values of $x^y=y^x$ for $x \neq y$, hopefully with straightforward explanation As I understand it, many people thought I am asking in a question with exactly this title about pairs of (x,y) such that $x^y=y^x$, or that I am looking for a function $f(x)$, such that $x^{f(x)}=(f(x))^x$, but I am not. I am asking for another function $g(x)=x^{f(x)}$, and where is it concave, convex, its minima, maxima, asymptotes etc. And more importantly, there are two such functions f(x) with the above property and their corresponding functions g(x). Because one $f(x)$ is simply $x$, so for $y=f(x)=x$ we get $g(x)=x^x$, but I am interested in the other function $g(x)$ that is yielded by the rather complicated function $f(x)$.
From the first time I asked the question, I already manage through guesses and observation on individual values come up with these possible properties:
So, with more of a thinking and wolframalpha, I found out (hopefuly this time correctly) that the function is defined only on the interval $(1,\infty)$, goes to infinity for $x$ going from the right to $1$ , it is decreasing  in a convex fashion all the way to $x=e$, where is global minimum $e^e$, and it is increasing towards infinity in rather a quite slow fashion, with a possible asymptote $x=y$, to which it is approaching from the left (from the above), which would indicate it is concave somewhere after $x=e$. Can anybody confirm or correct that?
Is anyone able to help me with this, now that I hope we understand each other?
EDIT: The asymptote part is incorrect, as proved by Ennar. Convexity and concavity are still debatable.

It's correct.
Let $y=kx$, where $k>1$.
Hence,
$$x^y=y^x$$ it's
$$x^{kx}=(kx)^x$$ or $$x=k^{\frac{1}{k-1}}$$ and from here $$y=k^{\frac{k}{k-1}}.$$
Thus, $$x^y=k^\frac{k^{\frac{k}{k-1}}}{k-1}.$$
Let $$f(k)=k^\frac{k^{\frac{k}{k-1}}}{k-1},$$ where $k>1.$ 
Hence,
$$f'(k)=\frac{f(k)\cdot k^\frac{k}{k-1}}{(k-1)^3}\cdot\left(\frac{k-1}{\sqrt k}+\ln{k}\right)\left(\frac{k-1}{\sqrt k}-\ln k\right)>0$$ 
and since $\lim\limits_{k\to1+}f(k)=e^e,$ we obtain $f(k)>e^e$ and we are done!
