To classify a stationary point, test the sign of the second derivative, or the sign of the gradient either side of the point. Positive d²y/dx² means a minimum, negative means a maximum, and zero means you must use the gradient test.
This lesson follows finding stationary points and feeds directly into forming an objective function, where you must show that your optimum really is a maximum or minimum.
Two tests, one idea
Both tests ask what the gradient does as you pass through the point.
At a minimum the gradient goes from negative to zero to positive, like a valley. At a maximum it goes from positive to zero to negative, like a hill. The second derivative d²y/dx² is the rate at which the gradient changes, so positive means the gradient is increasing, which matches a valley.
How to classify, step by step
- Find the stationary x-values from dy/dx = 0.
- Differentiate again to get d²y/dx².
- Substitute each x into d²y/dx². Positive gives a minimum, negative gives a maximum.
- If the result is zero, choose a value just left and just right of the point and test the sign of dy/dx.
- Write the conclusion with the point: “(1, 4) is a maximum”.
Worked example
Find and classify the stationary points of y = 2x³ − 9x² + 12x − 1.
Step 1, first derivative: dy/dx = 6x² − 18x + 12 = 6(x − 1)(x − 2). So x = 1 or x = 2.
Step 2, y-values:
When x = 1: y = 2 − 9 + 12 − 1 = 4.
When x = 2: y = 16 − 36 + 24 − 1 = 3.
Step 3, second derivative: d²y/dx² = 12x − 18.
Step 4, substitute:
At x = 1: 12 − 18 = −6, which is negative, so (1, 4) is a maximum.
At x = 2: 24 − 18 = 6, which is positive, so (2, 3) is a minimum.
The maximum value 4 is larger than the minimum value 3. That can look odd at first, but these are local highs and lows, and for this cubic the curve keeps climbing after x = 2.
The mistake to watch for
A common slip is to treat d²y/dx² = 0 as proof of an inflection.
Mistaken working: for y = x⁴, d²y/dx² = 12x² = 0 at x = 0, so x = 0 is a point of inflection.
The student assumed that a zero second derivative always means inflection.
The correction is to test the gradient.
Here dy/dx = 4x³. At x = −1 it is −4 (negative) and at x = 1 it is 4 (positive), so the gradient changes from negative to positive and x = 0 is a minimum. A true stationary inflection has the same sign of gradient on both sides.
Check yourself
Try these without a calculator, then open each answer.
1. Find and classify the stationary points of y = x³ − 3x.
Show answer
dy/dx = 3x² − 3 = 3(x − 1)(x + 1), so x = 1 or x = −1. d²y/dx² = 6x.
At x = 1: y = −2 and d²y/dx² = 6 > 0, so (1, −2) is a minimum.
At x = −1: y = 2 and d²y/dx² = −6 < 0, so (−1, 2) is a maximum.
2. For y = x⁴, show that x = 0 is a minimum using the gradient.
Show answer
dy/dx = 4x³. At x = −0.5 it is −0.5, negative. At x = 0.5 it is 0.5, positive. The gradient changes from negative to positive, so x = 0 is a minimum.
3. For y = x³, explain why (0, 0) is not a maximum or minimum.
Show answer
dy/dx = 3x², which is positive on both sides of x = 0 and zero only at x = 0. The gradient does not change sign, so (0, 0) is a stationary point of inflection.
Where this leads next
With classification secure, move to forming an objective function from a constraint. The quadratic structure explorer lets you see how the turning point of a quadratic relates to its second derivative.
If you can do the calculus but the choice of test still feels arbitrary, one-to-one Additional Mathematics tuition gives you a teacher to talk the decision through with.