What is the Domain of $f(x)=x^{\frac{1}{x}}$ Find the Domain of $$f(x)=x^{\frac{1}{x}}$$ I have Confusion on Negative Real numbers. if $x=-2$  then $f(x)$ is not Real, but if $x=-3$ ,$f(x)$ is Real. I am unable to figure out which set of negative real numbers come into the Domain.Books give the Domain as $\mathbb{R^+}$
 A: If you are thinking along the lines of $\sqrt[3]{-27}=-3$, that's not the same as $(-27)^{1/3}$. 
$(-27\pm i\varepsilon)^{1/3}$ are both nowhere near a negative real (where $\varepsilon$ is a small positive real), so if $(-27)^{1/3}$ had meaning, a desire to be working with continuous functions would mean $(-27)^{1/3}$ is not a negative real. 
These considerations should convince someone that it's not a good idea to say $\sqrt[3]{x}=x^{1/3}$ when $x$ is negative.
If you do insist that $\sqrt[3]{-27}=(-27)^{1/3}$, then why isn't it the same as $(-27)^{2/6}=\sqrt[6]{(-27)^2}$, which is a positive real? You run into trouble like this if you insist that $\sqrt[3]{x}=x^{1/3}$ when $x$ is negative.
A: General observation: given a real valued function $f$, we can write
$$
f(x)=\exp(\log(f(x)))
$$
only when $f>0$.
Hence in your case $\Bbb R^{+}$ is clearly contained in the domain of the function $f(x)=x^{\frac1 x}$.
It remains to show that no other point is admitted in the domain.
$0$ is clearly out.
Then if $x<0$ what you have is a power of a real negative number. But in the system of real numbers, the power is defined only for positive base. Hence no negative number could be taken in account.
Thus the domain of your function is given by all and only positive real numbers, $\Bbb R^{+}$.
A: @Alex.Jordan posted a nice answer but there is a domain that exists for ${x}^{1/x}$ if ${1/x}$ is reduced. Thus I don't believe his answer is proper to the OP's question.
When exponent of $x^{p/q}$ ,which is $p/q$, is reduced completely and $q$ is odd, $x^{p/q}$ satisfies the identities of $\sqrt[q]{x^p}$ and $\left(\sqrt[q]{x}\right)^{p}$ for the negative domain. Thus for ${x}^{1/x}$, we must find the values of $1/x$ where the output has an odd denominator and is completely reduced. 
Also one should also know that for $f(x)^{g(x)}$ that $f(x)^{g(x)}\neq{e^{\ln(f(x))g(x)}}$ when $x<0$. This because $\ln(x)$ for the real logarithm is undefined at negative numbers. So they're is no reason to use this identity for figuring out the negative domain. Infact we must analyze $x^{1/x}$ strictly for the real domain.
One example is for $e^{\ln{x}}=x$ for $x<0$. When $x=-1$ the two terms are not the same.
Using all of my previous points, one can find the true domain of $x^{\frac{1}{x}}$ is...
$(0,\infty)\bigcup{\left\{ {2n+1\over 2m+1}\ |\ n, m \in \Bbb Z\right\}}\bigcup\left\{ {2n+1\over 2m}\ |\ n, m \in \Bbb Z\right\}$
For $f(x)^{g(x)}$, when one tries to search for the output of $g(x)$ that has a fraction reduced with an odd denominator; one can find the domain of $f(x)^{g(x)}$ when $f(x)$ is negative.
Be sure to check:
Find the domain of $x^{2/3}$
What is the domain of $x^x$ when $ x<0$
Can the graph of $x^x$ have a real-valued plot below zero?
How can we describe the graph of $x^x$ for negative values?
