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4MA1/1H Jan 2019, Question 12: Gradient of a Curve at a Point

Pearson Edexcel International GCSE Mathematics A 4MA1 · Paper 1H (Higher) · Calculator · 5 marks

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The question, in short

Differentiate the cubic y = (1/3)x³ − 9x + 1, then determine the x-values where the curve's gradient is negative.

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

Part (a) · 2 marks

Find dy/dx of y = (1/3)x³ − 9x + 1.

  1. 1Differentiate each term: (1/3)x³ → 3 × (1/3)x² = x²
  2. 2−9x → −9, and the constant 1 → 0
  3. 3dy/dx = x² − 9

Answerx² − 9

Part (b) · 3 marks

Using the derivative, find the range of x values for which the gradient of C is negative.

  1. 1Negative gradient means dy/dx < 0: x² − 9 < 0
  2. 2x² < 9, so x is between the roots −3 and 3
  3. 3−3 < x < 3

Answer−3 < x < 3

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