A stationary point is where the gradient of a curve is zero, so you find it by solving f′(x) = 0 and substituting back into the curve for y. The second derivative, f″(x) or d²y/dx², classifies it: f″(x) > 0 means a minimum and f″(x) < 0 means a maximum. If f″(x) = 0 the test is inconclusive, so check the sign of the gradient either side.
Stationary points and the second derivative
- Stationary point: f′(x) = 0
- Second derivative: f″(x), also written d²y/dx²
- f″(x) > 0 at a stationary point: minimum
- f″(x) < 0 at a stationary point: maximum
- f″(x) = 0: inconclusive, so check the sign of f′(x) either side
How do you find stationary points?
Differentiate, set f′(x) = 0 and solve for x. Then substitute each x-value into the original equation, not the derivative, to get the y-coordinate. A curve can have several stationary points, so solve completely.
What is the second derivative?
Differentiating a second time gives the second derivative, f″(x) or d²y/dx². It measures how the gradient itself is changing. Rewrite roots and reciprocals as powers of x before each differentiation, exactly as for the first derivative.
Is a stationary point a maximum or a minimum?
Substitute the x-value of the stationary point into f″(x). A positive value means the curve bends upwards into a valley, so the point is a minimum. A negative value means it bends downwards into a peak, so the point is a maximum. Always state the value of f″(x) and the conclusion; that is where the marks are.
When the second derivative is zero
If f″(x) = 0 at a stationary point, the test fails: the point could be a maximum, a minimum or a point of inflection. Instead, find the sign of f′(x) just before and just after the point. A change from negative to positive means a minimum, positive to negative means a maximum, and no change means a point of inflection.
Optimisation
Optimisation finds the largest or smallest value of a quantity in context. Write the quantity as a function of one variable, using any given constraint to eliminate the other. Differentiate, set the derivative to zero and solve, confirm the type with the second derivative, then work out the maximum or minimum value and answer in the context of the question.
Worked examples
Example 1. Find the coordinates of the stationary point of y = x² − 4x + 1.
- f′(x) = 2x − 4. Setting 2x − 4 = 0 gives x = 2.
- y = 2² − 4(2) + 1 = −3.
Answer: (2, −3).
Marks: M1 for 2x − 4 = 0, A1 for the coordinates.
Example 2. Find d²y/dx² for y = x⁴ − 2x³ + 5x.
- f′(x) = 4x³ − 6x² + 5.
- f″(x) = 12x² − 12x.
Answer: d²y/dx² = 12x² − 12x.
Marks: M1 for the first derivative, A1 for the second.
Example 3. The curve y = x² − 4x + 1 has a stationary point at x = 2. Classify it.
- f′(x) = 2x − 4, so f″(x) = 2.
- f″(2) = 2 > 0.
Answer: A minimum.
Marks: M1 for f″(x), A1 for the conclusion.
Example 4. Show that y = x⁴ has a minimum at x = 0.
- f′(x) = 4x³ = 0 gives x = 0, and f″(x) = 12x², so f″(0) = 0: the test is inconclusive.
- Check the gradient either side: f′(−1) = −4 < 0 and f′(1) = 4 > 0.
- The gradient changes from negative to positive.
Answer: A minimum at x = 0.
Marks: M1 for noting f″(0) = 0, M1 for the sign test, A1 for the conclusion.
Example 5. A rectangle has area A = x(8 − x) cm². Find the value of x that maximises A, and the maximum area.
- A = 8x − x², so dA/dx = 8 − 2x. Setting 8 − 2x = 0 gives x = 4.
- d²A/dx² = −2 < 0, so this is a maximum.
- A = 4 × 4 = 16.
Answer: x = 4, giving a maximum area of 16 cm².
Marks: M1 for 8 − 2x = 0, A1 for x = 4, B1 for justifying the maximum, A1 for 16.
Common mistakes
- Finding y from the derivative. Substitute the x-value into the original equation to get the y-coordinate.
- Not justifying the nature. State the value or sign of f″(x) and the conclusion, such as “f″(2) = 2 > 0, so a minimum”.
- Reading f″(x) = 0 as a point of inflection. It only means the test is inconclusive; check the sign of f′(x) either side.
- Mixing up the signs. Positive f″(x) means a minimum, negative means a maximum.
- Not answering in context. In optimisation, give the maximum or minimum value of the quantity asked for, with units.
How stationary points are examined
Stationary points and the second derivative belong to Pure 2 (WMA12) in Edexcel International A-Level, and to Year 1 Pure in Edexcel A-Level (9MA0). Common question types:
- find the coordinates of the stationary points of a curve
- use the second derivative to classify each one
- handle a case where the second derivative is zero
- optimise a quantity, such as the surface area or volume of a box
The Pure 2 differentiation exam questions include full optimisation questions with mark schemes. For more practice, use the differentiation worksheet or the A-Level differentiation revision engine.
This builds on increasing and decreasing functions. In Pure 3 the same tests are applied with the chain rule and other rules.
Frequently asked questions
How do you find a stationary point?
Differentiate, set dy/dx = 0, solve for x, then substitute into the original equation to find y.
How does the second derivative show a maximum or minimum?
At a stationary point, f″(x) > 0 means a minimum and f″(x) < 0 means a maximum.
What if the second derivative is zero?
The test is inconclusive. Check the sign of the first derivative just before and after the point to decide.