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0580/42 Oct/Nov 2021, Question 10: Stationary Points and Turning Points

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 8 marks

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The question, in short

For the cubic y = x³ − 12x + 6, locate its turning points with full working, then classify each given turning point as maximum or minimum and justify the choice.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 5 marks

Differentiate y = x³ − 12x + 6, solve for the x-values where the gradient is zero, and state the coordinates of both turning points, showing all steps.

  1. 1dy/dx = 3x² − 12
  2. 23x² − 12 = 0, so x² = 4, x = 2 or x = −2
  3. 3x = 2: y = 8 − 24 + 6 = −10
  4. 4x = −2: y = −8 + 24 + 6 = 22
  5. 5Turning points (2, −10) and (−2, 22)

Answer(2, −10) and (−2, 22)

Part (b) · 3 marks

Using the turning points (2, −10) and (−2, 22), decide which is a maximum and which is a minimum, explaining the reasoning used.

  1. 1d²y/dx² = 6x
  2. 2At x = 2: d²y/dx² = 12 > 0, so (2, −10) is a minimum
  3. 3At x = −2: d²y/dx² = −12 < 0, so (−2, 22) is a maximum

Answer(2, −10) minimum with correct reason or sketch; (−2, 22) maximum with correct reason or sketch

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Also searched as: 0580/42/O/N/21 · 0580_w21_qp_42 · 0580 w21 42 · question 10.
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