What is the domain of $f(x)=x^{2/4}$

Question

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.

Answer 1

since $$x^{2/4}=x^{1/2}$$ we have $$x\geq 0$$

Answer 2

Let $f(x)=x^{\frac{2}{4}},$ $g(x)=\left(\sqrt[4]{x}\right)^2$, $h(x)=\sqrt[4]{x^2}.$

Thus, $d(f)=(0,+\infty),$ $d(g)=[0,+\infty)$, $d(h)=\mathbb R$.

Answer 3

Note $\sqrt{a^2}=|a|$ hence, $$f(x)=x^{2/4}=f(x)=\sqrt{|x|}$$

$Dom(f)=\Bbb R$

Answer 4

If you insist on taking powers of negative numbers then you need to work over the complex numbers, and then we find that some operations become multi-valued.

For example, what is $(-1)^{1/3}$ ?

Well, $(-1) \in \{ \mathrm e^{\pm\pi \mathrm i},\mathrm e^{\pm3\pi \mathrm i},\mathrm e^{\pm5\pi\mathrm i},\ldots\}$

Hence, $(-1)^{1/3} \in \{ \mathrm e^{\pm \pi \mathrm i/3},\mathrm e^{\pm\pi \mathrm i},\mathrm e^{\pm 5\pi\mathrm i/3},\ldots\}$

$$(-1)^{1/3} \in \left\{ -1 \ , \ \ \ \frac{1}{2} + \frac{\sqrt 3}{2}\mathrm i \ , \ \ \ \frac{1}{2} - \frac{\sqrt 3}{2}\mathrm i \ \right\} $$