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0580/42 Feb/Mar 2023, Question 9: Stationary Points and Turning Points

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

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The question, in short

For f(x) = x³ − 3x² − 4, find the gradient at x = 1, locate the turning points, and sketch the curve on the given axes.

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

Part (a) · 3 marks

Differentiate f(x) and evaluate the derivative at x = 1 to give the gradient.

  1. 1The gradient is the value of dy/dx
  2. 2x³ → 3x², −3x² → −6x, −4 → 0
  3. 3dy/dx = 3x² − 6x
  4. 4At x = 1: 3(1)² − 6(1) = 3 − 6 = −3

Answer−3

Part (b) · 4 marks

Solve f′(x) = 0, then find the y-coordinates to give both turning points.

  1. 13x² − 6x = 0
  2. 23x(x − 2) = 0, so x = 0 or x = 2
  3. 3f(0) = −4 and f(2) = 8 − 12 − 4 = −8
  4. 4(0, −4) and (2, −8)

Answer(0, −4) and (2, −8)

Part (c) · 2 marks

Sketch y = f(x) on the axes, showing the turning points and the y-intercept.

  1. 1Positive cubic: rises from bottom left, ends top right
  2. 2Maximum at (0, −4) on the negative y-axis, minimum at (2, −8) in the fourth quadrant
  3. 3Curve then rises and crosses the x-axis once, between x = 3 and x = 4, into the first quadrant

AnswerCorrect sketch (maximum on the negative y-axis, minimum in the correct quadrant, curve extends into the first quadrant)

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

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Also searched as: 0580/42/F/M/23 · 0580_m23_qp_42 · 0580 m23 42 · question 9.
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