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

What this dot point is asking

SEAB wants you to find the stationary points of a curve, the points where it momentarily levels off, and decide the nature of each: a maximum (a peak), a minimum (a trough), or occasionally a point of inflexion. Stationary points are where the gradient is zero, so you find them by setting the derivative to zero, then classify each one. This is the basis of all maximum and minimum (optimisation) problems.

The answer

What a stationary point is

A stationary point is where the curve has gradient zero, so the tangent is horizontal. Since the gradient is the derivative, stationary points occur where:

Finding the stationary points

1. Differentiate to get $\dfrac{dy}{dx}$.
2. Set it to zero and solve for $x$. Each solution is the $x$-coordinate of a stationary point.
3. Substitute each $x$ back into the original curve to get the $y$-coordinate.

Classifying with the second derivative

The fastest test is the second derivative $\dfrac{d^2y}{dx^2}$, found by differentiating again. At a stationary point:

• if $\dfrac{d^2y}{dx^2} > 0$, the curve is concave up, so it is a minimum,
• if $\dfrac{d^2y}{dx^2} < 0$, the curve is concave down, so it is a maximum,
• if $\dfrac{d^2y}{dx^2} = 0$, the test is inconclusive and you fall back on a sign test.

Classifying with a sign test

Alternatively, check the sign of $\dfrac{dy}{dx}$ just before and just after the point:

• changes from negative to positive: minimum (the curve falls then rises),
• changes from positive to negative: maximum (the curve rises then falls).

Either method is acceptable; the second derivative is usually quicker when it is easy to compute.

Examples in context

Example 1. Maximising an area. To find the largest rectangular area for a fixed length of fencing, write the area as a function of one side, differentiate, set to zero to find the stationary point, and confirm it is a maximum. Stationary points are the heart of every optimisation question.

Example 2. Minimum material for a container. A manufacturer minimising the metal used for a tin of fixed volume writes the surface area as a function, finds the stationary point, and checks it is a minimum with the second derivative. The same two-step routine, locate then classify, solves real design problems.

Try this

Q1. Find the $x$-coordinate of the stationary point of $y = x^2 - 10x$. [2 marks]

• Cue. $\dfrac{dy}{dx} = 2x - 10 = 0$, so $x = 5$.

Q2. A curve has $\dfrac{d^2y}{dx^2} = -4$ at its stationary point. State its nature. [1 mark]

• Cue. Negative second derivative means a maximum.

Q3. Find the stationary point of $y = x^2 + 2x + 3$ and state its nature. [3 marks]

• Cue. $2x + 2 = 0$ gives $x = -1$, $y = 2$; second derivative $2 > 0$, so $(-1, 2)$ is a minimum.