The second derivative measures the rate of change of the gradient and determines concavity and points of inflection.

How the second derivative determines whether a curve is concave up or down, locates points of inflection, and classifies stationary points.

Generated by Claude Opus 4.77 min answerUpdated 2026-05-29

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

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What this dot point is asking
Definition
Points of inflection
The second derivative test
Reading the signs
Why it matters

What this dot point is asking

You need to find and interpret $f''(x)$ , use its sign to describe concavity, locate points of inflection, and apply the second derivative test to stationary points.

Definition

The second derivative is found by differentiating the first derivative again:

f''(x)=\frac{d}{dx}\big[f'(x)\big],\qquad\text{also written}\quad \frac{d^2y}{dx^2}.

If $f'(x)$ is the rate of change of $y$ , then $f''(x)$ is the rate of change of the gradient.

Points of inflection

A point of inflection is a point where the curve changes concavity - from concave up to concave down or vice versa. At such a point $f''(x)=0$ and $f''$ changes sign across it.

The second derivative test

For a stationary point at $x=a$ (where $f'(a)=0$ ):

if $f''(a)>0$ , the point is a local minimum (concave up);
if $f''(a)<0$ , the point is a local maximum (concave down);
if $f''(a)=0$ , the test is inconclusive - use a sign diagram of $f'$ instead.

Classifying stationary points of $f(x)=x^3-3x$

Worked example

Find and classify the stationary points of $f(x)=x^3-3x$ .

First derivative: $f'(x)=3x^2-3$ . Setting $f'(x)=0$ gives $x^2=1$ , so $x=\pm 1$ .

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

At $x=1$ : $f''(1)=6>0$ , so a local minimum. $f(1)=1-3=-2$ , point $(1,-2)$ .

At $x=-1$ : $f''(-1)=-6<0$ , so a local maximum. $f(-1)=-1+3=2$ , point $(-1,2)$ .

Inflection: $f''(x)=6x=0$ at $x=0$ , and $f''$ changes from negative to positive there, so $(0,0)$ is a point of inflection.

Reading the signs

A small table helps connect everything:

$f'(x)$	$f''(x)$	Behaviour
$0$	$>0$	local minimum
$0$	$<0$	local maximum
any	$>0$	concave up
any	$<0$	concave down
$0$ (changes sign)	-	inflection (in $f''$ )

Why it matters

Concavity completes the shape information you need for accurate curve sketching: the first derivative gives where a curve rises and falls and locates stationary points, while the second derivative tells you how it bends and where it switches. In applications, $f''$ corresponds to acceleration (when $f$ is displacement) and to whether a rate is itself speeding up or slowing down.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2017 SACE Stage 22 marksFor the curve y = g(x) on 0 <= x <= 3, point A (at x = 1) is the only point of inflection. State whether g''(1) is negative, zero, or positive, and justify your answer.

Show worked answer →

The correct choice is g''(1) = 0.

Justification: a point of inflection is where the curve changes concavity (from concave down to concave up, or vice versa). At such a point the second derivative changes sign, and so it must pass through zero. Because A at x = 1 is stated to be a point of inflection, g''(1) = 0.

Marks: one for selecting g''(1) = 0, and one for the justification that concavity changes sign at an inflection point so the second derivative is zero there. (A full answer would also note that the second derivative changing sign, not merely being zero, is what confirms the inflection.)

2017 SACE Stage 22 marksFor the same curve y = g(x), point B is at x = 2, beyond the only inflection point at x = 1. State whether g''(2) is negative, zero, or positive, and justify your answer.

Show worked answer →

The curve has a single inflection point at x = 1, so the concavity is constant on the interval x > 1. From the shape of the graph the curve is concave down for x > 1 (it bends downwards as it approaches the maximum near B), which means g''(2) < 0.

Justification: concave down corresponds to a negative second derivative (the gradient is decreasing). Since x = 2 lies in the concave-down section beyond the inflection point, g''(2) < 0.

Marks: one for selecting g''(2) < 0 and one for justifying it from the concavity of the curve at that point. Reading concavity correctly off the diagram is the skill being tested.