## Calculus (3rd Edition)

f is increasing on $(0,\infty)$
$f(x)=x^{4}$ $f'(x)=4x^{3}$ f'(x)=0 when x=0. Therefore, x=0 is the critical point of f. When x<0, f'(x)<0. For example, $f'(-2)=4\times(-2)^{3}=-32$. $\implies\,$ f is decreasing on $(-\infty,0)$. When x>0, f'(x)>0 $\implies\,$ f is increasing on $(0,\infty)$