WAEC Further Mathematics 2018 Paper

To find the maximum value, we can use the second derivative test where, given $f(x)$, the second derivative < 0, makes it a maximum value.

$x(x+1)^2 = x(x^2+2x+1) = x^3+2x^2+x$

$\frac{dy}{dx} = 3x^2+4x+1=0$

Solving, we have $x = -\frac{1}{3}$ or $-1$.

$\frac{d^2y}{dx^2} = 6x+4$

When $x = -\frac{1}{3},\ \frac{d^2y}{dx^2} = 2 > 0$

When $x = -1,\ \frac{d^2y}{dx^2} = -2 < 0$

At maximum value of x being -1, $y = -1(-1+1)^2 = 0$