JAMB 1999 · UME · Q40

Find the value of $x$ for which the function $y = x^3 - x$ has a minimum value.

Worked solution (try it first)

At a turning point $\frac{dy}{dx} = 3x^2 - 1 = 0$ , so $x^2 = \frac13$ and $x = \pm\frac{1}{\sqrt3}$
$= \pm\frac{\sqrt3}{3}$ .
$\frac{d^2y}{dx^2} = 6x$ , which is positive at $x = \frac{\sqrt3}{3}$ , so that is the minimum.
So $x = \frac{\sqrt3}{3}$ , option C.