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Is it possible to have a null value at a given point of the derivative of a function while having this point not a local maximum or minimum ?

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    $\begingroup$ If by "null" you mean zero, consider $f(x)=x^3$ $\endgroup$
    – Joe
    May 23, 2021 at 16:26
  • $\begingroup$ @Joe : thank you. This is very interesting. But how is it possible ? I thought that monotonic function could not have a null derivative ? On wikipedia, it is said that : en.wikipedia.org/wiki/… : "A function f ( x ) is said to be absolutely monotonic over an interval if the derivatives of all orders of f are nonnegative or all nonpositive at all points on the interval" $\endgroup$ May 23, 2021 at 16:40
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    $\begingroup$ Non negative or non positive means $\ge 0$ or $\le 0$, not $>0$ or $<0$. A function has a local extremum at a point if & only if its derivative changes sign at that point. $\endgroup$
    – Bernard
    May 23, 2021 at 16:47
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    $\begingroup$ Mathieu-- Keep in mind that for example "nonnegative" just means not negative so in particular $0$ is nonnegative. $\endgroup$
    – coffeemath
    May 23, 2021 at 16:48
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    $\begingroup$ Actually, I believe the definition of a local maximum at $x_0$ is that there exists some $\epsilon>0$ such that for all $x$ with $|x-x_0|<\epsilon$, $f(x_0)\ge f(x)$. But a constant function satisfies that at every point, without the derivative changing sign, since it is identically zero. But if a function is continuously differentiable on an open interval containing the point, and the derivative changes sign at the point, then it is a local extremum. If the function is twice differentiable at the point, then the second derivative test may yield sufficient conditions. $\endgroup$
    – Joe
    May 23, 2021 at 17:22

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