Find the value of x at the minimum point of the... - JAMB Mathematics 2011 Question

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 maximum point is -1