Calculus (3rd Edition)

$f(x)$ is increasing on (-$\infty$,0) $\cup$ (0,$\infty$)
$f$($x$) =$x^{3}$ $f'(x)$=$3x^{2}$ $f'(x)=0$ when $x=0$ Therefore, $x=0$ is the critical point of $f(x)$ and the only point at which the function is not increasing. When $x<0$, $f'(x)>0$ When $x>0$, $f'(x)>0$ Therefore, when x $\ne$ 0, $f(x)$ is increasing.