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:

• IMATH_7 : local minimum (concave up).
• IMATH_8 : local maximum (concave down).
• IMATH_9 : 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).