How Do You Use the Second Derivative Test to Classify a Maximum or Minimum?

Use the second derivative test at a stationary point, where $f'(x)=0$. If $f''(x)<0$, the point is a local maximum; if $f''(x)>0$, it is a local minimum.

This is part of AHL 5.10: Second derivatives, testing for maximum and minimum, so it is Mathematics: Applications and Interpretation HL content.

The Mechanism

The first derivative gives the gradient. Solving $f'(x)=0$ identifies points where the tangent is horizontal, but this alone does not show whether each point is a maximum or minimum.

The second derivative describes the graph's concavity:

Why does the sign work? Near a stationary point, a positive second derivative means the gradient is increasing through zero, so the graph changes from decreasing to increasing. A negative second derivative means the gradient is decreasing through zero, so the graph changes from increasing to decreasing.

For example, let

$$f(x)=x^3-3x^2+2.$$

First find the stationary points:

$$f'(x)=3x^2-6x=3x(x-2).$$

Therefore, $f'(x)=0$ when $x=0$ or $x=2$. Now differentiate again:

$$f''(x)=6x-6.$$

At $x=0$, $f''(0)=-6<0$, so $(0,2)$ is a local maximum. At , , so is a local minimum.

The coordinate must include the function value, not only the x-value. Substituting each stationary x-value into $f$ gives the y-coordinate, which completes the classification.

A common misconception is that every solution of $f'(x)=0$ must be a maximum or minimum. This is false: a stationary point can also be a stationary point of inflexion.

Exam Technique

Show all three stages: solve $f'(x)=0$, evaluate (f''(x)) at each solution, and state the classification with coordinates. If the question asks for a global maximum or minimum on a closed interval, also check the endpoints; the second derivative test classifies local stationary points only.