The second derivative and concavity (SACE Stage 2 Mathematical Methods)

Syllabus dot point

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

8 min readUpdated 2 Jun 2026

Quick answer

The second derivative $f''(x)$ is the derivative of $f'(x)$ and measures how the gradient is changing. Where $f''(x)>0$ the curve is concave up; where $f''(x)<0$ it is concave down. A point of inflection is where concavity changes (so $f''$ changes sign). The second derivative also classifies stationary points via the second derivative test.

Jump to a section

What this dot point is asking
Definition
Points of inflection
The second derivative test
Reading the signs
A second worked classification
Concavity and motion
Why it matters
Exam-style questions

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.

Concavity

A curve is concave up on an interval if $f''(x)>0$ throughout (it curves like a cup, gradient increasing) and concave down if $f''(x)<0$ (it curves like a cap, gradient decreasing).

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.

f''=0

is necessary but not sufficient

$f''(x)=0$ alone does not guarantee an inflection. You must check that $f''$ actually changes sign. For example $f(x)=x^4$ has $f''(0)=0$ but $f''=12x^2\ge 0$ everywhere, so $x=0$ is not an inflection point - it is a minimum.

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

Find and classify all stationary points of $f(x)=x^3-3x$ , and locate any points of inflection.

Step 1: Differentiate and find where $f'(x)=0$

Stationary points occur where the gradient is zero. Differentiating using the power rule:

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

Setting $f'(x)=0$ gives $3x^2-3=0$ , so $x^2=1$ , and therefore $x=1$ or $x=-1$ .

Step 2: Find the second derivative

To classify each stationary point, differentiate $f'(x)$ again:

f''(x)=6x.

The sign of $f''$ at a stationary point tells us the concavity and hence the nature of that point.

Step 3: Classify the stationary point at $x=1$

Substituting $x=1$ into the second derivative: $f''(1)=6>0$ . A positive second derivative means the curve is concave up, so this is a local minimum. The $y$ -coordinate is $f(1)=1-3=-2$ , giving the point $(1,-2)$ .

Step 4: Classify the stationary point at $x=-1$

Substituting $x=-1$ into the second derivative: $f''(-1)=-6<0$ . A negative second derivative means the curve is concave down, so this is a local maximum. The $y$ -coordinate is $f(-1)=-1+3=2$ , giving the point $(-1,2)$ .

Step 5: Locate the point of inflection

A point of inflection is where $f''(x)=0$ and the sign of $f''$ changes. Setting $f''(x)=6x=0$ gives $x=0$ . For $x<0$ , $f''<0$ (concave down); for $x>0$ , $f''>0$ (concave up). The sign changes, confirming an inflection. The $y$ -coordinate is $f(0)=0$ , so $(0,0)$ is a point of inflection.

Final answer: local maximum at $(-1,2)$ , local minimum at $(1,-2)$ , and a point of inflection at $(0,0)$ .

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 $f''$ )

Confusing a stationary point with an inflection

A stationary point needs $f'(x)=0$ . A point of inflection needs $f''(x)=0$ with a sign change. These are different conditions. A curve can have an inflection where the gradient is not zero (a non-stationary inflection), and a stationary point that is a max or min has no inflection there.

A second worked classification

Using the second derivative test on

f(x)=2x^3-9x^2+12x

Find and classify all stationary points of $f(x)=2x^3-9x^2+12x$ , and locate any points of inflection.

Step 1: Find the first derivative and locate the stationary points

Differentiate term by term using the power rule:

f'(x)=6x^2-18x+12.

To solve $f'(x)=0$ , factor out $6$ to simplify the quadratic:

f'(x)=6(x^2-3x+2)=6(x-1)(x-2)=0,

giving stationary points at

x=1

and

x=2

Step 2: Find the second derivative

Differentiate $f'(x)$ to obtain the tool for classifying each stationary point:

f''(x)=12x-18.

The second derivative measures how the gradient is changing; its sign at a stationary point determines whether the curve bends up (minimum) or down (maximum).

Step 3: Classify the stationary point at $x=1$

Compute $f''(1)=12(1)-18=-6<0$ . Because the second derivative is negative, the curve is concave down at $x=1$ , so this is a local maximum. The $y$ -coordinate is $f(1)=2-9+12=5$ , giving the point $(1,5)$ .

Step 4: Classify the stationary point at $x=2$

Compute $f''(2)=12(2)-18=6>0$ . Because the second derivative is positive, the curve is concave up at $x=2$ , so this is a local minimum. The $y$ -coordinate is $f(2)=16-36+24=4$ , giving the point $(2,4)$ .

Step 5: Locate the point of inflection

Solve $f''(x)=0$ to find where concavity might change:

12x-18=0 \implies x=\tfrac{3}{2}.

Check the sign change: for $x<\tfrac{3}{2}$ , $f''<0$ (concave down); for $x>\tfrac{3}{2}$ , $f''>0$ (concave up). The sign does change, confirming an inflection. The $y$ -coordinate is $f\!\left(\tfrac{3}{2}\right)=2\cdot\tfrac{27}{8}-9\cdot\tfrac{9}{4}+12\cdot\tfrac{3}{2}=\tfrac{27}{4}-\tfrac{81}{4}+\tfrac{72}{4}=\tfrac{18}{4}=\tfrac{9}{2}$ , so the point of inflection is $\left(\tfrac{3}{2},\tfrac{9}{2}\right)$ , sitting exactly between the maximum and minimum.

Final answer: local maximum at $(1,5)$ , local minimum at $(2,4)$ , and a point of inflection at $\left(\tfrac{3}{2},\tfrac{9}{2}\right)$ .

Concavity and motion

When $f(t)$ is displacement, $f'(t)$ is velocity and $f''(t)$ is acceleration. Concave up ( $f''>0$ ) means velocity is increasing - the object is accelerating; concave down ( $f''<0$ ) means it is decelerating. The same idea applies to any rate-of-change model: the second derivative tells you whether the rate is itself speeding up or slowing down, which is exactly the language examiners use when asking you to interpret the behaviour of a model.

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 questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2022-style2 marks

Calculator-assumed. For the curve

y = g(x)

0 \le x \le 3

, the point

A

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 also notes that $g''$ changing sign, not merely being zero, is what confirms the inflection.

SACE 2022-style2 marks

Calculator-assumed. For the same curve

y = g(x)

, the point

B

is at

x = 2

, beyond the only inflection point at

x = 1

, where the curve is concave down. 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 $x > 1$ . The curve is concave down for $x > 1$ , 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.

SACE 2021-style3 marks

Calculator-assumed. For

f(x) = x^3 - 6x^2 + 5

, find

f''(x)

, hence find the coordinates of the point of inflection, and state the interval on which the curve is concave up.

Show worked answer →

Differentiate twice. $f'(x) = 3x^2 - 12x$ and $f''(x) = 6x - 12$ .

Set $f''(x) = 0$ : $6x - 12 = 0$ gives $x = 2$ . The second derivative changes from negative (for $x < 2$ ) to positive (for $x > 2$ ), confirming an inflection. Then $f(2) = 8 - 24 + 5 = -11$ , so the point of inflection is $(2, -11)$ .

The curve is concave up where $f''(x) > 0$ , that is for $x > 2$ .

Marks: one for $f''(x) = 6x - 12$ , one for the inflection point $(2, -11)$ with a sign-change check, and one for the concave-up interval $x > 2$ .

Practise this

Sources & how we know this

Stage 2 Mathematical Methods subject outline (2025 onwards) — SACE Board
Stage 2 Mathematical Methods: external assessment — SACE Board

What this dot point is asking

Definition

Points of inflection

The second derivative test

Step 1: Differentiate and find where f′(x)=0f'(x)=0f′(x)=0

Step 2: Find the second derivative

Step 3: Classify the stationary point at x=1x=1x=1

Step 4: Classify the stationary point at x=−1x=-1x=−1

Step 5: Locate the point of inflection

Reading the signs

A second worked classification

Step 1: Find the first derivative and locate the stationary points

Step 2: Find the second derivative

Step 3: Classify the stationary point at x=1x=1x=1

Step 4: Classify the stationary point at x=2x=2x=2

Step 5: Locate the point of inflection

Concavity and motion

Why it matters

Exam-style questions

Practise this

Sources & how we know this

Step 1: Differentiate and find where $f'(x)=0$

Step 3: Classify the stationary point at $x=1$

Step 4: Classify the stationary point at $x=-1$

Step 3: Classify the stationary point at $x=1$

Step 4: Classify the stationary point at $x=2$