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WAEC Further Mathematics 2018 Question 24

Question 24

Differentiation

Given that y = x(x + 1)2, calculate the maximum value of y.

  1. A.−2
  2. B.0
  3. C.1
  4. D.2
Show the answer and explanation

Answer: B (0)

Explanation

dy/dx = 3x² + 4x + 1 = (3x + 1)(x + 1) = 0. d²y/dx² = 6x + 4 is negative at x = −1, so the maximum value is y(−1) = 0.