Second Derivative Test

The Second Derivative Test determines whether a critical point is a local maximum or minimum using concavity. If $f'(c)=0$ and $f''(c)>0$ , then $c$ is a local minimum; if $f''(c)<0$ , it's a local maximum. When $f''(c)=0$ , the test is inconclusive.

The Second Derivative Test

The Second Derivative Test is a quick way to classify critical points using concavity information. It leverages the fact that $f''$ measures how the slope of $f$ is changing—in other words, the curve's concavity.

Understanding Concavity

Concave up ( $f''(x) > 0$ ): The curve shapes like a cup; tangent lines lie below it
Concave down ( $f''(x) < 0$ ): The curve shapes like a cap; tangent lines lie above it

When $f''(x)$ changes sign, the curve experiences an inflection point—a transition between concavities. Note that $f''(c)=0$ is necessary but not sufficient; you must verify the sign actually changes (e.g., $f(x)=x^4$ has $f''(0)=0$ but is concave up everywhere).

The Test

Suppose $f'(c) = 0$ (so $c$ is a critical point). Then:

Condition	Conclusion
$f''(c) > 0$	Local minimum at $c$
$f''(c) < 0$	Local maximum at $c$
$f''(c) = 0$	Test inconclusive

The intuition is geometric: if the tangent line is flat ( $f'(c)=0$ ) and the curve bends upward ( $f''(c)>0$ ), you're at the bottom of a cup—a minimum. Similarly, downward curvature with a flat tangent signals a maximum.

Example

For $f(x) = x^3 - 3x^2$ : $f'(x) = 3x(x-2)$ gives critical points $x=0$ and $x=2$ .

$f''(x) = 6x - 6$
$f''(0) = -6 < 0$ → local maximum at $(0, 0)$
$f''(2) = 6 > 0$ → local minimum at $(2, -4)$

Setting $f''(x)=0$ yields $x=1$ , where concavity switches from down to up—an inflection point at $(1, -2)$ .

Graph illustrating Second Derivative Test

When It Fails

If $f''(c)=0$ at a critical point, the test provides no information. Use the First Derivative Test instead by checking the sign of $f'$ on either side of $c$ .