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What does the second derivative tell us about the shape of a graph?

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 4.88 min answerUpdated 2026-06-02

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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

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.

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 practice questions

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

TCE 20245 marksCalculator-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.

TCE 20233 marksCalculator-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 $y''$ .