Stationary-point classification | Year 12

01 / LEARN

The idea, step by step

Find then classify

Stationary points occur where $f^{'} (x) = 0$ and the function is defined. Find all allowed solutions and use $f (x)$ for their coordinates.

A change in $f^{'}$ from positive to negative gives a local maximum. Negative to positive gives a local minimum. No sign change means the stationary point is neither.

A useful short test

At a stationary input a, $f^{''} (a) > 0$ gives a local minimum and $f^{''} (a) < 0$ gives a local maximum. When the second derivative is zero, this test is inconclusive.

Inconclusive does not mean inflection. Compare $x^{3}$ with $x^{4}$ : both have zero first and second derivatives at the origin, but their shapes differ.

A common trap

Using $f^{''} = 0$ as a classification.

Instead: It makes the second-derivative test inconclusive; use a sign chart or another justified argument.

02 / WORKED EXAMPLES

See the thinking

Example 1

Two stationary points

Classify the stationary points of $f (x) = x^{3} - 3 x$ .

$f^{'} = 3 (x - 1) (x + 1)$ , so $x = - 1, 1$ .
$f^{''} = 6 x$ is negative at $- 1$ and positive at 1.
The corresponding heights are 2 and $- 2$ .

Local maximum

(- 1, 2)

; local minimum

(1, - 2)

Example 2

When the second-derivative test is inconclusive

Classify the stationary point of $y = x^{4}$ .

The derivative $4 x^{3}$ is zero at 0 and the second derivative is also zero there.
The first derivative changes from negative to positive.

Local minimum

(0, 0)

Check your understanding

What does a change of $f^{'}$ from positive to negative show?

Show answer

A local maximum.

The original function rises into the point and falls away afterwards.

03 / WORKSHEET

Your turn

6 questions · 15 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find and classify the stationary point of $y = x^{2} - 6 x + 5$ .
2 marks
Find and classify the stationary point of $y = - x^{2} + 4 x$ .
2 marks
Classify the stationary points of $y = x^{3} - 3 x^{2}$ .
3 marks
Explain why the stationary point of $y = x^{3}$ is not an extremum.
2 marks
Classify the stationary point of $y = x e^{- x}$ .
3 marks
Can a local maximum fail to be an absolute maximum on an unbounded domain? Give an example.
3 marks

Read the worked solutions

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

Find and classify the stationary point of $y = x^{2} - 6 x + 5$ .
2 marks
The derivative is $2 x - 6$ , zero at 3.
The second derivative is 2; the height is $- 4$ .
Local minimum $(3, - 4)$ .
Find and classify the stationary point of $y = - x^{2} + 4 x$ .
2 marks
The derivative is $- 2 x + 4$ , zero at 2.
The second derivative is $- 2$ .
Local maximum $(2, 4)$ .
Classify the stationary points of $y = x^{3} - 3 x^{2}$ .
3 marks
The derivative is $3 x (x - 2)$ .
The second derivative $6 x - 6$ is negative at 0 and positive at 2.
Local maximum $(0, 0)$ ; local minimum $(2, - 4)$ .
Explain why the stationary point of $y = x^{3}$ is not an extremum.
2 marks
The derivative is $3 x^{2}$ .
There is no increasing/decreasing reversal.
The derivative is positive on both sides of zero.
Classify the stationary point of $y = x e^{- x}$ .
3 marks
The derivative is $e^{- x} (1 - x)$ .
Its sign changes from positive to negative at 1.
Local maximum $(1, e^{- 1})$ .
Can a local maximum fail to be an absolute maximum on an unbounded domain? Give an example.
3 marks
A local maximum compares nearby inputs only.
As $x$ grows, $x^{3} - 3 x$ grows without bound.
Yes: $x^{3} - 3 x$ has a local maximum at $(- 1, 2)$ but is unbounded above.

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Curriculum reference & lesson scope

SACE Stage 2 Mathematical Methods

Further differentiation and applications from the current SACE Methods outline. Logarithmic derivatives have their own chapter; integration follows later. Trigonometric calculus uses radians.

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