Given a stationary point, I was taught to test if it was a maximum or a minimum using the concavity test, i.e.

If $f''(x)>0$: concave up (thus a local minimum)

If $f''(x)<0$: concave down (thus a local maximum)

But if we used the concavity test on $f(x)=x^4$ at the stationary point $x=0$ we find that $f$ is concave up everywhere except at $x=0$ because $f''(0)=0$, which is horribly inconvenient.

So my question is: does that mean that point isn't a minimum, or is the definition a bit broader than I was taught? Logically, it looks like a minimum, so is the concavity test alone not fool proof?

Graph of $y=x^4$

enter image description here

  • $\begingroup$ Actually, the concavity text(= second derivative test) helps you in cases $f′′(x0)≠0$ [where $x_0$ is a point s.t $f'(x_0)=0$] If $f''(x_0)=0$ you can conclude nothing and you need to derive more till you find $f^{(n)}(x_0) \ne 0$ If $n$ is even, then your point is an extreme point, if $n$ is odd it's a saddle point. Good luck. $\endgroup$ – Galc127 Jan 13 '16 at 13:49
  • $\begingroup$ Ooh, good to know that test, thanks! $\endgroup$ – J-J Jan 13 '16 at 13:51

Usually when $f''(x_0) = 0$ we have to look at higher derivatives to argue about the nature of a point. In other words, if you limit yourself to the second derivative you have not enough informations to argue about the nature of that specific point.

The test is usually stated like this

The Second Derivative Test: Suppose that $c$ is a critical point at which $f'(c)=0$, that $f'(x)$ exists in a neighborhood of $c$, and that $f''(c)$ exists. Then $f$ has a relative maximum value at $c$ if $f''(c)<0$ and a relative minimum value at $c$ if $f''(c)>0$. If $f''(c)=0$, the test is not informative.

  • $\begingroup$ Thank you, that makes sense. So how do you test it if $f''(x)=0$? Is there a more rigorous test? $\endgroup$ – J-J Jan 13 '16 at 13:49
  • $\begingroup$ Keep going on until you find $f^{(i)}(x_0) \neq 0$. This tells you, essentially, that around that point $f(x) = f(x_0) + \frac{f^{(i)}(x_0)}{i!}(x-x_0)^i + o(x^{i+1})$, which means that if $i$ is odd you have a saddle point (changes sign). If $i$ is even you may have a maximum or a minimum, depending on the sign of $f^{(i)}(x_0)$ (like in the usual criteria). Note that this reasoning applies to the case $i=2$ too. $\endgroup$ – AnalysisStudent0414 Jan 13 '16 at 14:04
  • $\begingroup$ Thanks a lot :) Glad to have a better test. $\endgroup$ – J-J Jan 13 '16 at 15:37

Yes. The concavity test can only be used to determine whether a point is a max or min if $f'(x) = 0.$ In your case of $f(x) = x^{4},$ only one point has $f'(x) = 0,$ namely $x = 0.$ This is also called a critical point. Since no other points in $f(x) = x^{4}$ have $f'(x) = 0,$ you cannot use the concavity test to determine max or min, simply because there is no max or min at those points.

  • $\begingroup$ Are you saying the concavity test only applies when there is more than one stationary point? $\endgroup$ – J-J Jan 13 '16 at 13:44
  • $\begingroup$ It only applies at a stationary point. If a point is not a stationary point, the Second Derivative Test does not apply. $\endgroup$ – K. Jiang Jan 13 '16 at 14:35

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