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According to google, it's when $f'(x)$ doesn't exist. I was given the following functions:

\begin{align} y & = -\tfrac 1 3x^3 -4x + 16x \\[6pt] y & = xe^{-x/4} \\[6pt] y & = -\cos(x-4) \\[6pt] y & = -x^2 + 8x \end{align}

I was able to automatically rule out the first and last because they're polynomials. Then I was stuck with the middle two. But they both have first derivatives. AND second derivatives. So, is there any other way to determine when I can't use the second derivative test?

Michael Hardy
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Anna a
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    There are critical points where $f'=0$ and critical points where $f'$ DNE. In the latter case $f''$ also DNE, so you can't use the second derivative test. In the former case, you can attempt the second derivative test provided $f''$ exists at the critical point, but it will be inconclusive if $f''=0$. – Ian Feb 02 '21 at 18:00
  • Okay I kind of understand what you mean, but not really. Because I did find the first and second derivatives for all of the functions, and I never got 0. Am I supposed to plug in a value or something? – Anna a Feb 02 '21 at 18:09
  • In doing the second derivative test, you plug in the critical point. In all of these examples, $f'$ and $f''$ exist everywhere, so the only reason the second derivative test would fail would be if $f''$ were zero at some critical point. – Ian Feb 02 '21 at 18:25
  • Oh I'm so stupid I get It now. Thank you so much!! – Anna a Feb 02 '21 at 18:32
  • $y=x^3$ derivative is $y'=3x^2$ which is zero at $x=0$. Second derivative is $y''=6x$ and the test is not valid because $y''=0$ at $x=0$. same for $y=x^4$, for instance – Raffaele Feb 02 '21 at 19:17

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"I was able to automatically rule out the first and last because they're polynomials."

What does that mean? What question were you trying to answer?

What is usually called "the second derivative test" is used for identifying local extreme values – maxima and minima. Many polynomials do have local maxima and local minima. In particular, your fourth example has a local (and in this case, also global) maximum at a point where the first derivative is $0$ and the second derivative is negative.

And the same is true of your second example.

The third example has a global maximum that is attained at infinitely many points in the domain, all of them points where the first derivative is $0$ and the second is negative.

In the first example, I wonder whether there's a typo.

Michael Hardy
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  • I'm sorry, this is just so confusing to me. I was given the list of functions and I was asked to determine with which I am unable to use the second derivative test. The only information relating to that which I could find on google was when the first derivative DNE/ or = 0. I was hoping there are like other methods or ways to check because all the functions have first and second derivatives. And I checked that multiple times so I know they all do. – Anna a Feb 02 '21 at 18:14
  • You can use the second-derivative test at points where the first derivative is equal to $0$ and the second derivative exists and is not $0. \qquad$ – Michael Hardy Feb 02 '21 at 18:16