Find stationary points by setting the first derivative to zero and determine their nature using the first or second derivative test

What this dot point is asking

SEAB wants you to find the stationary points of a curve, where the gradient is zero, and to classify each as a maximum, a minimum or a point of inflexion. This is the basis of optimisation, where you find the greatest or least value of a quantity, and it appears in nearly every Paper 2 calculus question.

The answer

Finding stationary points

A stationary point is where the tangent is horizontal, so the gradient is zero:

Solve this equation for $x$, then substitute each solution into the original curve to find the $y$-coordinate. The pairs $(x, y)$ are the stationary points.

The second derivative test

The fastest classification uses the second derivative $\dfrac{d^2y}{dx^2}$ at each stationary point:

A positive second derivative means the curve bends upward (a valley); a negative one means it bends downward (a hill).

When the second derivative is zero

If $\dfrac{d^2y}{dx^2} = 0$, the test is inconclusive, and you fall back on the first derivative test.

The first derivative test

Examine the sign of $\dfrac{dy}{dx}$ just before and just after the point. Changing from positive to negative is a maximum; negative to positive is a minimum; no sign change is a point of inflexion (a stationary point that is neither a peak nor a trough).

Setting up an optimisation problem

The most valued application is optimisation, and the setup is where marks are won or lost. The routine is: write the quantity to be optimised as a function of one variable (using a constraint to eliminate any second variable), differentiate, set the derivative to zero to find the stationary point, then confirm it is the maximum or minimum required using the second derivative. For a box of fixed volume, you would express the surface area in terms of one dimension using the volume constraint, then minimise. The skill is reducing the problem to a single-variable function before differentiating, because the calculus only starts once the quantity is written in terms of one variable.

When to prefer the first derivative test

Although the second derivative test is usually quicker, the first derivative test is the better choice in two situations: when the second derivative is awkward to compute (for instance after a messy quotient rule), and when the second derivative comes out zero and the test is inconclusive. In those cases, checking the sign of $\tfrac{dy}{dx}$ just either side of the stationary point reliably gives the nature. A neat way to present this is a small sign table of the gradient across the point, which also distinguishes a genuine inflexion (no sign change) from a true maximum or minimum.

Examples in context

Example 1. Maximum enclosed area. Given a fixed length of fencing, expressing the enclosed area as a function of one side and finding its stationary point gives the dimensions of greatest area, the archetypal optimisation problem.

Example 2. Minimum material for a can. Writing the surface area of a cylindrical can of fixed volume in terms of its radius and minimising it finds the most economical shape, a direct industrial use of stationary points.

Try this

Q1. Find the stationary point of $y = x^2 - 4x + 1$. [3 marks]

• Cue. $\dfrac{dy}{dx} = 2x - 4 = 0$ gives $x = 2$, $y = -3$; point $(2, -3)$.

Q2. State the nature of a stationary point where $\dfrac{d^2y}{dx^2} = 5$. [1 mark]

• Cue. Positive, so a minimum.

Q3. Find and classify the stationary points of $y = x^3 - 12x$. [4 marks]

• Cue. $3x^2 - 12 = 0$ gives $x = \pm 2$; $\dfrac{d^2y}{dx^2} = 6x$, so $x = 2$ is a minimum and $x = -2$ is a maximum.