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0580/42 Oct/Nov 2023, Question 9: 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 curve y = 4x³ − x⁴ with x-axis crossings at O and A and maximum at B, differentiate, locate B, and find the gradient at A.

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

Part (a) · 2 marks

Differentiate the expression 4x³ − x⁴ with respect to x.

  1. 1Differentiate term by term: bring the power down, then reduce it by 1.
  2. 24x³ → 12x², −x⁴ → −4x³
  3. 3Answer: 12x² − 4x³

Answer12x² − 4x³

Part (b) · 3 marks

Find the coordinates of the maximum point B on the curve.

  1. 112x² − 4x³ = 0, 4x²(3 − x) = 0
  2. 2B is not at the origin, so x = 3
  3. 3y = 4 × 27 − 81 = 27, B = (3, 27)

Answer(3, 27)

Part (c) · 3 marks

Find the gradient of the curve at the point A where it crosses the x-axis.

  1. 14x³ − x⁴ = 0, x³(4 − x) = 0, so A has x = 4
  2. 2Gradient = 12 × 4² − 4 × 4³ = 192 − 256
  3. 3= −64

Answer−64

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Also searched as: 0580/42/O/N/23 · 0580_w23_qp_42 · 0580 w23 42 · question 9.
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