1.4.5 Stationary points when curve sketching

What you'll learn

How to find the coordinates of stationary points using differentiation.
How to classify a stationary point as a local maximum, local minimum or stationary point of inflection.
How stationary points help you produce an accurate curve sketch.
How to deal with cases where the second derivative test is inconclusive.

Prerequisite: gradients and derivatives

The gradient of a curve measures its steepness and direction. If a curve has equation $y = f (x)$ , its gradient is given by the derivative

dydx=f′(x).\frac{dy}{dx}=f'(x).

The sign of the derivative tells you how the curve is behaving:

If $f^{'} (x) > 0$ , the curve is increasing as you move from left to right.
If $f^{'} (x) < 0$ , the curve is decreasing as you move from left to right.
If $f^{'} (x) = 0$ , the curve has a horizontal tangent at that point.

Key Idea

Derivative as a shape detector

When sketching a curve, $f^{'} (x)$ tells you where the curve rises, where it falls and where it temporarily becomes horizontal.

What is a stationary point?

Definition

Stationary point

A stationary point on the curve $y = f (x)$ is a point where the derivative is zero:

f^{'} (x) = 0.

At a stationary point, the tangent to the curve is horizontal.

To find the coordinates of stationary points:

Differentiate the function.
Solve $f^{'} (x) = 0$ to find each stationary $x$ -coordinate.
Substitute each value of $x$ into the original equation to find the corresponding $y$ -coordinate.

Example

Finding stationary points

Find the stationary points of the curve

y=x^3-3x^2-9x+5.

Differentiate the equation:
$dydx=3x2−6x−9.\frac{dy}{dx}=3x^2-6x-9.$
Set the derivative equal to zero and factorise:
$3x2−6x−9=03(x2−2x−3)=03(x−3)(x+1)=0.\begin{aligned} 3x^2-6x-9&=0\\ 3(x^2-2x-3)&=0\\ 3(x-3)(x+1)&=0. \end{aligned}$
Therefore, $x = 3$ or $x = - 1$ .
Substitute $x = 3$ into the original equation:
$y=3^3-3(3^2)-9(3)+5=-22.$
This gives the stationary point $(3, - 22)$ .
Substitute $x = - 1$ into the original equation:
$y=(-1)^3-3(-1)^2-9(-1)+5=10.$
The other stationary point is $(- 1, 10)$ .

Common Mistake

Giving only the x-coordinates

Solving $f^{'} (x) = 0$ finds the stationary values of $x$ , not the complete points. Substitute into the original function to obtain the $y$ -coordinates.

Types of stationary point

Finding a stationary point does not tell you its shape. You must also classify it.

Local maximum

A local maximum occurs where the curve changes from increasing to decreasing. The derivative changes sign from positive to negative:

−.f'(x):\quad +\ \longrightarrow\ 0\ \longrightarrow\ -.

The word local means that the function value is greater than nearby values. It does not necessarily mean that this is the greatest value anywhere on the entire graph.

Local minimum

A local minimum occurs where the curve changes from decreasing to increasing. The derivative changes sign from negative to positive:

+.f'(x):\quad -\ \longrightarrow\ 0\ \longrightarrow\ +.

Stationary point of inflection

A point of inflection is a point where the curve changes its direction of curvature. A stationary point of inflection also has a horizontal tangent.

For example, the curve $y=x^3$ has

dydx=3x2,\frac{dy}{dx}=3x^2,

so the origin is stationary. However, the derivative is positive on both sides of the origin. The curve continues increasing, so the origin is not a maximum or minimum.

Key Idea

Stationary does not always mean turning

A stationary point is defined by $f^{'} (x) = 0$ . It is a turning point only if the derivative changes sign there.

Classifying by testing the derivative

The first derivative sign test examines the sign of $f^{'} (x)$ immediately before and after a stationary value.

You do not need the exact gradient on each side. Its sign is enough.

Example

Classifying stationary points using gradient signs

Classify the stationary points of

y=x^3-3x^2-9x+5.

The stationary values found earlier are $x = - 1$ and $x = 3$ , and

f^{'} (x) = 3 (x - 3) (x + 1) .

For $x < - 1$ , choose $x = - 2$ :
$f^{'} (- 2) = 3 (- 5) (- 1) > 0.$
The curve is increasing.
Between $x = - 1$ and $x = 3$ , choose $x = 0$ :
$f^{'} (0) = 3 (- 3) (1) < 0.$
The curve is decreasing. At $x = - 1$ , the sign changes from positive to negative, so $(- 1, 10)$ is a local maximum.
For $x > 3$ , choose $x = 4$ :
$f^{'} (4) = 3 (1) (5) > 0.$
At $x = 3$ , the sign changes from negative to positive, so $(3, - 22)$ is a local minimum.

Tip

Use factor signs efficiently

When the derivative is factorised, you can often determine its sign by checking whether each factor is positive or negative. You do not always need to calculate the full derivative value.

Classifying with the second derivative

Differentiating again gives the second derivative, written

d2ydx2=f′′(x).\frac{d^2y}{dx^2}=f''(x).

At a stationary point $x = a$ :

If $f^{''} (a) > 0$ , the stationary point is a local minimum.
If $f^{''} (a) < 0$ , the stationary point is a local maximum.
If $f^{''} (a) = 0$ , the test is inconclusive.

The result makes sense geometrically: near a minimum the gradients are increasing, while near a maximum they are decreasing.

Example

Using the second derivative test

Classify the stationary points of

y=x^3-3x^2-9x+5.

Differentiate the first derivative again:
$f^{''} (x) = 6 x - 6.$
At $x = - 1$ :
$f^{''} (- 1) = - 12 < 0.$
Therefore, $(- 1, 10)$ is a local maximum.
At $x = 3$ :
$f^{''} (3) = 12 > 0.$
Therefore, $(3, - 22)$ is a local minimum.

Common Mistake

When the second derivative is zero

If $f^{''} (a) = 0$ , you cannot conclude that the point is an inflection. For example, both $y=x^3$ and $y=x^4$ have $f^{'} (0) = f^{''} (0) = 0$ , but the origin is a stationary point of inflection for $y=x^3$ and a minimum for $y=x^4$ . Use the first derivative sign test or examine curvature more carefully.

Using stationary points in a curve sketch

A useful sketch should show the important mathematical features of the graph, even if it is not drawn exactly to scale.

For a polynomial curve, consider:

intercepts with the coordinate axes;
stationary points and their types;
intervals where the curve increases or decreases;
the behaviour as $x→∞x\to\infty$ and $x→−∞x\to-\infty$ .

The highest-power term usually determines the end behaviour of a polynomial.

Example

Building a curve sketch from key features

Sketch the main features of

y=x^3-3x^2-9x+5.

Use the stationary-point analysis: the curve has a local maximum at $(- 1, 10)$ and a local minimum at $(3, - 22)$ .
Find the vertical-axis intercept by setting $x = 0$ :
$y = 5.$
Therefore, the curve crosses the $y$ -axis at $(0, 5)$ .
Use the leading term $x^3$ to determine the end behaviour:
$x→−∞⇒y→−∞,x→∞⇒y→∞.x\to-\infty\Rightarrow y\to-\infty, \qquad x\to\infty\Rightarrow y\to\infty.$
Combine the gradient information. The curve rises to $(- 1, 10)$ , falls through $(0, 5)$ to $(3, - 22)$ , and then rises again. Because the maximum is above the $x$ -axis and the minimum is below it, continuity shows that the cubic crosses the $x$ -axis three times.

Tip

Sketch the behaviour before joining the curve

Mark stationary points and intercepts first. Then add small direction arrows showing where the curve increases or decreases before drawing one smooth curve through the features.

Common Mistake

Joining points with straight lines

A curve sketch must remain smooth where the function is differentiable. Stationary points have horizontal tangents, so the curve should flatten there rather than form sharp corners.

Stationary points in modelling

In a model, a stationary point may represent a greatest or least value, such as maximum profit or minimum cost. Classification matters because solving $f^{'} (x) = 0$ only identifies candidates.

You must also check that the value lies within the model’s permitted domain. If the domain is a closed interval, an overall maximum or minimum could occur at an endpoint rather than at a stationary point.

Exam technique

In the exam

Differentiate accurately, solve $f^{'} (x) = 0$ , and use the original function to find complete coordinates.
Classify every stationary point using derivative signs or the second derivative test; do not rely only on the appearance of your sketch.
Mark intercepts and stationary points before drawing a smooth curve with correct end behaviour.
In modelling questions, check the domain and compare relevant stationary values with endpoint values.

Self review

Check yourself

How can the signs of $f^{'} (x)$ distinguish a maximum, a minimum and a stationary point of inflection?
Why does $f^{''} (a) = 0$ not classify a stationary point?
What features would you find before sketching a polynomial curve?

What you'll learn

How to find the coordinates of stationary points using differentiation.
How to classify a stationary point as a local maximum, local minimum or stationary point of inflection.
How stationary points help you produce an accurate curve sketch.
How to deal with cases where the second derivative test is inconclusive.

Prerequisite: gradients and derivatives

The gradient of a curve measures its steepness and direction. If a curve has equation $y = f (x)$ , its gradient is given by the derivative

dydx=f′(x).\frac{dy}{dx}=f'(x).

The sign of the derivative tells you how the curve is behaving:

If $f^{'} (x) > 0$ , the curve is increasing as you move from left to right.
If $f^{'} (x) < 0$ , the curve is decreasing as you move from left to right.
If $f^{'} (x) = 0$ , the curve has a horizontal tangent at that point.

Key Idea

Derivative as a shape detector

When sketching a curve, $f^{'} (x)$ tells you where the curve rises, where it falls and where it temporarily becomes horizontal.

What is a stationary point?

Definition

Stationary point

A stationary point on the curve $y = f (x)$ is a point where the derivative is zero:

f^{'} (x) = 0.

At a stationary point, the tangent to the curve is horizontal.

To find the coordinates of stationary points:

Differentiate the function.
Solve $f^{'} (x) = 0$ to find each stationary $x$ -coordinate.
Substitute each value of $x$ into the original equation to find the corresponding $y$ -coordinate.

Example

Finding stationary points

Find the stationary points of the curve

y=x^3-3x^2-9x+5.

Differentiate the equation:
$dydx=3x2−6x−9.\frac{dy}{dx}=3x^2-6x-9.$
Set the derivative equal to zero and factorise:
$3x2−6x−9=03(x2−2x−3)=03(x−3)(x+1)=0.\begin{aligned} 3x^2-6x-9&=0\\ 3(x^2-2x-3)&=0\\ 3(x-3)(x+1)&=0. \end{aligned}$
Therefore, $x = 3$ or $x = - 1$ .
Substitute $x = 3$ into the original equation:
$y=3^3-3(3^2)-9(3)+5=-22.$
This gives the stationary point $(3, - 22)$ .
Substitute $x = - 1$ into the original equation:
$y=(-1)^3-3(-1)^2-9(-1)+5=10.$
The other stationary point is $(- 1, 10)$ .

Common Mistake

Giving only the x-coordinates

Solving $f^{'} (x) = 0$ finds the stationary values of $x$ , not the complete points. Substitute into the original function to obtain the $y$ -coordinates.

Types of stationary point

Finding a stationary point does not tell you its shape. You must also classify it.

Local maximum

A local maximum occurs where the curve changes from increasing to decreasing. The derivative changes sign from positive to negative:

−.f'(x):\quad +\ \longrightarrow\ 0\ \longrightarrow\ -.

The word local means that the function value is greater than nearby values. It does not necessarily mean that this is the greatest value anywhere on the entire graph.

Local minimum

A local minimum occurs where the curve changes from decreasing to increasing. The derivative changes sign from negative to positive:

+.f'(x):\quad -\ \longrightarrow\ 0\ \longrightarrow\ +.

Stationary point of inflection

A point of inflection is a point where the curve changes its direction of curvature. A stationary point of inflection also has a horizontal tangent.

For example, the curve $y=x^3$ has

dydx=3x2,\frac{dy}{dx}=3x^2,

so the origin is stationary. However, the derivative is positive on both sides of the origin. The curve continues increasing, so the origin is not a maximum or minimum.

Key Idea

Stationary does not always mean turning

A stationary point is defined by $f^{'} (x) = 0$ . It is a turning point only if the derivative changes sign there.

Classifying by testing the derivative

The first derivative sign test examines the sign of $f^{'} (x)$ immediately before and after a stationary value.

You do not need the exact gradient on each side. Its sign is enough.

Example

Classifying stationary points using gradient signs

Classify the stationary points of

y=x^3-3x^2-9x+5.

The stationary values found earlier are $x = - 1$ and $x = 3$ , and

f^{'} (x) = 3 (x - 3) (x + 1) .

For $x < - 1$ , choose $x = - 2$ :
$f^{'} (- 2) = 3 (- 5) (- 1) > 0.$
The curve is increasing.
Between $x = - 1$ and $x = 3$ , choose $x = 0$ :
$f^{'} (0) = 3 (- 3) (1) < 0.$
The curve is decreasing. At $x = - 1$ , the sign changes from positive to negative, so $(- 1, 10)$ is a local maximum.
For $x > 3$ , choose $x = 4$ :
$f^{'} (4) = 3 (1) (5) > 0.$
At $x = 3$ , the sign changes from negative to positive, so $(3, - 22)$ is a local minimum.

Tip

Use factor signs efficiently

When the derivative is factorised, you can often determine its sign by checking whether each factor is positive or negative. You do not always need to calculate the full derivative value.

Classifying with the second derivative

Differentiating again gives the second derivative, written

d2ydx2=f′′(x).\frac{d^2y}{dx^2}=f''(x).

At a stationary point $x = a$ :

If $f^{''} (a) > 0$ , the stationary point is a local minimum.
If $f^{''} (a) < 0$ , the stationary point is a local maximum.
If $f^{''} (a) = 0$ , the test is inconclusive.

The result makes sense geometrically: near a minimum the gradients are increasing, while near a maximum they are decreasing.

Example

Using the second derivative test

Classify the stationary points of

y=x^3-3x^2-9x+5.

Differentiate the first derivative again:
$f^{''} (x) = 6 x - 6.$
At $x = - 1$ :
$f^{''} (- 1) = - 12 < 0.$
Therefore, $(- 1, 10)$ is a local maximum.
At $x = 3$ :
$f^{''} (3) = 12 > 0.$
Therefore, $(3, - 22)$ is a local minimum.

Common Mistake

When the second derivative is zero

Using stationary points in a curve sketch

A useful sketch should show the important mathematical features of the graph, even if it is not drawn exactly to scale.

For a polynomial curve, consider:

intercepts with the coordinate axes;
stationary points and their types;
intervals where the curve increases or decreases;
the behaviour as $x→∞x\to\infty$ and $x→−∞x\to-\infty$ .

The highest-power term usually determines the end behaviour of a polynomial.

Example

Building a curve sketch from key features

Sketch the main features of

y=x^3-3x^2-9x+5.

Use the stationary-point analysis: the curve has a local maximum at $(- 1, 10)$ and a local minimum at $(3, - 22)$ .
Find the vertical-axis intercept by setting $x = 0$ :
$y = 5.$
Therefore, the curve crosses the $y$ -axis at $(0, 5)$ .
Use the leading term $x^3$ to determine the end behaviour:
$x→−∞⇒y→−∞,x→∞⇒y→∞.x\to-\infty\Rightarrow y\to-\infty, \qquad x\to\infty\Rightarrow y\to\infty.$
Combine the gradient information. The curve rises to $(- 1, 10)$ , falls through $(0, 5)$ to $(3, - 22)$ , and then rises again. Because the maximum is above the $x$ -axis and the minimum is below it, continuity shows that the cubic crosses the $x$ -axis three times.

Tip

Sketch the behaviour before joining the curve

Mark stationary points and intercepts first. Then add small direction arrows showing where the curve increases or decreases before drawing one smooth curve through the features.

Common Mistake

Joining points with straight lines

A curve sketch must remain smooth where the function is differentiable. Stationary points have horizontal tangents, so the curve should flatten there rather than form sharp corners.

Stationary points in modelling

In a model, a stationary point may represent a greatest or least value, such as maximum profit or minimum cost. Classification matters because solving $f^{'} (x) = 0$ only identifies candidates.

Exam technique

In the exam

Differentiate accurately, solve $f^{'} (x) = 0$ , and use the original function to find complete coordinates.
Classify every stationary point using derivative signs or the second derivative test; do not rely only on the appearance of your sketch.
Mark intercepts and stationary points before drawing a smooth curve with correct end behaviour.
In modelling questions, check the domain and compare relevant stationary values with endpoint values.

Self review

Check yourself

How can the signs of $f^{'} (x)$ distinguish a maximum, a minimum and a stationary point of inflection?
Why does $f^{''} (a) = 0$ not classify a stationary point?
What features would you find before sketching a polynomial curve?

What you'll learn

Prerequisite: gradients and derivatives

What is a stationary point?

Types of stationary point

Local maximum

Local minimum

Stationary point of inflection

Classifying by testing the derivative

Classifying with the second derivative

Using stationary points in a curve sketch

Stationary points in modelling

Review 1.4.5 Stationary points when curve sketching by teaching Genie

1.4.5 Stationary points when curve sketching

What you'll learn

Prerequisite: gradients and derivatives

What is a stationary point?

Types of stationary point

Local maximum

Local minimum

Stationary point of inflection

Classifying by testing the derivative

Classifying with the second derivative

Using stationary points in a curve sketch

Stationary points in modelling

Review 1.4.5 Stationary points when curve sketching by teaching Genie

1.4.5 Stationary points when curve sketching

What you'll learn

Prerequisite: gradients and derivatives

What is a stationary point?

Types of stationary point

Local maximum

Local minimum

Stationary point of inflection

Classifying by testing the derivative

Classifying with the second derivative

Using stationary points in a curve sketch

Stationary points in modelling

Review 1.4.5 Stationary points when curve sketching by teaching Genie

1.4.5 Stationary points when curve sketching Revision Guide

1.4.5 Stationary points when curve sketching

What you'll learn

Prerequisite: gradients and derivatives

What is a stationary point?

Types of stationary point

Local maximum

Local minimum

Stationary point of inflection

Classifying by testing the derivative

Classifying with the second derivative

Using stationary points in a curve sketch

Stationary points in modelling

Review 1.4.5 Stationary points when curve sketching by teaching Genie

1.4.5 Stationary points when curve sketching Revision Guide