $F(x)$ as defined be by nanchangjian.
 Obviously, $F(x)$ has its maximum. WLOG: $F(c)=max(F(x))(c \in [0;\frac{1}{2}])$.
If $ 0<c<\frac{1}{2}$, the conclusion becomes obvious.
If $c=\frac{1}{2}$, thus $ F'(\frac{1}{2}) \ge 0 \Leftrightarrow f(0) \ge f(\frac{1}{2})$
$ \Rightarrow F(0) \ge F(1/2) \Rightarrow QED$
If $c=0$ , we just have to consider the minimum $F(d)$ of $F(x)$.
A: Suppose,contrary to our claim，Use the  Darboux theorem： http://en.wikipedia.org/wiki/Darboux's_theorem_(analysis)
Without loss of , we Assmue that
$$g(x)=f'(x)-2x[f(x)-f(0)]>0,0<x<1/2$$
then
$$F(x)=e^{-x^2}[f(x)-f(0)]\Longrightarrow F'(x)=e^{-x^2}[f'(x)-2x(f(x)-f(0))]>0$$
so
$$F(x)>F(0)=0$$
so
$$f'(x)-2x(f(x)-f(0))>0,0<x\le\dfrac{1}{2}$$
let
$x=1/2$
then
$$g(1/2)=f'(1/2)-(f(1/2)-f(0))=-(f(1/2)-f(0))<0$$
a contradiction.
