# Let $f$ a differentiable function on an interval $I$ and $a,b,c$ are elements of this interval such that $a<c<b \text{ and } f(a)<0 ;f(c)>0 ;f(b)<0$

Let $f$ a differentiable function on an interval $I$ and $a,b,c$ are elements of this interval such that $a<c<b, f(a)<0 ;f(c)>0 ;f(b)<0$

Prove that: $$\exists c\in ]a,b[ ;f'(c)=0$$

I tried to use Rolle but i didn't find a good result

I think you mean $\exists k\in[a,b]$ such that $f'(k) = 0$?
If $f(a) = f(b)$ the result is obvious by Rolle's theorem.
Suppose not. We suppose $f(a) > f(b)$. Then by Intermediate value theorem there exists $d\in (c,b)$ such that $f(d) = f(a)$. Then Rolle's theorem gives the desired result.
The argument is the same if $f(a) < f(b)$.
You have to understand clearly the conditions of the question. It says that $f$ changes sign as it moves from $a$ to $c$ and again changes sign when it moves from $c$ to $b$. Therefore by Intermediate Value Theorem $f$ vanishes once (say at $p$) in $(a, c)$ and once more (say at $q$) in $(c, b)$. Now by Rolle's Theorem $f'$ vanishes once in the interval $(p, q)$.