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Let $f\colon[a,b]\to\mathbb{R}$ be a differentiable function and $x\in[a,b]$ with $f'(x)=0$. Is there a counterexample with respect to the equivalence

$x$ is not a local extremum of $f$

$\Leftrightarrow$

There is an $\epsilon>0$ such that $f'$ is monotonically increasing on $[x-\epsilon;x]$ and monotonically decreasing on $[x;x+\epsilon]$ or

there is an $\epsilon>0$ such that $f'$ is monotonically decreasing on $[x-\epsilon;x]$ and monotonically increasing on $[x;x+\epsilon]$.

In other words: Are the two definitions of a saddle point

$x$ is a saddle point of a differentiable function $f$ iff $x$ is a stationary point which is not an local extremum.

and

$x$ is a saddle point of a differentiable function $f$ iff $x$ is a stationary inflection point

really equivalent?

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The two definitions are not equivalent. Look at the graph of the differentiable function $$x\mapsto x^2 \sin\frac1x \qquad (x\neq 0); \qquad 0\mapsto 0$$

enter image description here

$0$ is a stationary point, but neither a local extremum nor a stationary inflection point.

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    $\begingroup$ For completeness, note how $f'(x)$ is non-monotonic on both of the intervals $[-\epsilon, 0]$ and $[0, \epsilon]$ for any $\epsilon>0$, making it a direct counterexample for the proposed equivalence. $\endgroup$ – Arthur Feb 20 '18 at 11:20

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