1.7.6 Convex, concave and points of inflection (A-level only)

What you'll learn

How the second derivative describes the shape of a curve.
How to identify intervals where a curve is convex or concave.
How to find and verify points of inflection.
Why solving $f^{''} (x) = 0$ is not enough on its own.

Prerequisite: the first and second derivatives

For a function $y = f (x)$ , the first derivative $f^{'} (x)$ gives the gradient of the curve at each value of $x$ .

The second derivative is obtained by differentiating again:

f′′(x)=ddx(f′(x))f''(x)=\frac{\mathrm{d}}{\mathrm{d}x}\left(f'(x)\right)

It tells you how the gradient itself is changing.

If $f^{''} (x) > 0$ , the gradients are increasing as you move from left to right.
If $f^{''} (x) < 0$ , the gradients are decreasing as you move from left to right.
If $f^{''} (x) = 0$ , the gradient is momentarily not changing. This gives a possible boundary between different types of curvature.

Key Idea

Meaning of the second derivative

The sign of $f^{''} (x)$ describes how the curve bends, because it tells you whether the tangent gradients are increasing or decreasing.

Example

Calculating a second derivative

Find the first and second derivatives of $f(x)=2x^4-5x^3+3x$ .

Differentiate each term to find the gradient function:
$f'(x)=8x^3-15x^2+3$
Differentiate the gradient function:
$f''(x)=24x^2-30x$
The expression $24x^2-30x$ can now be used to determine where the graph is convex or concave.

Convex curves

Definition

Convex

A section of a curve is convex when its tangent gradients increase as $x$ increases. For a twice-differentiable function, this occurs where $f^{''} (x) > 0$ .

A convex curve bends upwards, like a bowl. The gradients may still be negative, but they must be becoming more positive.

For example, on the left side of $y=x^2$ , the gradient changes from a large negative value towards zero. On the right side, it changes from zero to increasingly positive values. In both cases, the gradient is increasing.

Concave curves

Definition

Concave

A section of a curve is concave when its tangent gradients decrease as $x$ increases. For a twice-differentiable function, this occurs where $f^{''} (x) < 0$ .

A concave curve bends downwards, like an upside-down bowl. The gradients might be positive or negative; what matters is that they are decreasing.

Graphs showing convex and concave curves through changing tangent gradients, and a cubic curve changing concavity at a point of inflection

Tip

A quick visual check

Convex means “bowl-shaped” and $f^{''} (x) > 0$ ; concave means “cap-shaped” and $f^{''} (x) < 0$ .

Example

Finding convex and concave intervals

Determine where $f(x)=x^3-3x^2-9x$ is convex and where it is concave.

Differentiate twice:
$f′(x)=3x2−6x−9,f′′(x)=6x−6.\begin{aligned} f'(x)&=3x^2-6x-9,\\ f''(x)&=6x-6. \end{aligned}$
Find where the second derivative could change sign:
$\Rightarrow x=1$
For $x < 1$ , choose a test value such as $x = 0$ :
$f^{''} (0) = - 6 < 0$
Therefore, the curve is concave when $x < 1$ .
For $x > 1$ , choose a test value such as $x = 2$ :
$f^{''} (2) = 6 > 0$
Therefore, the curve is convex when $x > 1$ .

Common Mistake

Confusing gradient with curvature

The sign of $f^{'} (x)$ tells you whether the function is increasing or decreasing. The sign of $f^{''} (x)$ tells you whether the gradient is increasing or decreasing, and hence whether the curve is convex or concave.

Points of inflection

Definition

Point of inflection

A point of inflection is a point on a curve where the concavity changes: from convex to concave, or from concave to convex.

If $f^{''} (x)$ is continuous near the point, a point of inflection is usually found by:

solving $f^{''} (x) = 0$ to obtain candidate values;
checking the sign of $f^{''} (x)$ on either side;
confirming that the sign changes;
substituting into $f (x)$ to find the coordinate.

The direction of the sign change tells you how the curvature changes:

Negative to positive: concave to convex.
Positive to negative: convex to concave.

Key Idea

The essential test

A point is an inflection point only if the concavity changes there. The equation $f^{''} (x) = 0$ identifies candidates, not guaranteed points of inflection.

Example

Finding two points of inflection

Find the points of inflection of $f(x)=x^4-4x^3$ and state the concavity on each interval.

Differentiate twice:
$f′(x)=4x3−12x2,f′′(x)=12x2−24x=12x(x−2).\begin{aligned} f'(x)&=4x^3-12x^2,\\ f''(x)&=12x^2-24x\\ &=12x(x-2). \end{aligned}$
Find the candidate values by solving $f^{''} (x) = 0$ :
$12 x (x - 2) = 0$
Therefore, $x = 0$ or $x = 2$ .
Check the sign of $f^{''} (x)$ in the three intervals:
- For $x < 0$ , use $x = - 1$ : $f^{''} (- 1) = 36 > 0$ , so the curve is convex.
- For $0 < x < 2$ , use $x = 1$ : $f^{''} (1) = - 12 < 0$ , so the curve is concave.
- For $x > 2$ , use $x = 3$ : $f^{''} (3) = 36 > 0$ , so the curve is convex.
The sign changes at both candidate values:
- At $x = 0$ , the sign changes from positive to negative.
- At $x = 2$ , the sign changes from negative to positive.
Both values therefore give points of inflection.
Find the corresponding coordinates:
$f (0) = 0$
and
$f(2)=2^4-4(2^3)=-16.$
The points of inflection are $(0, 0)$ and $(2, - 16)$ .

When $f^{''} (x) = 0$ is not enough

Consider $f(x)=x^4$ . Its second derivative is

f''(x)=12x^2.

Solving $f^{''} (x) = 0$ gives $x = 0$ . However, $12x^2$ is positive on both sides of zero, so the curve remains convex.

Therefore, the origin is not a point of inflection.

Common Mistake

Stopping after solving the equation

Do not conclude that every solution of $f^{''} (x) = 0$ is a point of inflection. Check that $f^{''} (x)$ changes sign across the candidate value.

Common Mistake

Second derivative may be undefined

In some functions, concavity can change at a point where $f^{''} (x)$ does not exist. The definition is based on a change in concavity, not solely on the equation $f^{''} (x) = 0$ .

Stationary and non-stationary inflection points

A stationary point is a point where $f^{'} (x) = 0$ , so its tangent is horizontal.

A point of inflection may be either stationary or non-stationary:

A stationary point of inflection has changing concavity and $f^{'} (x) = 0$ .
A non-stationary point of inflection has changing concavity but $f′(x)≠0f'(x)\neq0$ .

Example

Classifying an inflection point

Classify the point of inflection of $f(x)=x^3+x$ .

Differentiate twice:
$f′(x)=3x2+1,f′′(x)=6x.f'(x)=3x^2+1,\qquad f''(x)=6x.$
Solve $f^{''} (x) = 0$ , giving $x = 0$ . Since $6 x$ is negative for $x < 0$ and positive for $x > 0$ , the concavity changes and there is a point of inflection.
Evaluate the first derivative:
$f′(0)=1≠0.f'(0)=1\neq0.$
The tangent is not horizontal, so this is a non-stationary point of inflection at $(0, 0)$ .

Exam technique

In the exam

Differentiate twice and solve $f^{''} (x) = 0$ to find candidate values.
Test the sign of $f^{''} (x)$ on both sides of every candidate; show these tests clearly.
State the convex and concave intervals using inequalities, then substitute into $f (x)$ if point coordinates are required.
If asked to classify an inflection point, also calculate $f^{'} (x)$ to decide whether it is stationary.

Self review

Check yourself

What does the sign of $f^{''} (x)$ tell you about the tangent gradients and the shape of the curve?
Why does $f^{''} (a) = 0$ not necessarily mean that $x = a$ gives a point of inflection?
How would you distinguish a stationary point of inflection from a non-stationary one?

What you'll learn

How the second derivative describes the shape of a curve.
How to identify intervals where a curve is convex or concave.
How to find and verify points of inflection.
Why solving $f^{''} (x) = 0$ is not enough on its own.

Prerequisite: the first and second derivatives

For a function $y = f (x)$ , the first derivative $f^{'} (x)$ gives the gradient of the curve at each value of $x$ .

The second derivative is obtained by differentiating again:

f′′(x)=ddx(f′(x))f''(x)=\frac{\mathrm{d}}{\mathrm{d}x}\left(f'(x)\right)

It tells you how the gradient itself is changing.

If $f^{''} (x) > 0$ , the gradients are increasing as you move from left to right.
If $f^{''} (x) < 0$ , the gradients are decreasing as you move from left to right.
If $f^{''} (x) = 0$ , the gradient is momentarily not changing. This gives a possible boundary between different types of curvature.

Key Idea

Meaning of the second derivative

The sign of $f^{''} (x)$ describes how the curve bends, because it tells you whether the tangent gradients are increasing or decreasing.

Example

Calculating a second derivative

Find the first and second derivatives of $f(x)=2x^4-5x^3+3x$ .

Differentiate each term to find the gradient function:
$f'(x)=8x^3-15x^2+3$
Differentiate the gradient function:
$f''(x)=24x^2-30x$
The expression $24x^2-30x$ can now be used to determine where the graph is convex or concave.

Convex curves

Definition

Convex

A section of a curve is convex when its tangent gradients increase as $x$ increases. For a twice-differentiable function, this occurs where $f^{''} (x) > 0$ .

A convex curve bends upwards, like a bowl. The gradients may still be negative, but they must be becoming more positive.

Concave curves

Definition

Concave

A section of a curve is concave when its tangent gradients decrease as $x$ increases. For a twice-differentiable function, this occurs where $f^{''} (x) < 0$ .

A concave curve bends downwards, like an upside-down bowl. The gradients might be positive or negative; what matters is that they are decreasing.

Graphs showing convex and concave curves through changing tangent gradients, and a cubic curve changing concavity at a point of inflection

Tip

A quick visual check

Convex means “bowl-shaped” and $f^{''} (x) > 0$ ; concave means “cap-shaped” and $f^{''} (x) < 0$ .

Example

Finding convex and concave intervals

Determine where $f(x)=x^3-3x^2-9x$ is convex and where it is concave.

Differentiate twice:
$f′(x)=3x2−6x−9,f′′(x)=6x−6.\begin{aligned} f'(x)&=3x^2-6x-9,\\ f''(x)&=6x-6. \end{aligned}$
Find where the second derivative could change sign:
$\Rightarrow x=1$
For $x < 1$ , choose a test value such as $x = 0$ :
$f^{''} (0) = - 6 < 0$
Therefore, the curve is concave when $x < 1$ .
For $x > 1$ , choose a test value such as $x = 2$ :
$f^{''} (2) = 6 > 0$
Therefore, the curve is convex when $x > 1$ .

Common Mistake

Confusing gradient with curvature

Points of inflection

Definition

Point of inflection

A point of inflection is a point on a curve where the concavity changes: from convex to concave, or from concave to convex.

If $f^{''} (x)$ is continuous near the point, a point of inflection is usually found by:

solving $f^{''} (x) = 0$ to obtain candidate values;
checking the sign of $f^{''} (x)$ on either side;
confirming that the sign changes;
substituting into $f (x)$ to find the coordinate.

The direction of the sign change tells you how the curvature changes:

Negative to positive: concave to convex.
Positive to negative: convex to concave.

Key Idea

The essential test

A point is an inflection point only if the concavity changes there. The equation $f^{''} (x) = 0$ identifies candidates, not guaranteed points of inflection.

Example

Finding two points of inflection

Find the points of inflection of $f(x)=x^4-4x^3$ and state the concavity on each interval.

Differentiate twice:
$f′(x)=4x3−12x2,f′′(x)=12x2−24x=12x(x−2).\begin{aligned} f'(x)&=4x^3-12x^2,\\ f''(x)&=12x^2-24x\\ &=12x(x-2). \end{aligned}$
Find the candidate values by solving $f^{''} (x) = 0$ :
$12 x (x - 2) = 0$
Therefore, $x = 0$ or $x = 2$ .
Check the sign of $f^{''} (x)$ in the three intervals:
- For $x < 0$ , use $x = - 1$ : $f^{''} (- 1) = 36 > 0$ , so the curve is convex.
- For $0 < x < 2$ , use $x = 1$ : $f^{''} (1) = - 12 < 0$ , so the curve is concave.
- For $x > 2$ , use $x = 3$ : $f^{''} (3) = 36 > 0$ , so the curve is convex.
The sign changes at both candidate values:
- At $x = 0$ , the sign changes from positive to negative.
- At $x = 2$ , the sign changes from negative to positive.
Both values therefore give points of inflection.
Find the corresponding coordinates:
$f (0) = 0$
and
$f(2)=2^4-4(2^3)=-16.$
The points of inflection are $(0, 0)$ and $(2, - 16)$ .

When $f^{''} (x) = 0$ is not enough

Consider $f(x)=x^4$ . Its second derivative is

f''(x)=12x^2.

Solving $f^{''} (x) = 0$ gives $x = 0$ . However, $12x^2$ is positive on both sides of zero, so the curve remains convex.

Therefore, the origin is not a point of inflection.

Common Mistake

Stopping after solving the equation

Do not conclude that every solution of $f^{''} (x) = 0$ is a point of inflection. Check that $f^{''} (x)$ changes sign across the candidate value.

Common Mistake

Second derivative may be undefined

In some functions, concavity can change at a point where $f^{''} (x)$ does not exist. The definition is based on a change in concavity, not solely on the equation $f^{''} (x) = 0$ .

Stationary and non-stationary inflection points

A stationary point is a point where $f^{'} (x) = 0$ , so its tangent is horizontal.

A point of inflection may be either stationary or non-stationary:

A stationary point of inflection has changing concavity and $f^{'} (x) = 0$ .
A non-stationary point of inflection has changing concavity but $f′(x)≠0f'(x)\neq0$ .

Example

Classifying an inflection point

Classify the point of inflection of $f(x)=x^3+x$ .

Differentiate twice:
$f′(x)=3x2+1,f′′(x)=6x.f'(x)=3x^2+1,\qquad f''(x)=6x.$
Solve $f^{''} (x) = 0$ , giving $x = 0$ . Since $6 x$ is negative for $x < 0$ and positive for $x > 0$ , the concavity changes and there is a point of inflection.
Evaluate the first derivative:
$f′(0)=1≠0.f'(0)=1\neq0.$
The tangent is not horizontal, so this is a non-stationary point of inflection at $(0, 0)$ .

Exam technique

In the exam

Differentiate twice and solve $f^{''} (x) = 0$ to find candidate values.
Test the sign of $f^{''} (x)$ on both sides of every candidate; show these tests clearly.
State the convex and concave intervals using inequalities, then substitute into $f (x)$ if point coordinates are required.
If asked to classify an inflection point, also calculate $f^{'} (x)$ to decide whether it is stationary.

Self review

Check yourself

What does the sign of $f^{''} (x)$ tell you about the tangent gradients and the shape of the curve?
Why does $f^{''} (a) = 0$ not necessarily mean that $x = a$ gives a point of inflection?
How would you distinguish a stationary point of inflection from a non-stationary one?

What you'll learn

Prerequisite: the first and second derivatives

Convex curves

Concave curves

Points of inflection

When $f^{''} (x) = 0$ is not enough

Stationary and non-stationary inflection points

Review 1.7.6 Convex, concave and points of inflection (A-level only) by teaching Genie

1.7.6 Convex, concave and points of inflection (A-level only)

What you'll learn

Prerequisite: the first and second derivatives

Convex curves

Concave curves

Points of inflection

When $f^{''} (x) = 0$ is not enough

Stationary and non-stationary inflection points

Review 1.7.6 Convex, concave and points of inflection (A-level only) by teaching Genie

1.7.6 Convex, concave and points of inflection (A-level only)

What you'll learn

Prerequisite: the first and second derivatives

Convex curves

Concave curves

Points of inflection

When f′′(x)=0f''(x)=0f′′(x)=0 is not enough

Stationary and non-stationary inflection points

Review 1.7.6 Convex, concave and points of inflection (A-level only) by teaching Genie

1.7.6 Convex, concave and points of inflection (A-level only) Revision Guide

1.7.6 Convex, concave and points of inflection (A-level only)

What you'll learn

Prerequisite: the first and second derivatives

Convex curves

Concave curves

Points of inflection

When f′′(x)=0f''(x)=0f′′(x)=0 is not enough

Stationary and non-stationary inflection points

Review 1.7.6 Convex, concave and points of inflection (A-level only) by teaching Genie

1.7.6 Convex, concave and points of inflection (A-level only) Revision Guide

When $f^{''} (x) = 0$ is not enough

When $f^{''} (x) = 0$ is not enough