Answer
f is increasing on $(0,\infty)$
Work Step by Step
$f(x)=x^{4}$
$f'(x)=4x^{3}$
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 $(0,\infty)$
