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If (y = x^{3} – x^{2} – x + 6), find the values of x…

If (y = x^{3} – x^{2} – x + 6), find the values of x at the turning point.

  • A.
    (frac{1}{2}, 3)
  • B.
    (frac{1}{3}, -frac{1}{2})
  • C.
    (1, -frac{1}{3})
  • D.
    (1, frac{1}{3})
Correct Answer: Option C
Explanation

At turning point, (frac{mathrm d y}{mathrm d x} = 0).

Given (x^{3} – x^{2} – x + 6 )

(frac{mathrm d y}{mathrm d x} = 3x^{2} – 2x – 1 = 0 )

(3x^{2} – 3x + x – 1 = 0 implies (3x + 1)(x – 1) = 0)

(x = frac{-1}{3}, 1)