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JAMB 2011 · UTME · Q38

Find the value of xx at the minimum point of the curve y=x3+x2−x+1y = x^3 + x^2 - x + 1.

Worked solution (try it first)
  1. At a turning point dydx=3x2+2x−1=0\frac{dy}{dx} = 3x^2 + 2x - 1 = 0.
  2. Factorise: (3x−1)(x+1)=0(3x - 1)(x + 1) = 0, so x=13x = \frac13 or x=−1x = -1.
  3. d2ydx2=6x+2\frac{d^2y}{dx^2} = 6x + 2.
  4. At x=13x = \frac13 it is 4>04 > 0 (minimum).
  5. At x=−1x = -1 it is −4<0-4 < 0 (maximum).
  6. So the minimum point is at x=13x = \frac13, option A.

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