The second derivative, concavity and points of inflection (TCE Mathematics Methods, Tasmania)

Syllabus dot point

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

8 min readUpdated 2 Jun 2026

Quick answer

The second derivative $f''(x)$ is the rate of change of the gradient. 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, requiring $f''(x)=0$ and a genuine sign change. The second derivative test then classifies a stationary point as a maximum or minimum.

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

Concave up and concave down

A curve is concave up on an interval if it bends upward like a cup, so the gradient is increasing and $f''(x)>0$ . It is concave down if it bends downward like a cap, so the gradient is decreasing and $f''(x)<0$ .

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.

Locating a point of inflection

At a point of inflection $f''(x)=0$ , but that alone is not enough. You must also confirm that $f''$ changes sign across the point. If $f''$ touches zero without changing sign, the concavity does not switch and it is not an inflection point.

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.

The second derivative test

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

if $f''(a)>0$ , the point is a local minimum;

if $f''(a)<0$ , the point is a local maximum;

if $f''(a)=0$ , the test is inconclusive and you must use the sign of $f'$ on each side.

Worked example

Concavity, inflection and nature of stationary points

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

Step 1: Find the stationary points by solving $f'(x) = 0$

Differentiate once and factor the result. Stationary points are where the gradient is zero:

f'(x) = 3x^2 - 6x = 3x(x-2) = 0 \quad\Rightarrow\quad x = 0 \text{ and } x = 2.

The $y$ -values are $f(0) = 0$ and $f(2) = 8 - 12 = -4$ , giving stationary points $(0, 0)$ and $(2, -4)$ .

Step 2: Compute the second derivative

Differentiate $f'$ to get $f''$ . Its sign at each stationary point will tell us whether the curve is concave up (minimum) or concave down (maximum) there:

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

Step 3: Classify each stationary point using $f''$

Substitute each stationary $x$ -value into $f''$ . A positive value means the curve bends upward like a bowl (local minimum); a negative value means it bends downward like a cap (local maximum):

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.

Step 4: Locate the point of inflection

A point of inflection is where $f''(x) = 0$ and the sign of $f''$ changes. Set $f''(x) = 0$ to find the candidate, then check the sign on each side:

6x - 6 = 0 \quad\Rightarrow\quad x = 1.

Since $f''(0) = -6 < 0$ and $f''(2) = 6 > 0$ , the concavity changes from down to up at $x = 1$ . The $y$ -value is $f(1) = 1 - 3 = -2$ , giving a point of inflection at $(1, -2)$ .

Final answer: Local maximum at $(0, 0)$ , local minimum at $(2, -4)$ , and point of inflection at $(1, -2)$ .

Choosing between the two classification tests

You have two ways to classify a stationary point, and knowing when to use each saves time. The second derivative test is fastest when $f''$ is easy to evaluate, such as for polynomials: just substitute the stationary $x$ -value into $f''$ . The first derivative sign test, where you check the sign of $f'$ just to the left and right of the point, is the fallback when $f''(a) = 0$ (inconclusive) or when $f''$ is awkward to compute. For a stationary point of inflection, the second derivative test is inconclusive by design, so you must use the first derivative sign test, which will show $f'$ keeping the same sign on both sides.

A stationary point of inflection

Problem

Show that $f(x) = x^{3}$ has a stationary point of inflection at the origin.

Stationary point

$f'(x) = 3x^{2}$ , which is zero at $x = 0$ , so $(0, 0)$ is a stationary point.

Second derivative test is inconclusive

$f''(x) = 6x$ , so $f''(0) = 0$ . The test gives no decision, so switch to the first derivative sign test.

First derivative sign test

$f'(x) = 3x^{2} \ge 0$ for all $x$ , so the curve is increasing on both sides of $x = 0$ . The gradient does not change sign, so the point is neither a maximum nor a minimum. Since $f''$ changes sign (negative for $x < 0$ , positive for $x > 0$ ), the concavity switches, confirming a stationary point of inflection at the origin.

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.

Assuming

f''(x)=0

always gives an inflection

Setting $f''(x)=0$ only finds candidates. Consider $f(x)=x^4$ : here $f''(x)=12x^2$ , which is zero at $x=0$ , yet $f''$ stays positive on both sides, so the curve is concave up throughout and $x=0$ is a minimum, not an inflection. Always verify the sign change of $f''$ before declaring a point of inflection.

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.

Exam-style questions

Questions in the style of TASC 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.

Practice5 marks

Calculator-free. Consider

f(x) = x^{3} - 3x^{2} - 9x + 5

. (a) Find the coordinates of the stationary points and classify each using the second derivative test. (b) Find the coordinates of the point of inflection.

Show worked answer →

(a) Differentiate: $f'(x) = 3x^{2} - 6x - 9 = 3(x - 3)(x + 1)$ . Stationary points at $x = 3$ and $x = -1$ .
$f(3) = 27 - 27 - 27 + 5 = -22$ and $f(-1) = -1 - 3 + 9 + 5 = 10$ .
Second derivative: $f''(x) = 6x - 6$ .
At $x = 3$ : $f''(3) = 12 > 0$ , so $(3, -22)$ is a local minimum.
At $x = -1$ : $f''(-1) = -12 < 0$ , so $(-1, 10)$ is a local maximum.

(b) Set $f''(x) = 0$ : $6x - 6 = 0$ , so $x = 1$ . The sign of $f''$ changes from negative to positive there, confirming an inflection. $f(1) = 1 - 3 - 9 + 5 = -6$ , so the point of inflection is $(1, -6)$ . One mark per stationary point with classification, one for setting $f'' = 0$ , one for the verified inflection coordinates.

Practice3 marks

Calculator-assumed. The curve

y = 2x^{3} + bx^{2} + 4

has a point of inflection at

x = 2

. (a) Find the value of

b

. (b) State the concavity of the curve for

x < 2

Show worked answer →

(a) Differentiate twice: $y' = 6x^{2} + 2bx$ and $y'' = 12x + 2b$ . A point of inflection requires $y'' = 0$ at $x = 2$ :

12(2) + 2b = 0 \quad\Rightarrow\quad 24 + 2b = 0 \quad\Rightarrow\quad b = -12.

(b) With $b = -12$ , $y'' = 12x - 24 = 12(x - 2)$ . For $x < 2$ , $y'' < 0$ , so the curve is concave down. Markers reward setting $y'' = 0$ at the given point, solving for $b$ , and the correct concavity from the sign of $y''$ .

Practise this

Sources & how we know this

Mathematics Methods Level 4 (MTM415117) course document (2017 onwards) — TASC

What this dot point is asking

Concavity

Points of inflection

The second derivative test

Worked example

Step 1: Find the stationary points by solving f′(x)=0f'(x) = 0f′(x)=0

Step 2: Compute the second derivative

Step 3: Classify each stationary point using f′′f''f′′

Step 4: Locate the point of inflection

Choosing between the two classification tests

Problem

Stationary point

Second derivative test is inconclusive

First derivative sign test

Acceleration as a second derivative

Summary

Exam-style questions

Practise this

Sources & how we know this

Step 1: Find the stationary points by solving $f'(x) = 0$

Step 3: Classify each stationary point using $f''$