QLDMath MethodsSyllabus dot point

Topic 3: Introduction to differential calculus

Use the derivative to find stationary points of a polynomial function and classify them, and apply differentiation to simple optimisation problems

A focused answer to the QCE Math Methods Unit 2 dot point on stationary points. Locates stationary points by solving $f'(x) = 0$ , classifies them as maxima, minima or stationary points of inflection using the first-derivative sign test, and works the QCAA-style optimisation problem (maximising the area of a fenced rectangle).

Generated by Claude Opus 4.87 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Stationary points
Finding stationary points
Classifying with the first-derivative sign test
Local versus global extrema
Optimisation procedure
The second-derivative test

What this dot point is asking

QCAA wants you to locate stationary points of polynomial functions (where $f'(x) = 0$ ), classify them as local maxima, local minima or stationary points of inflection, and apply the technique to simple worded optimisation problems.

Stationary points

A stationary point is a point on the curve where $f'(x) = 0$ , that is, the tangent line is horizontal. Stationary points come in three types:

Local maximum: $f'$ changes from positive (rising) to negative (falling) as $x$ passes through the point.
Local minimum: $f'$ changes from negative (falling) to positive (rising).
Stationary point of inflection: $f'$ does not change sign (positive on both sides, or negative on both sides).

Finding stationary points

Procedure:

Differentiate the function to get $f'(x)$ .
Solve $f'(x) = 0$ for $x$ to find the $x$ -coordinates of stationary points.
Substitute each $x$ -value back into $f(x)$ to find the $y$ -coordinate.

Classifying with the first-derivative sign test

Pick a point just to the left and just to the right of the stationary point, and compute the sign of $f'$ at each.

Sign of $f'$ before	Sign of $f'$ after	Type
+	-	Local maximum
-	+	Local minimum
+	+	Stationary point of inflection (rising)
-	-	Stationary point of inflection (falling)

Local versus global extrema

A stationary point gives a local extremum, the largest or smallest value in its immediate neighbourhood. Over a restricted domain (common in optimisation, where a length cannot be negative) the global maximum or minimum may instead occur at an endpoint of the domain rather than at a stationary point. The complete method therefore compares the values at every stationary point with the values at the domain endpoints, and the largest of these is the global maximum. Forgetting the endpoints is a frequent way to miss the true optimum in a worded problem.

Optimisation procedure

For a worded problem asking for a maximum or minimum:

Identify the quantity to optimise (area, volume, cost).
Express it as a function of a single variable (use the constraint to eliminate other variables).
Differentiate; set $f'(x) = 0$ ; solve for $x$ .
Verify with a sign test (or check endpoints of the feasible region).
State the answer in words with units.

Finding and classifying stationary points

Find and classify all stationary points of $f(x) = x^3 - 3x^2 + 4$ .

Step 1: Differentiate to get the gradient function

Apply the power rule term by term:

f'(x) = 3x^2 - 6x.

This expression gives the gradient of the tangent at any value of $x$ .

Step 2: Solve $f'(x) = 0$ to locate the stationary points

Stationary points occur where the tangent is horizontal. Factor the derivative and apply the null factor law:

3x^2 - 6x = 3x(x - 2) = 0 \;\Rightarrow\; x = 0 \text{ or } x = 2.

Step 3: Find the $y$ -coordinates

Substitute each $x$ -value back into the original function (not the derivative):

f(0) = 0 - 0 + 4 = 4, \qquad f(2) = 8 - 12 + 4 = 0.

The stationary points are $(0, 4)$ and $(2, 0)$ .

Step 4: Classify with the first-derivative sign test

Choose a test value on each side of each stationary point and check whether $f'$ is positive or negative.

Near $x = 0$ : $f'(-1) = 3 + 6 = 9 > 0$ , then $f'(1) = 3 - 6 = -3 < 0$ . The sign changes from $+$ to $-$ , so the function rises then falls: $(0, 4)$ is a local maximum.

Near $x = 2$ : $f'(1) = -3 < 0$ , then $f'(3) = 27 - 18 = 9 > 0$ . The sign changes from $-$ to $+$ , so the function falls then rises: $(2, 0)$ is a local minimum.

Final answer: Local maximum at $(0, 4)$ ; local minimum at $(2, 0)$ .

The second-derivative test

An alternative to the sign test classifies a stationary point using the second derivative: if $f''(x) > 0$ at the point the curve is concave up and it is a local minimum, while $f''(x) < 0$ means concave down and a local maximum. If $f''(x) = 0$ the test is inconclusive and you fall back on the first-derivative sign test. The second-derivative test is quicker when the second derivative is easy to compute, which is why it is worth knowing alongside the sign diagram.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

QCAA 20225 marksPaper 2 (complex familiar). A farmer has

40

m of fence to enclose a rectangular paddock against a straight river (no fence along the river). Determine the dimensions that maximise the enclosed area, and state that maximum area.

Show worked answer →

Let the sides perpendicular to the river be $x$ and the side parallel be $y$ . The fence gives $2x + y = 40$ , so $y = 40 - 2x$ .

Area $A(x) = x(40 - 2x) = 40x - 2x^2$ , so $A'(x) = 40 - 4x$ . Setting $A'(x) = 0$ gives $x = 10$ .

Sign test: $A'(9) = 4 > 0$ and $A'(11) = -4 < 0$ , confirming a maximum. Then $y = 40 - 20 = 20$ and $A = 200$ m $^2$ .

Markers reward defining variables, the constraint, the area function, the sign test, and the dimensions plus area.

QCAA 20235 marksPaper 2 (complex familiar). An open-top box is made from a square sheet of side

24

cm by cutting squares of side

x

from each corner and folding up the sides. (a) Express the volume as a function of

x

. (b) Determine the value of

x

that maximises the volume.

Show worked answer →

(a) The base is $(24 - 2x)$ by $(24 - 2x)$ and the height is $x$ , so $V(x) = x(24 - 2x)^2$ for $0 < x < 12$ .

(b) Expand: $V = x(576 - 96x + 4x^2) = 576x - 96x^2 + 4x^3$ , so $V'(x) = 576 - 192x + 12x^2 = 12(x^2 - 16x + 48) = 12(x - 4)(x - 12)$ . In the domain, $x = 4$ (since $x = 12$ gives zero volume). A sign test confirms a maximum at $x = 4$ cm.

Markers reward the volume model with its domain, the derivative, and selecting the valid root with a maximum check.