To find stationary points, differentiate, set dy/dx = 0, solve for x, then substitute each x into the original equation to get y. The answer is a pair of coordinates, not just an x-value.
This skill opens stationary points and optimisation and is needed in every later lesson. It relies on the rules in differentiation techniques.
What makes a point stationary?
The derivative dy/dx measures the gradient of the curve. Where the gradient is zero, the tangent is horizontal and the curve is momentarily neither rising nor falling.
Picture the crest of a hill or the floor of a valley. At that instant you are walking on level ground. The height there is what optimisation questions eventually ask for.
How to find them, step by step
- Differentiate y with respect to x, writing negative or fractional powers in index form first if needed.
- Set dy/dx = 0 and factorise or solve. A cubic’s derivative is a quadratic, so factorising usually works.
- Solve for every x. A quadratic derivative can give two values.
- Substitute each x into y, the original equation.
- Write each answer as a point, such as (1, 6).
Worked example
Find the stationary points of y = x³ − 6x² + 9x + 2.
Step 1, differentiate: dy/dx = 3x² − 12x + 9.
Step 2, set to zero: 3x² − 12x + 9 = 0. Divide by 3 to get x² − 4x + 3 = 0.
Step 3, solve: (x − 1)(x − 3) = 0, so x = 1 or x = 3.
Step 4, substitute into y:
When x = 1: y = 1 − 6 + 9 + 2 = 6.
When x = 3: y = 27 − 54 + 27 + 2 = 2.
Step 5, answer: the stationary points are (1, 6) and (3, 2).
A quick check is to see that the two x-values sit either side of the middle of the curve and that 6 > 2, which suits a cubic with a positive x³ term rising, falling, then rising again.
The mistake to watch for
A common slip is to substitute the x-values into dy/dx, or to set y = 0 at the start.
Mistaken working: x = 1 gives 3(1) − 12 + 9 = 0, so the point is (1, 0).
The student reused the derivative, which is zero by construction, and reported that as the y-coordinate.
The correction is to remember what each equation measures. The derivative gives the gradient and the original equation gives the height. Once you have solved dy/dx = 0, every x goes back into y.
Check yourself
Try these without a calculator, then open each answer.
1. Find the stationary point of y = x² − 8x + 3.
Show answer
dy/dx = 2x − 8 = 0, so x = 4. Then y = 16 − 32 + 3 = −13.
(4, −13)
2. Find the stationary points of y = 2x³ − 3x² − 12x.
Show answer
dy/dx = 6x² − 6x − 12 = 6(x² − x − 2) = 6(x − 2)(x + 1), so x = 2 or x = −1.
When x = 2: y = 16 − 12 − 24 = −20. When x = −1: y = −2 − 3 + 12 = 7.
(2, −20) and (−1, 7)
3. Find the stationary points of y = x + 4/x, where x ≠ 0.
Show answer
Write y = x + 4x⁻¹, so dy/dx = 1 − 4x⁻² = 1 − 4/x². Setting this to zero gives x² = 4, so x = 2 or x = −2.
When x = 2: y = 2 + 2 = 4. When x = −2: y = −2 − 2 = −4.
(2, 4) and (−2, −4)
Where this leads next
Finding the points is half the job. Next, classify stationary behaviour using an appropriate test to decide whether each one is a maximum, a minimum or neither. The non-calculator working trainer is handy for keeping the substitution arithmetic accurate.
Some students follow each step here but freeze when the derivative is not a neat quadratic. A teacher in online one-to-one Additional Mathematics tuition can see exactly which step you hesitate at.