Stationary points | Year 11

01 / LEARN

The idea, step by step

Stationary versus turning

At a stationary point f′(x)=0. Solve this equation for the input, then substitute into f to find its height. The original function and its derivative have different jobs.

A derivative changing from positive to negative gives a local maximum. A change from negative to positive gives a local minimum. If the sign is unchanged, the horizontal tangent is not a turn.

Support the classification

Use a derivative sign chart or an equivalent justified argument. Listing zeros of the derivative does not classify them. For x³, the derivative is positive on both sides of 0, so (0,0) is stationary but neither a maximum nor a minimum.

Local describes nearby values only. A local maximum need not be the greatest value in the full domain. Endpoints can also give extrema in restricted practical domains without being stationary.

A common trap

Calling every derivative zero a turning point.

Instead: Check signs on each side; a horizontal tangent can occur without a turn.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example 1

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

$f^{'} (x) = 2 x - 4 = 0$ gives x=2.
$f (2) = 1$ . The derivative changes − to +.

Minimum (2,1).

Example 2

Worked example 2

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

$g^{'} (x) = 3 (x - 1) (x - 3)$ gives x=1,3.
$g (1) = 5, g (3) = 1$ .
Derivative signs +, −, + give a maximum then a minimum.

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

Check your understanding

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

Show answer

Local minimum.

The graph falls towards the point and rises away from it.

03 / WORKSHEET

Your turn

6 questions · 21 marks · Work at your own pace.

Worksheet PDF 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 ↗

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