# Counterexample to second derivative test when f''(x) is not continuously differentiable

When looking over true/false questions on previous midterms, one of my conscientious students said:

"If f is defined on an open interval containing c, f'(c)=0, and f''(c)>0, then c is a local min of f"

was false because one of the hypotheses for the second derivative test (at least in Stewart) is that the second derivative is continuous in a neighborhood of c.

Can anyone think of a counterexample for this statement (in one real variable)? It's apparently been too long since I've taken an analysis class to come up with something clever.

• I don't think continuity is actually required. The first derivative has to exist in a neighborhood for the second derivative to exist. The assumption that $\lim_{h\to 0} \frac{f'(c+h)-f'(c)}{h} > 0$ combined with $f'(c) = 0$ implies that $f'(c+h) > 0$ for $h>0$ and $h$ small enough and $< 0$ for $h < 0$ and $h$ small enough. From here it's a consequence of the mean value theorem. – Chris Janjigian Oct 31 '13 at 20:51
• Ah thanks. I didn't consider that Stewart would just add an unnecessary hypothesis to the statement of a theorem when it doesn't even bother to prove it. I won't feel too bad about not finding a counterexample then! – user16544 Oct 31 '13 at 21:22
• I suppose the hypothesis is an artifact of thinking of the multivariable case, where one wants equality of mixed partials. – user16544 Oct 31 '13 at 21:27

As pointed out in the comments, the fact that $f'(c)=0$ is enough; the continuity of the second derivative is not required for a single variable case.