(a) Find the maximum and minimum points of the curve (y = 2x^{3} – 3x^{2} – 12x + 4).
(b) Sketch the curve in (a) above.
Explanation
(a) (y = 2x^{3} – 3x^{2} – 12x + 4)
At stationary points, (frac{mathrm d y}{mathrm d x} = 0).
(frac{mathrm d y}{mathrm d x} = 6x^{2} – 6x – 12 = 0)
(implies x^{2} – x – 2 = 0)
(x^{2} – 2x + x – 2 = 0 implies x(x – 2) + 1(x – 2) = 0)
(x = -1 ; x = 2)
When (x = -1), (y = 2(-1)^{3} – 3(-1)^{2} – 12(-1) + 4 = 11)
(frac{mathrm d^{2} y}{mathrm d x^{2}} = 12x – 6 = 12(-1) – 6 = -18 < 0)
(therefore (-1, 11) = text{maximum point})
When (x = 2), (y = 2(2)^{3} – 3(2)^{2} – 12(2) + 4 = -16)
(frac{mathrm d^{2} y}{mathrm d x^{2}} = 12x – 6 = 12(2) – 6 = 18 > 0)
(therefore (2, -16) = text{minimum point})
(b)