Stationary points | Year 11

01 / LEARN

Understand the idea.

Stationary versus turning

A stationary point is a point on the curve with zero derivative. Solve $f^{'} (x) = 0$ for its input, then substitute into $f (x)$ for its height. Writing only the derivative root does not give the point’s full coordinates.

Classify the point by following the graph from left to right. A positive-to-negative derivative change gives a local maximum; a negative-to-positive change gives a local minimum. With no sign change, a horizontal tangent is not a turning point.

The graph of y equals x cubed minus three x, shown from x minus two to two and y minus three to three. Turning points are visible without coordinate annotations.

Support the classification

Use the factored derivative to justify signs on both sides of each stationary input. For polynomial functions, its roots divide the real line into intervals of fixed derivative sign. This supports classification across whole intervals rather than relying only on the sketch.

Local extrema compare nearby values. They need not be greatest or least across the whole domain. On a restricted interval, an endpoint may beat every stationary value. Also remember that $x^{3}$ has a stationary point at zero without either a maximum or a minimum.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Classify the stationary point of $f (x) = x^{2} - 4 x + 5$ .

Worked solution

1
Find the stationary input.
$\begin{aligned} f^{'} (x) & = 2 x - 4 \\ 2 x - 4 = 0 & ⟹ x = 2 \end{aligned}$
Stationary points have zero tangent gradient.
2
Calculate the corresponding height.
$f (2) = 2^{2} - 4 (2) + 5 = 1$
The original function supplies the vertical coordinate, giving $(2, 1)$ .
3
Check the derivative on both sides.
$\begin{aligned} x < 2 : & f^{'} (x) = 2 (x - 2) < 0 \\ x > 2 : & f^{'} (x) = 2 (x - 2) > 0 \end{aligned}$
The graph falls towards $(2, 1)$ and rises away from it, so the point is a minimum.

Answer

Minimum (2,1).

EXAMPLE 02

Worked example 2

Classify stationary points of $g (x) = x^{3} - 6 x^{2} + 9 x + 1$ .

Worked solution

1
Differentiate and factor.
$g^{'} (x) = 3 x^{2} - 12 x + 9 = 3 (x - 1) (x - 3)$
Factoring identifies all roots of the quadratic derivative.
2
Find the stationary inputs.
$3 (x - 1) (x - 3) = 0 ⟹ x = 1 or x = 3$
A product is zero when at least one factor is zero.
3
Calculate both stationary heights.
$\begin{aligned} g (1) & = 1 - 6 + 9 + 1 = 5 \\ g (3) & = 27 - 54 + 27 + 1 = 1 \end{aligned}$
Substituting into the original cubic gives the points $(1, 5)$ and $(3, 1)$ .
4
Use derivative signs to classify the points.
$\begin{aligned} x < 1 : & g^{'} (x) > 0 \\ 1 < x < 3 : & g^{'} (x) < 0 \\ x > 3 : & g^{'} (x) > 0 \end{aligned}$
The factors give a positive-to-negative change at 1 and a negative-to-positive change at 3: local maximum $(1, 5)$ and local minimum $(3, 1)$ .

Answer

Local maximum (1,5); local minimum (3,1).

03 / YOUR TURN

Try it yourself.

Check your understanding

A derivative changes from negative to positive at x=4. What type of stationary point is there?

Show answer & reasoning

Local minimum.

The derivative is negative before $x = 4$ and positive after it. The graph therefore falls towards the stationary point and rises away, which identifies a local minimum.

Keep practising

6 worksheet questions · 21 marks

Open worksheet Worked solutions

Read the questions online

Find and classify the stationary point of $f (x) = x^{2} + 2 x + 4$ .
3 marks
Classify the stationary point of $g (x) = - x^{2} + 6 x - 1$ .
3 marks
Describe the stationary point of $h (x) = x^{3} + 2$ .
3 marks
Classify all stationary points of $p (x) = x^{3} - 12 x$ .
4 marks
Classify all stationary points of $q (x) = x^{4} - 8 x^{2}$ .
5 marks
Explain why $f (x) = x^{4} + 1$ has a stationary minimum at 0.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find and classify the stationary point of $f (x) = x^{2} + 2 x + 4$ .
3 marks
$f^{'} (x) = 2 x + 2 = 0$ at $- 1$ .
$f (- 1) = 3$ ; signs change − to +.
Minimum $(- 1, 3)$
Classify the stationary point of $g (x) = - x^{2} + 6 x - 1$ .
3 marks
$g^{'} (x) = - 2 x + 6 = 0$ at 3.
$g (3) = 8$ ; signs change + to −.
Maximum $(3, 8)$
Describe the stationary point of $h (x) = x^{3} + 2$ .
3 marks
$h^{'} (x) = 3 x^{2}$ is zero at 0.
The derivative is positive on both sides, so there is no turn.
$(0, 2)$ , neither maximum nor minimum.
Classify all stationary points of $p (x) = x^{3} - 12 x$ .
4 marks
$p^{'} (x) = 3 (x - 2) (x + 2)$ .
Signs +, −, + give the types; evaluate $p$ at $- 2$ and 2.
Maximum $(- 2, 16)$ ; minimum $(2, - 16)$ .
Classify all stationary points of $q (x) = x^{4} - 8 x^{2}$ .
5 marks
$q^{'} (x) = 4 x (x - 2) (x + 2)$ .
Signs across $- 2$ , 0 and 2 are −, +, −, +.
Maximum $(0, 0)$ ; minima $(- 2, - 16)$ and $(2, - 16)$ .
Explain why $f (x) = x^{4} + 1$ has a stationary minimum at 0.
3 marks
$f^{'} (x) = 4 x^{3}$ is negative before 0 and positive after 0.
The point is $(0, 1)$ ; the derivative changes − to +.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics topic 6: rates of change, derivatives and applications. Introductory polynomial calculus with explicit modelling and interpretation.

Official curriculum source ↗

In this lesson

Understand the idea.

Stationary versus turning

Support the classification

See the method in action.

Worked example 1

Find the stationary input.

Calculate the corresponding height.

Check the derivative on both sides.

Worked example 2

Differentiate and factor.

Find the stationary inputs.

Calculate both stationary heights.

Use derivative signs to classify the points.

Try it yourself.

Check your understanding

Keep practising