Use the second derivative to determine concavity, locate points of inflection, and apply the second derivative test.

How the sign of the second derivative gives concavity, how to find and confirm points of inflection, and how the second derivative test classifies stationary points for TCE Mathematics Methods Unit 3.

Generated by Claude Opus 4.78 min answerUpdated 2026-05-30

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

What this dot point is asking

The first derivative tells you whether a curve is rising or falling. The second derivative, found by differentiating again, tells you about its curvature, which is the missing piece you need to describe a graph fully.

Concavity

A helpful image: on a concave up section the tangent lines lie below the curve, and on a concave down section they lie above it. Because $f''$ measures how the gradient changes, its sign is exactly the test for which way the curve bends.

Points of inflection

A point of inflection is where the concavity switches from up to down or from down to up.

A point of inflection need not be stationary. A stationary point of inflection has $f'(x)=0$ as well (a flat spot where the curve levels off then continues in the same direction), while a non-stationary inflection has a nonzero gradient.

The second derivative test

Once you have found a stationary point by solving $f'(x)=0$ , the second derivative gives a quick classification.

Worked example

Concavity, inflection and nature of stationary points

Problem

For $f(x) = x^3 - 3x^2$ , find the stationary points and classify them, and find the point of inflection.

Solution

Differentiate: $f'(x) = 3x^2 - 6x = 3x(x-2)$ , so stationary points are at $x=0$ and $x=2$ .

Second derivative: $f''(x) = 6x - 6$ .

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

At $x=2$ : $f''(2) = 6 > 0$ , so $(2,-4)$ is a local minimum.

For the inflection, set $f''(x)=0$ : $6x - 6 = 0$ gives $x=1$ . Since $f''$ is negative for $x<1$ and positive for $x>1$ , the concavity changes, confirming a point of inflection at $(1, -2)$ .

Acceleration as a second derivative

In motion problems the second derivative has a physical meaning. If $x(t)$ is position, then $x'(t)$ is velocity and $x''(t)$ is acceleration. Positive acceleration (concave up displacement graph) means the velocity is increasing, and negative acceleration means it is decreasing.

Summary

Differentiate twice. The sign of $f''$ gives concavity: positive is concave up, negative is concave down. Points of inflection occur where $f''=0$ and the sign genuinely changes. For a stationary point, $f''>0$ means a minimum and $f''<0$ means a maximum, falling back to a first-derivative sign test only when $f''=0$ . These curvature facts let you sketch a curve completely and underpin acceleration in kinematics.