SEAB Original-style practice: Find the coordinates and nature of the stationary points of y = x^3 - 3x^2 - 9x + 5.

dydx = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1). Stationary where dydx = 0: x = 3 or x = -1. At x = 3: y = 27 - 27 - 27 + 5 = -22. At x = -1: y = -1 - 3 + 9 + 5 = 10. Second derivative: d^2 ydx^2 = 6x - 6. At x = 3: 12 > 0, a minimum (3, -22). At x = -1: -12 < 0, a maximum (-1, 10). Markers reward solving dydx = 0, both points, and using the second derivative to classify each.

SEAB Original-style practice: A closed cylindrical can has volume 500\ cm^3. Find the radius that minimises the surface area.

Volume: π r^2 h = 500, so h = 500π r^2. Surface area: A = 2π r^2 + 2π r h = 2π r^2 + 2π r · 500π r^2 = 2π r^2 + 1000r. dAdr = 4π r - 1000r^2 = 0 4π r^3 = 1000 r^3 = 250π r = (250π)^1/3 ≈ 4.30\ cm. d^2 Adr^2 = 4π + 2000r^3 > 0, confirming a minimum. Markers reward expressing A in one variable using the constraint, differentiating, solving, and the second-derivative check.

SingaporeMathsSyllabus dot point

How do we locate and classify stationary points and solve optimisation problems?

Find and classify stationary points, determine increasing and decreasing intervals and concavity, and solve optimisation problems in context

A focused answer to the H2 Mathematics outcome on applications of differentiation. Finding stationary points, classifying them with the first and second derivative tests, concavity and points of inflexion, and optimisation.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-06

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

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What this dot point is asking
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Try this

What this dot point is asking

SEAB wants you to find stationary points, classify them as maxima, minima or points of inflexion, determine where a function is increasing or decreasing and its concavity, and solve optimisation problems by reducing to one variable and applying calculus.

The answer

Stationary points

A stationary point occurs where $\dfrac{dy}{dx} = 0$ : the tangent is horizontal. Solving this equation gives the $x$ -coordinates; substitute back for the $y$ -coordinates.

Increasing and decreasing

A function is increasing where $\dfrac{dy}{dx} > 0$ and decreasing where $\dfrac{dy}{dx} < 0$ . The sign of the first derivative on each side of a stationary point classifies it.

The two classification tests

First derivative test: check the sign of $\dfrac{dy}{dx}$ just before and after the point. Positive then negative is a maximum; negative then positive is a minimum.
Second derivative test: at a stationary point, $\dfrac{d^2y}{dx^2} > 0$ indicates a minimum, $\dfrac{d^2y}{dx^2} < 0$ a maximum. If $\dfrac{d^2y}{dx^2} = 0$ the test is inconclusive and you fall back on the first derivative test.

Concavity and points of inflexion

$\dfrac{d^2y}{dx^2} > 0$ means concave up; $< 0$ means concave down. A point of inflexion is where concavity changes, requiring $\dfrac{d^2y}{dx^2} = 0$ and a sign change in the second derivative there.

Optimisation

To optimise a quantity:

Write the quantity to optimise.
Use the constraint to express it as a function of one variable.
Differentiate, set to zero, and solve.
Confirm it is a maximum or minimum (second derivative or context).

Worked example

Find and classify the stationary points of $y = 2x^3 - 3x^2 - 12x + 1$ , and state the intervals where the curve is increasing.

Step 1: Differentiate and set to zero

\frac{dy}{dx} = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x - 2)(x + 1) = 0

so $x = 2$ or $x = -1$ .

Step 2: Find the y-coordinates

At $x = 2$ : $y = 16 - 12 - 24 + 1 = -19$ . At $x = -1$ : $y = -2 - 3 + 12 + 1 = 8$ .

Step 3: Classify with the second derivative

$\dfrac{d^2y}{dx^2} = 12x - 6$ . At $x = 2$ : $18 > 0$ , minimum $(2, -19)$ . At $x = -1$ : $-18 < 0$ , maximum $(-1, 8)$ .

Step 4: Increasing intervals

$\dfrac{dy}{dx} > 0$ outside the roots, that is for $x < -1$ or $x > 2$ . The curve increases on $(-\infty, -1)$ and $(2, \infty)$ .

Examples in context

Example 1. Maximum projectile range. Expressing range as a function of launch angle and differentiating shows the maximum range occurs at $45^\circ$ , a classic optimisation where the constraint (fixed launch speed) reduces the problem to one variable.

Example 2. Least-cost packaging. Minimising the surface area of a fixed-volume container, as in the worked can example, finds the most material-efficient shape, the everyday industrial use of stationary-point analysis.

Try this

Q1. Find the stationary point of $y = x^2 - 6x + 5$ and state its nature. [3 marks]

Cue. $\frac{dy}{dx} = 2x - 6 = 0$ at $x = 3$ , $y = -4$ ; $\frac{d^2y}{dx^2} = 2 > 0$ , a minimum.

Q2. State the condition for a function to be increasing. [1 mark]

Cue. $\frac{dy}{dx} > 0$ .

Q3. What two conditions identify a point of inflexion? [2 marks]

Cue. $\frac{d^2y}{dx^2} = 0$ and the second derivative changes sign there.

Exam-style practice questions

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

Original5 marksFind the coordinates and nature of the stationary points of

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

Show worked answer →

$\dfrac{dy}{dx} = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1)$ .

Stationary where $\dfrac{dy}{dx} = 0$ : $x = 3$ or $x = -1$ .

At $x = 3$ : $y = 27 - 27 - 27 + 5 = -22$ . At $x = -1$ : $y = -1 - 3 + 9 + 5 = 10$ .

Second derivative: $\dfrac{d^2 y}{dx^2} = 6x - 6$ . At $x = 3$ : $12 > 0$ , a minimum $(3, -22)$ . At $x = -1$ : $-12 < 0$ , a maximum $(-1, 10)$ .

Markers reward solving $\dfrac{dy}{dx} = 0$ , both points, and using the second derivative to classify each.

Original6 marksA closed cylindrical can has volume

500\ \text{cm}^3

. Find the radius that minimises the surface area.

Show worked answer →

Volume: $\pi r^2 h = 500$ , so $h = \dfrac{500}{\pi r^2}$ .

Surface area: $A = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r \cdot \dfrac{500}{\pi r^2} = 2\pi r^2 + \dfrac{1000}{r}$ .

$\dfrac{dA}{dr} = 4\pi r - \dfrac{1000}{r^2} = 0 \Rightarrow 4\pi r^3 = 1000 \Rightarrow r^3 = \dfrac{250}{\pi} \Rightarrow r = \left(\dfrac{250}{\pi}\right)^{1/3} \approx 4.30\ \text{cm}$ .

$\dfrac{d^2 A}{dr^2} = 4\pi + \dfrac{2000}{r^3} > 0$ , confirming a minimum.

Markers reward expressing $A$ in one variable using the constraint, differentiating, solving, and the second-derivative check.