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0580/43 May/Jun 2023, Question 12: Stationary Points and Turning Points

Cambridge IGCSE Mathematics 0580 · Paper 43 (Extended) · Calculator · 8 marks

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The question, in short

For the curve y = x⁴ − 8x² + 5, find its derivative, locate the three turning points with full working, and identify which turning point is a maximum with justification.

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

Part (a) · 2 marks

Differentiate y = x⁴ − 8x² + 5 to obtain dy/dx.

  1. 1Multiply by the power, then take 1 off the power: axⁿ → naxⁿ⁻¹
  2. 2x⁴ → 4x³
  3. 3−8x² → 2 × (−8)x = −16x
  4. 45 → 0
  5. 5dy/dx = 4x³ − 16x

Answer4x³ − 16x

Part (b) · 4 marks

Solve dy/dx = 0 and find the coordinates of all three turning points, showing all working.

  1. 1dy/dx = 0: 4x³ − 16x = 0
  2. 24x(x − 2)(x + 2) = 0, so x = 0, 2, −2
  3. 3x = 0: y = 5
  4. 4x = ±2: y = 16 − 32 + 5 = −11
  5. 5(0, 5), (−2, −11), (2, −11)

Answer(0, 5), (−2, −11), (2, −11)

Part (c) · 2 marks

Given turning points (0, 5), (−2, −11) and (2, −11), decide which is a maximum and justify your choice.

  1. 1d²y/dx² = 12x² − 16
  2. 2At x = 0: d²y/dx² = −16 < 0
  3. 3So (0, 5) is a maximum

Answer(0, 5) with correct reasoning, e.g. d²y/dx² = 12x² − 16 = −16 < 0 at x = 0

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

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Also searched as: 0580/43/M/J/23 · 0580_s23_qp_43 · 0580 s23 43 · question 12.
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