y = 3x\(^3\) - 9x\(^2\) \(\frac{dy}{dx}\) = 9x\(^2\) - 18x As \(\frac{dy}{dx}\) = 0 9x\(^2\) - 18x = 0 9x(x-2) = 0 9x = 0 which implies x = 0 x - 2 = 0 implies x = 2 \(\frac{d^2y}{dx^2}\) = 18x - 18 when x = 0 \(\frac{d^2y}{dx^2}\) > 0 ∴ x is minimum when x = 2, \(\frac{d^2y}{dx^2}\) = 18 ∴ the value > 0 x is minimum