Use differentiation to analyse the behaviour of functions, including locating and classifying stationary points, finding tangent and normal equations, and solving optimisation problems

What this dot point is asking

VCAA wants you to apply differentiation to function analysis: finding stationary points and classifying them, deriving tangent and normal equations, and solving optimisation problems by constructing a single-variable function and finding its extremum.

Stationary points

A point on $y = f(x)$ where $f'(x) = 0$. Three types:

• Local maximum: $f'$ changes from positive to negative.
• Local minimum: $f'$ changes from negative to positive.
• Stationary point of inflection: $f'$ does not change sign.

First-derivative sign test

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

Second-derivative test

If $f'(a) = 0$ and:

• $f''(a) > 0$: local minimum (concave up).
• $f''(a) < 0$: local maximum (concave down).
• $f''(a) = 0$: test is inconclusive; use the first-derivative sign test instead.

Tangent and normal lines

Tangent at $x = a$: $y - f(a) = f'(a)(x - a)$.

Normal at $x = a$: $y - f(a) = -\dfrac{1}{f'(a)}(x - a)$ (perpendicular to the tangent; provided $f'(a) \neq 0$).

If $f'(a) = 0$ the tangent is horizontal and the normal is vertical ($x = a$).

Optimisation procedure

1. Identify the quantity to optimise (area, volume, cost, distance).
2. Express it as a function of one variable using the constraints.
3. Differentiate; set $f'(x) = 0$; solve.
4. Verify with a sign test or the second-derivative test.
5. Check the domain boundaries (the global extremum might be at an endpoint).
6. State the answer in words with units.

Worked example (stationary points)

For $f(x) = x^3 - 6x^2 + 9x + 2$:

$f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3)$.

Stationary points: $x = 1, x = 3$.

$y$-coordinates: $f(1) = 1 - 6 + 9 + 2 = 6$; $f(3) = 27 - 54 + 27 + 2 = 2$.

Second-derivative test: $f''(x) = 6x - 12$. $f''(1) = -6 < 0$, so local max at $(1, 6)$. $f''(3) = 6 > 0$, so local min at $(3, 2)$.

Common traps

Forgetting to find $y$-coordinates

A stationary point is a point; both $x$ and $y$ are needed.

Skipping the classification

$f'(x) = 0$ identifies candidates; the test tells you which type.

Treating local extrema as global

Always check the domain endpoints.

Domain restriction in optimisation

A rectangle cannot have negative or zero side lengths. Enforce the practical domain.

In one sentence

Differentiation locates stationary points by setting $f'(x) = 0$, classifies them by the first-derivative sign test or the sign of $f''(x)$, gives tangent ($m = f'(a)$) and normal ($m = -1/f'(a)$) line equations, and solves optimisation problems by modelling the quantity to optimise as a single-variable function and finding its extremum (with domain endpoints checked).

Examples in context

Example 1. Minimising fencing. A farmer encloses a rectangular paddock of area $200 \text{ m}^2$ against a straight river (no fence needed along the river). If $x$ is the side perpendicular to the river, the depth is $\frac{200}{x}$ and fencing used is $L = 2x + \frac{200}{x}$. Then $L'(x) = 2 - \frac{200}{x^2} = 0$ gives $x^2 = 100$, so $x = 10$ m. Since $L''(x) = \frac{400}{x^3} > 0$, this is a minimum. Minimum fencing $L = 20 + 20 = 40$ m.

Example 2. Maximum profit. A company's profit (thousands of dollars) from producing $x$ hundred units is $P(x) = -x^3 + 12x^2 - 36x + 50$. Then $P'(x) = -3x^2 + 24x - 36 = -3(x - 2)(x - 6)$. Stationary points at $x = 2$ and $x = 6$. As $P''(x) = -6x + 24$, $P''(6) = -12 < 0$, so $x = 6$ is the local maximum: P(6) = -216 + 432 - 216 + 50 = \50{,}000$.

Try this

Q1. Find and classify the stationary points of $f(x) = x^3 - 3x^2$. [3 marks]

• Cue. $f'(x) = 3x^2 - 6x = 3x(x - 2)$; $x = 0$ (local max, $f''(0) = -6$) and $x = 2$ (local min, $f''(2) = 6$).

Q2. Find the equation of the tangent to $y = x^2 - 4x$ at $x = 3$. [3 marks]

• Cue. $f(3) = -3$, $f'(x) = 2x - 4$, $f'(3) = 2$; tangent $y + 3 = 2(x - 3)$, so $y = 2x - 9$.

Q3. A rectangle has perimeter $40$ cm. (a) Express its area $A$ in terms of one side $x$. (b) Find the dimensions giving maximum area. [2+2 marks]

• Cue. (a) $A = x(20 - x) = 20x - x^2$. (b) $A'(x) = 20 - 2x = 0 \Rightarrow x = 10$; a $10 \times 10$ square, area $100 \text{ cm}^2$.