AP Calculus AB and BC

When the Second Derivative Test Is Inconclusive

The Second Derivative Test needs the second derivative to be nonzero at the critical point. When it is zero the test is silent, not negative: the point may be a maximum, a minimum, or neither, and you have to fall back on the First Derivative Test.

Second Derivative Test

f'(c) = 0 \text{ and } f''(c) > 0 \implies \text{local minimum at } c

At a critical point where the second derivative is positive there is a local minimum, and where it is negative there is a local maximum.

The hypotheses, and what each one buys

A theorem is only as strong as its conditions. Below, each hypothesis is dropped on its own while every other one is kept, so you can see precisely what it was holding up.

1
The second derivative at c is nonzero
$f''(c) \ne 0$
The sign of $f''(c)$ is the entire content of the test. At zero there is no sign to read, and the concavity can do anything nearby. The test does not return a negative verdict in that case; it returns no verdict at all, which is a different thing and the source of most lost points.
Drop it and the theorem fails
$f(x) = x^{3}, \qquad f'(0) = f''(0) = 0$
f(x) = x cubed at the origin, where f'' is also zero
Here $f'(0) = 0$ and $f''(0) = 0$ , so the test has nothing to say. And it is right to say nothing: the check confirms the origin is not an extremum at all. Compare $f(x) = x^4$ , where $f''(0) = 0$ as well and the origin IS a minimum. Same second derivative reading, opposite answers, which is exactly why the test has to stay silent.
2
c is a critical point, so f'(c) = 0
$f'(c) = 0$
The test only classifies points that are already candidates. Concavity at a non-critical point tells you about the shape of the graph, never about an extremum, because the graph is still going somewhere. Applying the test to a point where the first derivative is nonzero is a category error rather than a wrong answer, so no counterexample is displayed.
No counterexample built from an elementary formula breaks this one on its own, so none is claimed here.

The counterexample above is checked numerically on every build: each function is evaluated and the conclusion is confirmed to fail. A counterexample that stopped working would fail the build rather than sit here misleading you.

Why it is true

Suppose $f'(c) = 0$ and $f''(c) > 0$ . By definition $f''(c)$ is the limit of $\frac{f'(c+h) - f'(c)}{h} = \frac{f'(c+h)}{h}$ .
A positive limit forces the quotient to be positive for all small $h$ , so $f'(c+h)$ has the same sign as $h$ .
That means $f'$ is negative just left of $c$ and positive just right of it, so $f$ decreases into $c$ and increases out of it.
That is a local minimum. When $f''(c) = 0$ the limit gives no sign information, and the argument produces nothing, which is precisely why the test is inconclusive there.

What it does not say

Inconclusive means it is not an extremum.

It means the test cannot tell. $x^4$ at 0 has a zero second derivative and is a genuine minimum. $x^3$ at 0 has a zero second derivative and is not an extremum. The test cannot distinguish them, so it declines to.

The Second Derivative Test always works when the First Derivative Test does.

The First Derivative Test is strictly more general. It needs only a sign change in $f'$ , which exists at corners and at points where $f''$ does not exist at all.

f''(c) = 0 means there is an inflection point at c.

Only if the concavity actually changes sign there. $f(x) = x^4$ has $f''(0) = 0$ and stays concave up on both sides, so the origin is not an inflection point.

Frequently asked questions

What do I do when the Second Derivative Test is inconclusive?

Fall back on the First Derivative Test: examine the sign of $f'$ on both sides of the critical point. A change from positive to negative is a local maximum, negative to positive is a local minimum, and no change is neither.

Why is the Second Derivative Test taught if it can fail?

Because when it applies it is faster, needing one evaluation rather than a sign analysis on both sides. It is a shortcut with a stated precondition, and checking the precondition is part of using it.