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0580/43 May/Jun 2024, Question 10: Stationary Points and Turning Points

Cambridge IGCSE Mathematics 0580 · Paper 43 (Extended) · Calculator · 11 marks

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The question, in short

For y = x⁷ − 7x⁶, differentiate, find the tangent line at x = −1 in y = mx + c form, and locate both turning points showing working.

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

Part (a) · 2 marks

Differentiate y = x⁷ − 7x⁶ with respect to x.

  1. 1Multiply by the power, then take 1 off the power: axⁿ → naxⁿ⁻¹
  2. 2x⁷ → 7x⁶
  3. 3−7x⁶ → 6 × (−7)x⁵ = −42x⁵
  4. 4dy/dx = 7x⁶ − 42x⁵

Answer7x⁶ − 42x⁵

Part (b) · 4 marks

Find the tangent to y = x⁷ − 7x⁶ at x = −1, giving the equation as y = mx + c.

  1. 1y = (−1)⁷ − 7(−1)⁶ = −1 − 7 = −8
  2. 2m = 7(−1)⁶ − 42(−1)⁵ = 7 + 42 = 49
  3. 3−8 = 49(−1) + c, so c = 41
  4. 4y = 49x + 41

Answery = 49x + 41

Part (c) · 5 marks

Find the coordinates of the two turning points of y = x⁷ − 7x⁶, showing all working.

  1. 17x⁶ − 42x⁵ = 0
  2. 27x⁵(x − 6) = 0
  3. 3x = 0 or x = 6
  4. 4x = 0: y = 0
  5. 5x = 6: y = 6⁷ − 7 × 6⁶ = 6⁶(6 − 7) = −46656

Answer(0, 0) and (6, −46656)

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Also searched as: 0580/43/M/J/24 · 0580_s24_qp_43 · 0580 s24 43 · question 10.
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