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JAMB 2018 · UTME · Q28

Find the value of xx for which the function f(x)=2x3−x2−4x+4f(x) = 2x^3 - x^2 - 4x + 4 has a maximum value.

Worked solution (try it first)
  1. At a turning point f′(x)=6x2−2x−4=0f'(x) = 6x^2 - 2x - 4 = 0.
  2. Factorise: 2(3x+2)(x−1)=02(3x + 2)(x - 1) = 0, so x=−23x = -\frac23 or x=1x = 1.
  3. f′′(x)=12x−2f''(x) = 12x - 2.
  4. At x=−23x = -\frac23 it is −10<0-10 < 0 (maximum).
  5. At x=1x = 1 it is 10>010 > 0 (minimum).
  6. So the maximum is at x=−23x = -\frac23, option C.

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