(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) 