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Additional Mathematics · Differentiation

Finding stationary points and testing their nature

You can find where the gradient is zero, but deciding maximum or minimum feels like a guess.

A stationary point is where the gradient is zero. To find it, solve dy/dx = 0 for x, then substitute into the curve for y. To decide the nature, use the second derivative.

This lesson is part of SPM Additional Mathematics differentiation. It uses the rules from differentiating powers, products and quotients.

The method in four steps

  1. Differentiate: find dy/dx.
  2. Solve dy/dx = 0 to get the x-values.
  3. Substitute each x into y to get the coordinates.
  4. Find d²y/dx² and test each x: negative means maximum, positive means minimum.

Worked example

Find the stationary points of y = x³ − 6x² + 9x + 2 and state their nature.

Gradient. dy/dx = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3).

Stationary x-values. 3(x − 1)(x − 3) = 0 gives x = 1 or x = 3.

Coordinates. At x = 1: y = 1 − 6 + 9 + 2 = 6, so the point is (1, 6). At x = 3: y = 27 − 54 + 27 + 2 = 2, so the point is (3, 2).

Nature. d²y/dx² = 6x − 12. At x = 1 it is −6, which is negative, so (1, 6) is a maximum. At x = 3 it is 6, which is positive, so (3, 2) is a minimum.

Why the second derivative works

The second derivative tells you how the gradient is changing. A negative value means the gradient is falling, so it goes from positive through zero to negative, and the curve peaks. A positive value is the reverse, and the curve bottoms out.

The mistake that costs marks

The usual slip is to stop after finding x and leave the nature untested, or to state the nature without evidence.

Step Weak answer Full answer
Point x = 1 (1, 6)
Nature “It is a maximum” d²y/dx² = −6 < 0, so maximum

When d²y/dx² = 0, the test is silent. For y = x³ at x = 0, the gradient 3x² is positive on both sides, so the point is an inflexion and not a turning point.

Check yourself

Find the stationary points of y = 2x³ − 9x² + 12x − 1 and state their nature.

Answer

dy/dx = 6x² − 18x + 12 = 6(x − 1)(x − 2), so x = 1 or x = 2.

At x = 1: y = 2 − 9 + 12 − 1 = 4. At x = 2: y = 16 − 36 + 24 − 1 = 3.

d²y/dx² = 12x − 18. At x = 1 it is −6, so (1, 4) is a maximum. At x = 2 it is 6, so (2, 3) is a minimum.

What to study next

Stationary points lead straight into optimisation and rates of change. If a question limits x to an interval, read comparing endpoints and stationary points in a constrained optimisation.

To have a teacher read your working and tell you which line loses the mark, see online one-to-one Additional Mathematics tuition.

Common questions

What is a stationary point?

A stationary point is a point on a curve where the gradient is zero, so dy/dx = 0. It can be a maximum, a minimum or, for some curves, a point of inflexion. The value of x comes from solving dy/dx = 0, and y comes from substituting back into the curve.

How do I tell a maximum from a minimum?

Find d²y/dx² and substitute each x-value. If the result is negative, the point is a maximum. If it is positive, the point is a minimum.

What if the second derivative is zero?

The test does not decide. Check the sign of dy/dx just before and just after the point. A change from positive to negative means a maximum, negative to positive means a minimum, and no change in sign means a point of inflexion.

Do I give the x-value or the coordinates?

Give the coordinates unless the question asks only for x. A full answer states both the point and its nature, such as a minimum point at (3, 2).

If you find the stationary points but lose marks on the nature or the coordinates, a one-to-one teacher can check your layout and show the full-mark version.

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