I'm studying about stationary points of a function. First, in order to get all critical (or stationary points) we equate $f'(x)$ to zero and get all possible critical values. To check it's nature whether its a minima or maxima we plug that specific $x$ value (say $c$) to $f''(x)$. Its clear to me that if $f''(c)>0$ its definitely a minima and if $f''(c)<0$ its definitely a maxima But what if we encounter with $f''(c)=0$?? from what I found in the internet it can be an inflection point.

My problem is can it still be a minima or maxima if $f ''(c)=0$?

Can it be any other shape apart from inflection shape in the last case?

I would like to see an example if $f ''(c)=0$ and its some other shape apart from inflection point.

Your help is highly appreciated.

  • $\begingroup$ If the first non-zero derivative has even order, then it is a maximum or a minimum. If the first non-zero derivative has odd order, then it is an inflection point. $\endgroup$
    – robjohn
    Aug 26, 2021 at 11:15

2 Answers 2


Take $f(x)=x^4$. Note that $f''(0)=0$ and $0$ is a point of global minima of $f$.

The actual statement (given in this link) which is most helpful for you to determine whether it is an inflection point or one of minima, maxima requires you to find as many derivatives as you can and then depends on the sign of the even-order derivative.

The method of classification of the critical point, given in the link works so far as a non-zero derivative is reached at the critical point $c$ you want to investigate. (A non-trivial function for which every derivative at $x=c_0$ is $0$, which still achieves a global minima at $x=c_0$ is given by $$f(x)=\left\{\begin{matrix} \exp((x-c_0)^{-2}) & \text{if } x\ne c_0 \\ 0 & \text{if } x=c_0\end{matrix}\right\}$$ and this is a very useful/popular function with this property.)


Yes it can be either a maximum/minimum or it could be an inflection point.

Let consider for example

$$f(x)=x^3 \implies f'(x)=3x^2\quad f''(x)=6x$$

and in this case $x=0$ is an inflection point.

For the case

$$f(x)=x^4 \implies f'(x)=4x^3\quad f''(x)=12x^2$$

to check the minimum by derivatives we need to consider the fourth derivative which is indeed positive.

More in general the nature of the point is given by the sign of the first even derivative not equal to zero at that point.

  • $\begingroup$ Oh right so in $x^4$ its a minimum point right? $\endgroup$
    – emil
    Aug 30, 2020 at 8:31
  • $\begingroup$ @emil Let consider $y=x^4$ for the minimum, $y=-x^4$ for the maximum and $y=x^3$ for the inflection point. $\endgroup$
    – user
    Aug 30, 2020 at 8:33
  • $\begingroup$ Thank you very much! $\endgroup$
    – emil
    Aug 30, 2020 at 8:34
  • 1
    $\begingroup$ Recall that when second derivative test fails we can take a look to the sign of $f'(x)$ to determine the nature of the point. You are welcome! Bye $\endgroup$
    – user
    Aug 30, 2020 at 8:34
  • $\begingroup$ This answer seems to imply that a critical point can be both a relative extrema and an inflection point. However, neither example does that. The first example, x^3 has no extrema, increasing everywhere, except x=0. The 2nd example, has no inflection points, concave up everywhere. @user has a better explanation, when 2nd deriv. test is inconclusive use the first derivative test to determine extrema. $\endgroup$
    – nickalh
    Aug 26, 2021 at 5:59

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