Let $f$ a function which is differentiable on $[a,b]$ such that $f(a)=f(b)=f^\prime(a)=0$ Let $f$ a function  which is differentiable on $[a,b]$ such that $f(a)=f(b)=f^\prime(a)=0$
Prove that: $\exists c\in ]a,b[$ : $$f^\prime(c)=\frac{f(c)}{c-a}$$
I tried to use Rolle theorem but i can't find the result
 A: Let's rewrite this a bit.
$$
f'(c) = \frac{f(c)}{c-a}\\
f'(c)(c-a) = f(c)\\
f'(c)(c-a) - f(c) = 0
$$
Here comes a little trick: Look at the function $g(x) = \frac{f(x)}{x-a}$. Its derivative is $g'(x) = \frac{f'(x)(x-a) - f(x)}{(x-a)^2}$. Do you recognize that numerator? (This might look like pure magic if you're not used to it, but to me, since I've seen and done the trick a few times before, $f'(x)(x-a) - f(x)$ practically screams "$u'v - uv'$" and the quotient rule. There is, of course, a very similar trick we could've done had we had $f'(x)(x-a) + f(x)$ instead.) 
It turns out that a $c$ such as the one we're looking for is exactly an extremum of $g(x)$. Why should such a $c$ exist in $]a, b[$?
Small problem: $g$ is only defined on $]a, b]$, because of the denominator, and Rolle's theorem demands all of $[a, b]$. What can be done about that? Well, $\lim_{x \to a^+}g(x) = 0$, because of $f'(a) = 0$ (it's easiest to invoke l'Hôpital on this one, if you've heard of that rule). So we can expand the definition of $g$ to the following:
$$
g(x) = \cases{0 & if $x = a$\\\frac{f(x)}{x-a} & otherwise}
$$
and by the last paragraph, this new $g$ is stil continuous.
But now we clearly have $g(a) = g(b) = 0$. Rolle's theorem then tells us that there is a $c \in ]a, b[$ such that $g'(c) = 0$. Inserting that into our formula for $g'(x)$, this means that
$$
0 = g'(c) = \frac{f'(c)(c-a) - f(c)}{(c-a)^2}\\
0 = f'(c)(c-a) - f(c)\\
f'(c) = \frac{f(c)}{c-a}
$$
and we're done.
