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Let $f:I\to\Bbb{R}$ be differentiable in open $I$. A critical point of $f$ is a pont $c\in I$ s.t. $f'(c)=0$ and a critic $c$ is a non-degenerate point if exists $f''(c)$ and $f''(c)\neq0$. Prove that if $c$ is a non-degenerate critical point, then exists $\delta>0$ st $f$ has no other critical points in $(c-\delta, c+\delta)$.

I think I need a hint to start this question. What am I missing?

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    I am not sure about how to write a formal proof, but think that $f''(c)\neq0$ means that, around $c$, the first derivative is not constant, because $f''$ is the change in $f'(x)$ along $x$. So, around $c$, $f'(x)$ changes from $f'(c)$, so it cannot take the same value as $f'(c)=0$. – 2017-02-08
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    I would recommend assuming (WLOG) that $f''(c)>0$. – 2017-02-08

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Hint for $f''(c) > 0$:

Show $f'(c+x) > 0$ and $f'(c-x) < 0$ for $x>0$ sufficiently small. Then, $f$ has a strict local minimum at $c$.

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    I showed it, but I didn't understand where I use the fact that $f''(c)>0$. – 2017-02-08
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    @LucasBarbiere if $f''(c)\ne 0$ it is either $>0$ or $<0$. How can you reduce the $<0$ case to the $>0$ case? – 2017-02-08
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Hint: If the result fails, then there exists a sequence $x_n \to c,$ with $x_n \ne c$ for all $n,$ such that $f'(x_n) = 0.$

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    I liked your proof, but I do not know how to handle it. – 2017-02-08
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    Note $(f'(x_n) -f'(c))/(x_n-c) \to f''(c).$ – 2017-02-08