What is the domain of $f(x)=x^{2/4}$ What is the domain of $f(x)=x^{2/4}$
Is $f(x)= (\sqrt[4]x)^2 $ with $\operatorname{dom} (f)  = [0,\infty) $ 
or $\sqrt[4]{x^2}$ with $\operatorname{dom} (f) = (-\infty,\infty)$ 
or is $f(x)=\sqrt x$ with $\operatorname{dom} (f)=[0,\infty)$
I have tried searching the internet but couldn't find anything.

Let $f(x)=x^{\frac{2}{4}},$ $g(x)=\left(\sqrt[4]{x}\right)^2$, $h(x)=\sqrt[4]{x^2}.$
Thus, $d(f)=(0,+\infty),$ $d(g)=[0,+\infty)$, $d(h)=\mathbb R$.
