# Does $f'(x)=0$ implies Max/min or point of inflection

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

• 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. – fleablood Dec 21 '18 at 17:15
• @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. – Ted Shifrin Dec 21 '18 at 17:19
• I think you need to check your definitions carefully, @fleablood. – Ted Shifrin Dec 21 '18 at 17:23
• Actually you're right. I was thinking of the term "critical points". – fleablood Dec 21 '18 at 17:25
• @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. – N. S. Dec 21 '18 at 17:37

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 :)

• Thanks it helps a lot – Cloud JR Dec 21 '18 at 17:38
• @CloudJR Check my P.S. too, the claim is actually true for "nice" functions. – N. S. Dec 21 '18 at 17:41
• @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 $$f'(c)=....f^{(n-1)}(c)=0 \\f^{(n)}(c) \neq 0$$ – N. S. Dec 21 '18 at 17:46
• Where can I find more about this....like proof and more ..any book? – Cloud JR Dec 21 '18 at 17:46
• @CloudJR see the new edit. – N. S. Dec 21 '18 at 18:06

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.

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