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Let $f$ be a differentiable function on $(a,b)$ such that $f'(c)=0$ for some $c\in (a,b)$. Also there is no maximum or minimum at $c$ .Does this means $c$ is point of inflection?

Consider $f(x)=x^3$ , this is true. But I want to know if it is true in general. I try to find counter example but can't get any. I hope this is true. If it is please give a hint to begin a proof

Edit : hint this question by maxima / minima, I mean strict maxima or minima

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  • $\begingroup$ If it's differential (big if) then $f'(c) =0$ means is an extreme point. Geometrically this means the slope is zero. This is either a local min or max or inflection point. As you know it's neither of two of the options, it must be the third option. $\endgroup$ – fleablood Dec 21 '18 at 17:15
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    $\begingroup$ @fleablood: I don't believe you're correct. A (local) extreme point is by definition a (local) maximum or minimum. So $f'(c)=0$ does NOT imply that $c$ is a local extreme point. $\endgroup$ – Ted Shifrin Dec 21 '18 at 17:19
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    $\begingroup$ I think you need to check your definitions carefully, @fleablood. $\endgroup$ – Ted Shifrin Dec 21 '18 at 17:23
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    $\begingroup$ Actually you're right. I was thinking of the term "critical points". $\endgroup$ – fleablood Dec 21 '18 at 17:25
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    $\begingroup$ @fleablood It is true that local extremum implies critical point, I think you confuse this with the converese which is not true. And inflection point is typically defined as a point where the concavity changes. $\endgroup$ – N. S. Dec 21 '18 at 17:37
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Consider the function $$g(x)=x\sin(\frac{1}{x})$$ with $g(0)=0$. This function is continuous on $\mathbb R$.

Let $F(x)$ be any antiderivative of $g(x)$. Then $F'(0)=0$.

Now, since $g(x)$ is an even function, $F$ is odd, and hence it cannot have a local max/min at $x=0$.

Moreover, $F(x)$ cannot have an inflection point at $x=0$, since this would imply that for some $(0,a)$ the function $g(x)$ would be monotonic.

Added: if $n\geq 2$, $f$ is $n$ times differentiable, $f^{(n)}$ is continuous at $c$ and $$f'(c)=...=f^{(n-1)}(c)=0 \\ f^{(n)}(c) \neq 0$$ then

  • If $n$ is odd $c$ is an inflection point
  • If $n$ is even and $f^{(n)}(c) >0$ then $c$ is a local min.
    • If $n$ is even and $f^{(n)}(c) <0$ then $c$ is a local max.

Sketch Proof: Since $f^{(n)}(c) \neq 0$ it is either positive or negative.

If $f^{(n)}(c)<0$ then replace $f$ by $-f$. Note that in the case $n$ even thiwill change local max to min.

By continuity, there exists some $a>0$ such that $f^{(n)}>0$ on $(c-a, c+a)$.

Now you do an inductive argument. $f^{(n)}>0$ on $(c-a, c+a)$ means $f^{(n-1)}$ is strictly increasing on $(c-a, c+a)$ and thus, since $f^{(n-1)}(c)=0$ you get $$f^{n-1}(x) <0 \forall x \in (c-a,c) \\ f^{n-1}(x) >0 \forall x \in (c,c-a) (*)\\$$

This gives that $f^{n-2}$ is decreasing on (c-a,c)$ and increasing on $(c,c+a)$. Therefore $$f^{n-2}(x) >0 \forall x \in (c-a,c) \ f^{n-2}(x) >0 \forall x \in (c,c-a) (**)\$$

Then, same argument shows that $$f^{n-3}(x) <0 \forall x \in (c-a,c) \\ f^{n-3}(x) >0 \forall x \in (c,c-a) (*)\\$$

and so on, with (*) and (**)alternating. Based of $n$ being odd or even, either $f'$ or $f''$ will satisfy $(*)$ [this is why we need $n \geq 2$].

Now, if $f'$ satisfies $(*)$ then it is easy to see that $c$ is a local max.

If $f''$ satisfies $(*)$ theb it is easy to see that $c$ is an inflection point.

QED

P.P.S. The key for the proof is the existence and continuity of the first derivative which doesn't vanish at $c$. The above counterexample fails this :)

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  • $\begingroup$ Thanks it helps a lot $\endgroup$ – Cloud JR Dec 21 '18 at 17:38
  • $\begingroup$ @CloudJR Check my P.S. too, the claim is actually true for "nice" functions. $\endgroup$ – N. S. Dec 21 '18 at 17:41
  • $\begingroup$ @CloudJR If you need the details I can sketch the idea of proof for analytic functions..."Analyticity" can actually be replaced by $f$ is $n$ times differentiable, $f^{(n)}$ is continuous and $$f'(c)=....f^{(n-1)}(c)=0 \\f^{(n)}(c) \neq 0$$ $\endgroup$ – N. S. Dec 21 '18 at 17:46
  • $\begingroup$ Where can I find more about this....like proof and more ..any book? $\endgroup$ – Cloud JR Dec 21 '18 at 17:46
  • $\begingroup$ @CloudJR see the new edit. $\endgroup$ – N. S. Dec 21 '18 at 18:06
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The simplest counter example is the constant function $$ f\equiv 0 $$ on any interval $(a,b)$. Clearly $f'(c)=0$ for any $c\in(a,b)$ but it's not a point of inflection.

Edit: I interpreted the words max/min to mean strict minima/maxima, if the question refers to nonstrict ones then this is not a counter example.

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  • $\begingroup$ But every point is a local min/max. $\endgroup$ – N. S. Dec 21 '18 at 17:17
  • $\begingroup$ But every point is maximum or minimum $\endgroup$ – Cloud JR Dec 21 '18 at 17:17
  • $\begingroup$ Yeah, I just realized that I probably misinterpreted the question. Sorry for that. $\endgroup$ – BigbearZzz Dec 21 '18 at 17:18

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