y = x3 + x2 - x + 1 \(\frac{dy}{dx}\) = \(\frac{d(x^3)}{dx}\) + \(\frac{d(x^2)}{dx}\) - \(\frac{d(x)}{dx}\) + \(\frac{d(1)}{dx}\) \(\frac{dy}{dx}\) = 3x2 + 2x - 1 = 0 \(\frac{dy}{dx}\) = 3x2 + 2x - 1 At the maximum point \(\frac{dy}{dx}\) = 0 3x2 + 2x - 1 = 0 (3x2 + 3x) - (x - 1) = 0 3x(x + 1) -1(x + 1) = 0 (3x - 1)(x + 1) = 0 therefore x = \(\frac{1}{3}\) or -1 For the maximum point \(\frac{d^2y}{dx^2}\) < 0 \(\frac{d^2y}{dx^2}\) 6x + 2 when x = \(\frac{1}{3}\) \(\frac{dx^2}{dx^2}\) = 6(\(\frac{1}{3}\)) + 2 = 2 + 2 = 4 \(\frac{d^2y}{dx^2}\) > o which is the minimum point when x = -1 \(\frac{d^2y}{dx^2}\) = 6(-1) + 2 = -6 + 2 = -4 -4 < 0 therefore, \(\frac{d^2y}{dx^2}\) < 0 the minimum point is 1/3