Can anyone find a function $f$ with an $x$ such that $f'(x)=0$, $f(x)$ is either a relative max/min and $(x,f(x))$ is an inflection point? In other words, suppose you're using the second derivative test, see that $f''(x)=0$ and then note that it's actually an inflection point. Can you conclude that $(x,f(x))$ is not a relative extremum?

It's easy to find an example where $f''(x)=0$ and there is an extremum--take $f(x)=x^4$, but in that case, there isn't actually an inflection point at 0.

I'm certainly open to the possibility that an inflection point can never be an extremum, though it seems like something I ought to have discovered by now, were it true.

  • $\begingroup$ Isn't an inflection point, by definition, a critical point which is not a local extremum? $\endgroup$
    – Zhen Lin
    Nov 9, 2011 at 22:17
  • $\begingroup$ @ZhenLin: No, an inflection point is a point where the concavity changes. You can have inflection points that are local extremes. $\endgroup$ Nov 9, 2011 at 22:20

3 Answers 3


It is certainly possible to have an inflection point that is also a (local) extreme: for example, take $$y(x) = \left\{\begin{array}{ll} x^2 &\text{if }x\leq 0;\\ x^{2/3}&\text{if }x\geq 0. \end{array}\right.$$ Then $y(x)$ has a global minimum at $0$. In addition, $y$ is concave up on $x\lt 0$, and concave down on $x\gt 0$ (the second derivative is $2$ for $x\lt 0$, and $-\frac{2}{9}x^{-4/3}$ for $x\gt 0$).

However, this function does not satisfy your original conditions, since the critical point at $0$ is not a stationary point, but rather a point where the function is not differentiable.

But say we have $f'(a)=0$, $f$ twice differentiable in a neighborhood of $a$, and $f(x)$ has an inflection point at $a$. Then the derivative is increasing before $a$ and decreasing after, or else $f'$ is decreasing before $a$ and increasing after. That means that $f$ does not have a local extreme at $a$ by the First Derivative Test: in the first case, $f'$ is negative before $a$ and also negative after; in the second it is positive both before and after. So in this situation, you can conclude that it is not a local extreme.

  • 1
    $\begingroup$ Is there a $C^\infty$ example of such a function? $\endgroup$ Nov 9, 2011 at 22:30
  • $\begingroup$ For the amount of time I and some others spent trying to construct counterexamples, that's a depressingly simple reason. But I guess the point is we were having fun, not thinking about it. $\endgroup$
    – hoyland
    Nov 9, 2011 at 22:31
  • $\begingroup$ @Zev: Of what kind of function? $\endgroup$ Nov 9, 2011 at 22:44
  • 1
    $\begingroup$ I believe Zev means a smooth function which has an inflection point which is also a local extreme. $\endgroup$
    – anon
    Nov 9, 2011 at 23:24
  • 1
    $\begingroup$ @anon: If that's what he meant, then the last paragraph answers it: if $f$ has continuous derivatives of all orders, and $f'(a)=0$ where $a$ is an inflection point, then $f'(x)$ has the same sign on both sides of $a$, so $f$ does not have a local extreme at $a$; that's the content of my last paragraph. That's why I'm not clear on what he means. $\endgroup$ Nov 9, 2011 at 23:28

From the question: "In other words, suppose you're using the second derivative test, see that $f''(x)=0$ and then note that it's actually an inflection point."

Hoyland, it seems you may be under the impression that all points where the second derivative is $0$ are inflection points. That's not true. For example, if $f(x)=x^4$, then $f''(0)=0$, but that's not an inflection point because $f''$ does not changes signs there: $f''$ is positive on both sides of $0$. And notice that that is an absolute minimum point. So if your question is whether a maximum or minimum point can occurr where $f''$ is $0$, the answer is "yes". But that doesn't mean there's an inflection point there.

  • $\begingroup$ I'm not sure that interpretation necessarily follows. I read the "and then note" as saying: suppose you are using the second derivative test, find $f''(a)=0$, and then, though some other method, you determine that $f$ has an inflection point at $a$... $\endgroup$ Nov 9, 2011 at 23:20
  • $\begingroup$ In fact... isn't your example exactly what Hoyland says in his second paragraph? $\endgroup$ Nov 9, 2011 at 23:33
  • $\begingroup$ Yeah, Arturo's reading is what I meant (hence the $x^4$ example). $\endgroup$
    – hoyland
    Nov 9, 2011 at 23:35

If using the non-strict definitions, then the function $f(x)=0$ has a local minimum (even a global minimum) and an inflection point everywhere:

Be $x_0$ arbitrary.

  • At $x_0$ there's a global, and thus also local, minimum, since for any $x$, $f(x)\ge f(x_0)$.

  • At $x_0$ there's an inflection point:

    • In the range $x<x_0$, $f$ is convex, since for any $a<b<x_0$ and $0<\lambda<1$ we have $f(\lambda a + (1-\lambda)b)\le \lambda f(a) + (1-\lambda)f(b)$.

    • In the range $x>x_0$, $f$ is concave, since for any $x_0<a<b$ and $0<\lambda<1$ we have $f(\lambda a + (1-\lambda)b)\ge \lambda f(a) + (1-\lambda)f(b)$.

Also note that everywhere $f''(x)=0$. Moreover note that $f$ has a local and global maximum everywhere.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .