Given the following function:

$f(x) = \ln(x^4 + 27)$

I found $f'(x) =$ $\frac{4x^3}{x^4+27}$

I set the $f'(x) = 0$ and found that $x = 0$ The interval would thus be: $(-\infty, 0), (0, \infty)$

Next I'm told to find the following:

Find the local minimum values. Find the local maximum values. Find the inflection points. Find when the graph is concave upward. Find when the graph is concave downward.

So I know that a local minimum occurs when $f'' > 0$ and a local maximum occurs when $f'' < 0$. So given my function, I don't have a local maximum, but I do have a local minimum. My guess is that it occurs at zero since $f'$ changes from negative to positive. Am I wrong?


You are not wrong. It is correct that $f(x)$ will have a local minimum at $x=0$ since $f'$ changes from negative to positive at $x=0$.

  • $\begingroup$ The confusion stems from the fact that $f'$ has saddle point in $x = 0$, therefore $f''(0) = 0$. Still the sign-change-argument you presented holds. $\endgroup$ – Viktor Glombik Dec 24 '18 at 0:40

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