Classify stationary points with derivative tests
Use sign changes in the first derivative or the second derivative at a stationary input to justify a local maximum or minimum.
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does f″(a) < 0 show at a stationary input a?
What you will learn
- Use the first derivative test to classify a stationary point
- Use the second derivative test when it is conclusive
- Give full justification and recognise an inconclusive test
A sign change in f′ shows the direction of the curve
If f′ changes from positive to negative at a stationary input, the curve changes from increasing to decreasing and has a local maximum. A change from negative to positive gives a local minimum.
If f′ does not change sign, the stationary point is neither a local maximum nor minimum. Inspect signs on both sides rather than guessing from the shape of the formula.
The second derivative can shorten the classification
At a stationary input a, if f″(a) < 0, the curve is locally concave down and has a maximum; if f″(a) > 0 it is locally concave up and has a minimum.
If f″(a) = 0, this test is inconclusive. Use a first-derivative sign test instead. Points of inflexion are not assessed in this syllabus, and conclusions about maxima or minima still need full justification.
Pause and explain
If f″(a) = 0 at a stationary point, what should you do?
Worked example
Classify the stationary points of f(x) = x³ − 3x² − 9x + 1 using the second derivative.
Show the worked solution
- From f′(x) = 3(x − 3)(x + 1), the stationary inputs are −1 and 3.
- f″(x) = 6x − 6. Thus f″(−1) = −12 < 0, so (−1, 6) is a local maximum.
- f″(3) = 12 > 0, so (3, −26) is a local minimum.
Answer Local maximum at (−1, 6); local minimum at (3, −26).
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
For f(x) = x³ − 3x, classify its stationary points at x = −1 and x = 1 using the first derivative test.
Give me a hint
Factor f′ and check its sign in the intervals around each root.
Compare my reasoning
- f′(x) = 3(x − 1)(x + 1).
- At x = −1, f′ changes positive to negative, so there is a local maximum; f(−1) = 2.
- At x = 1, f′ changes negative to positive, so there is a local minimum; f(1) = −2.
Local maximum (−1, 2); local minimum (1, −2).
Look for these in your work
- Derivative sign intervals checked
- Sign changes interpreted correctly
- Coordinates included
Classify the stationary point of f(x) = x⁴ at x = 0. Explain why the second derivative test alone is not enough.
Give me a hint
Check f′ on both sides because f″(0) = 0.
Compare my reasoning
- f′(x) = 4x³ and f″(0) = 0, so the second derivative test is inconclusive.
- For x < 0, f′(x) < 0; for x > 0, f′(x) > 0.
- The curve changes from decreasing to increasing, so (0, 0) is a local minimum.
Local minimum at (0, 0); the first derivative test resolves the inconclusive second derivative test.
Look for these in your work
- Inconclusive case identified
- Signs checked on both sides
- Classification justified
A height model has stationary points at t = 2 and t = 5, with v(t) changing from positive to negative at 2 and negative to positive at 5. Classify each point in context.
Give me a hint
Use the sign of the rate of change on either side.
Compare my reasoning
- At t = 2, height changes from increasing to decreasing.
- This is a local maximum height.
- At t = 5, height changes from decreasing to increasing, so it is a local minimum height.
A local maximum occurs at t = 2; a local minimum occurs at t = 5.
Look for these in your work
- Rate sign interpreted as motion
- Maximum transition identified
- Minimum transition identified
Common mistakes
- Classifying from f′(a) = 0 alone
- Reversing the positive-to-negative and negative-to-positive cases
- Calling f″(a) = 0 a stationary point of inflexion automatically
- Giving a classification without evaluating the derivative sign or second derivative
What can you explain now?
What does f′ changing from negative to positive indicate?
Compare with the explanation
A local minimum
The curve changes from decreasing to increasing.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
How do the first- and second-derivative tests justify a stationary-point classification?
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.