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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

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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 $f(a) < f(b)$.

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  • $\begingroup$ I didn't understand $\endgroup$ – user281932 Nov 3 '16 at 14:33
  • $\begingroup$ What does Rolle's theorem say, and do you know the intermediate value theorem? $\endgroup$ – Camille Nov 3 '16 at 18:48
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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 $(p, q)$.

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