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0580/21 May/Jun 2023, Question 21: Gradient of a Curve at a Point

Cambridge IGCSE Mathematics 0580 · Paper 21 (Extended) · Calculator · 6 marks

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The question, in short

For the cubic y = x³ − 12x, work out the slope at the point (1, −11) and locate all stationary points, giving their coordinates.

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Part (a) · 3 marks

Differentiate the cubic and evaluate the derivative at x = 1 to obtain the slope at (1, −11).

  1. 1dy/dx = 3x² − 12
  2. 2At x = 1: 3 × 1² − 12 = −9
  3. 3Gradient = −9

Answer−9

Part (b) · 3 marks

Set the derivative to zero, solve for x, then substitute back into y = x³ − 12x to get both turning point coordinates.

  1. 1At turning points dy/dx = 0: 3x² − 12 = 0
  2. 2x² = 4, so x = 2 or x = −2
  3. 3x = −2: y = (−2)³ − 12 × (−2) = −8 + 24 = 16
  4. 4x = 2: y = 8 − 24 = −16
  5. 5(−2, 16) and (2, −16)

Answer(−2, 16) and (2, −16)

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