What is the Domain of $f(x) = x^{1/x}$

Question

looking on wolphram alpha for the function $$f(x) = x^{1/x}$$ it says that the domain is for each $x>0$ but for example if I make $x=-3$ it becomes $$ (-3)^{1/(-3)} = \left(-\frac{1}{3}\right)^{1/3} = -0.69336.... $$

So where am I wrong?

Answer 1

As a real function, $x^y$ is defined as $$x^y\overset{\text{def}}{=}\mathrm e^{y\ln x}, \enspace\text{ so }\quad x^{\tfrac 1x}= \mathrm e^{\tfrac{\ln x}x},$$ which requires $x>0$ to be defined.

Answer 2

Its

$$(-3)^{1/(-3)}$$

not

$$-(3^{1/(-3)})$$

Answer 3

$f(x) =x^{\frac{1}{x}} =e^{\frac{1}{x} \ln(x)} $

So :

The Domain is $ ] 0,+\infty [$

Because the domain of $ \ln(x) $ is $ ] 0,+\infty [$

And the domain of $\frac{1}{x}$ is $\mathbb{R^*} $

And the domain of $e $ is $\mathbb{R} $