x to the power of an irrational number If one were to graph the function of 
$$ f(x) = x^e $$
How would this look? (With explanation as to why) Particularly in the case of negative x values. 
 A: I don't know how rigorus my rational is or if it's even correct, but I like to use the infinite sums of certain irrational or trancendental numbers to graph them as they can be related to fractional powers. One thing about fractional powers is that if the denominator has 2 as a factor(is "even"), then it is not defined for negative numbers, e.g $x^\frac{1}{2}$ or $x^\frac{1}{6} = (x^\frac{1}{2})^\frac{1}{3}$ So this analogy can sort of be used to explain why $x^e$ cannot be defined for negative numbers(when I say not defined I mean that the result is complex). Recall the Series
\begin{equation}
e = \sum_{k = 0}^\infty \frac{1}{k!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots = \frac{(1!2!3!\ldots) + (0!2!3!\dots )+ (0!1!3! \ldots) + \ldots}{0!1!2!3! \ldots}
\end{equation} 
You can see that the denominator contains a 2, which means when factored out
\begin{equation}
x^e = x^{\frac{(1!2!3!\ldots) + (0!2!3!\dots )+ (0!1!3! \ldots) + \ldots}{0!1!2!3! \ldots}} = (x^{\frac{1}{2}})^\frac{(1!2!3!\ldots) + (0!2!3!\dots )+ (0!1!3! \ldots) + \ldots}{0!1!3! \ldots}
\end{equation}
This means $x^e$ would be complex for negative values. When I enter $(-1)^e$ on google it returns a complex number which supports the idea, though I am sure there is a more rigorus and mathematically sound way to show this.
