For which negative real $x$, does the expression $x^{1/x}$ gives real values? I started to think about the function $f(x) =x^{1/x} $ and how it behaves. How would you show for which negative real values of x that $f(x)$ gives real values?

I tried by rewritting, set $y=-x$ and we get $\displaystyle-\frac{(-1)^{((y-1)/y)} }{y^{(1/y)} } $. The nominator will be complex if $y&gt;1$. Is there any other way to show this? By using euler's formula maybe? Since it is basically just the finding the roots for all negative x.
This is real when $x$ is negative only when $x$ is also an odd integer or a rational with odd numerator, for if $x$ were an even integer, we have complex values. If $x$ is rational, then it should have the form $$\frac{\text{odd integer}}{\text{nonzero integer}},$$ for the same reason as in the integer case. If $x$ is irrational, then we might want to write the expression as $$e^{\log x^{1/x}}=e^{\frac{\log x}{x}},$$ which is not then real.

Thus, in summary, if $x&lt;0,$ then it should either be an odd integer, or a rational with odd numerator as well.
If you define $x^{1/x}$ to be $(-x)^{1/x}(-1)^{1/x}$ you get reals only when the exponent of $-1$ is an integer, i.e.

$$x=-\frac1n.$$

Anyway, this resolution is questionable.