Locate stationary points and determine their nature
Solve the zero-gradient equation, evaluate the original curve and use a valid second-derivative or sign test to justify a sketch or an optimum.
Section 1.7 · Differentiation
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Use an informal understanding of limits; formal general differentiation from first principles and points of inflexion are not required in section 1.7. Written practice is self-checked.
Before you begin
- Differentiate polynomials twice and solve factored equations.
- Interpret derivative signs and the difference between a local feature and a whole-domain maximum.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which equation locates a stationary input on a differentiable curve?
What you will learn
- Find stationary point coordinates and identify local maxima or minima.
- Use the second derivative or a derivative sign change, and compare endpoints for a constrained optimum.
First locate the point; then justify its nature
A stationary point on y = f(x) has finite first derivative f′(x) = 0. Solve that equation for the input and substitute into the original function for the height. If f″(a) > 0, a stationary point is a local minimum; if f″(a) < 0, it is a local maximum. A positive-to-negative first-derivative sign change also gives a maximum, and negative-to-positive gives a minimum.
For f(x) = x³ − 6x² + 9x + 1, f′(x) = 3(x − 1)(x − 3), so the stationary inputs are 1 and 3. Their points are (1, 5) and (3, 1). Since f″(x) = 6x − 12, its values −6 and 6 show a local maximum at (1, 5) and a local minimum at (3, 1). Together with the y-intercept (0, 1), the derivative signs and the positive cubic leading term, these facts support a rising–falling–rising sketch.
A zero second derivative is inconclusive
For y = (x − 2)⁴ + 1, the first derivative is 4(x − 2)³ and the second is 12(x − 2)². Both vanish at x = 2, so the second-derivative test alone gives no decision. However the first derivative is negative to the left and positive to the right, proving a minimum at (2, 1). Use another permitted test when none is specified, rather than labelling every zero second derivative a maximum or minimum.
A local maximum need not be the greatest value over the whole domain. The cubic example grows without bound to the right even though it has a local maximum. On a closed model interval, compare stationary values with both endpoint values to find the greatest or least permitted result. Points of inflexion are outside this section's stated requirements; the focus here is stationary points, justified extrema and relevant graph features.
Pause and explain
At a stationary point f″(a) = 0. What can this test alone conclude?
Worked example
Locate and classify the stationary points of y = x³ − 6x² + 9x + 1.
Show the worked solution
- Set y′ = 3(x − 1)(x − 3) = 0, giving x = 1 or 3.
- The original curve gives y = 5 and 1 respectively, so the points are (1, 5) and (3, 1).
- Use y″ = 6x − 12: y″(1) = −6 gives a local maximum and y″(3) = 6 gives a local minimum; the derivative signs confirm both.
Answer Local maximum (1, 5); local minimum (3, 1).
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find and classify the stationary point of y = x² − 4x + 7.
Give me a hint
Solve the first derivative, find the curve height and then use the second derivative.
Compare my reasoning
- y′ = 2x − 4 is zero at x = 2.
- The original curve gives y(2) = 3, so the point is (2, 3).
- y″ = 2 > 0, proving a minimum; completing the square to (x − 2)² + 3 independently confirms it.
Minimum at (2, 3).
Look for these in your work
- Find both stationary coordinates.
- Justify the nature with a valid positive second derivative.
Find and classify the stationary point of y = (x − 2)⁴ + 1 without making a conclusion from its zero second derivative alone.
Give me a hint
A first-derivative sign test works even when the second test is inconclusive.
Compare my reasoning
- y′ = 4(x − 2)³ is zero only at x = 2, where y = 1.
- y″ = 12(x − 2)² is zero there, so that test alone is inconclusive.
- The first derivative is negative for x < 2 and positive for x > 2, proving a minimum at (2, 1).
Minimum at (2, 1), by a negative-to-positive first-derivative sign change.
Look for these in your work
- Recognise an inconclusive test rather than treating it as a classification.
- Use both neighbouring derivative signs to justify the minimum.
A supplied output model is R(x) = −x² + 10x + 4 for 0 ≤ x ≤ 8. Find its greatest value on the permitted interval and justify it.
Give me a hint
Find any interior stationary input, classify it and compare with both included endpoints.
Compare my reasoning
- R′(x) = −2x + 10 vanishes at x = 5, within the domain; R″ = −2 < 0 gives a local maximum.
- Evaluate R(5) = 29, R(0) = 4 and R(8) = 20.
- The model increases before 5 and decreases after it; together with the endpoint comparison, its greatest permitted value is 29 at x = 5.
Greatest permitted value 29 at x = 5; endpoint values 4 and 20 are smaller.
Look for these in your work
- Retain only stationary candidates in the permitted domain.
- Compare the interior value with both closed endpoints for a whole-interval conclusion.
Common mistakes
- A stationary-point answer needs both coordinates and a justified nature.
- A zero second derivative makes that test inconclusive; it does not prove the point's nature.
- A constrained optimum must include permitted endpoint comparisons.
What can you explain now?
For y = (x − 2)⁴ + 1, classify its stationary point when the second derivative is zero there.
Compare with the explanation
Minimum at (2, 1), justified by a first-derivative change from negative to positive.
The second-derivative test is inconclusive, but 4(x − 2)³ has the required sign change.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Describe the full stationary-point routine from f′ = 0 to a labelled sketch. Explain what to do when f″ = 0 and when a model has closed endpoints.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.