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0580/42 Oct/Nov 2024, Question 10: Stationary Points and Turning Points

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 9 marks

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The question, in short

For the curve y = x³ − 9x² − 48x, find its derivative, locate its turning points, and classify each turning point as a maximum or minimum with justification.

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

Part (a) · 2 marks

Work out the derivative of x³ − 9x² − 48x.

  1. 1Rule: xⁿ differentiates to nxⁿ⁻¹.
  2. 2x³ → 3x², 9x² → 18x, 48x → 48
  3. 3dy/dx = 3x² − 18x − 48

Answer3x² − 18x − 48

Part (b) · 4 marks

Calculate the coordinates of the stationary points on y = x³ − 9x² − 48x, showing every step.

  1. 1At turning points 3x² − 18x − 48 = 0
  2. 2x² − 6x − 16 = 0
  3. 3(x − 8)(x + 2) = 0, so x = −2 or x = 8
  4. 4x = −2: y = −8 − 36 + 96 = 52; x = 8: y = 512 − 576 − 384 = −448
  5. 5(−2, 52) and (8, −448)

Answer(−2, 52) and (8, −448)

Part (c) · 3 marks

Given turning points (−2, 52) and (8, −448), decide which is a maximum and which is a minimum, explaining your reasoning.

  1. 1d²y/dx² = 6x − 18
  2. 2At x = −2: 6(−2) − 18 = −30 < 0, so (−2, 52) is a maximum
  3. 3At x = 8: 6(8) − 18 = 30 > 0, so (8, −448) is a minimum

Answer(−2, 52) maximum with reason and (8, −448) minimum with reason and no incorrect statement

Stuck on a step? Read the free lesson: Stationary Points and Turning Points

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Also searched as: 0580/42/O/N/24 · 0580_w24_qp_42 · 0580 w24 42 · question 10.
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