Any intuitive way to understand the graph for $x^x$? Why does it switch directions at around $x=0.36787$? I'm asking more specifically about why it's such an arbitrary number. Is this number treated the same as the other famous mathematical numbers such as $\pi$, $e$ or the golden ratio?
I realize that as $x\rightarrow0 \implies x^x \rightarrow x^0$, and $x^0=1$, which means there has to be a global minimum when $0\lt x \lt1$. But I was wondering if there was a more intuitive way of understanding why this is.
 A: Given $y = x^x$ we can use logarithmic differentiation to answer your question:
$$\ln(y) = x\ln(x).$$  Differentiating, then, we get $$\frac{y'}{y} = \ln(x) + 1$$ and so $$y' = y\bigl(\ln(x) + 1\bigr) = x^{x}\bigl(\ln(x)+1\bigr) = x^x\ln(ex).$$
So, we have $y' = 0 \iff x = \frac{1}{e}$, which is where the global minimum occurs.
A: The more intuitive reason why the minimum occurs at $x=1/e$ is that you have a nice family of rational solutions to $x^x=y^y$ given by
$$
x=x_n=\left(\frac{n}{n+1}\right)^n,\quad
y=y_n=\left(\frac{n}{n+1}\right)^{n+1}.
$$
and it is elementary that $x_n\downarrow 1/e$ while $y_n\uparrow 1/e$.
How to get this family: Take log of both sides of $x^x=y^y$ and rearrange, we have $\log_y x=\frac{y}{x}=\frac{p}{q}$ in lowest form.  So $x=\frac{qy}{p}$ and $x^x=y^y$ gives $y^y=(qy/p)^{qy/p}$, which after taking $y$-th root of both sides, reduces to $y=(p/q)^{q/(q-p)}$.  A easy way to make sure $x$ is rational is to make $p=q-1$, so $y=((q-1)/q)^q$ and $x=((q-1)/q)^p$, now relabel $p$ as $n$, $q$ as $n+1$.
A: The reason is because of it's derivative
$$x^x\times(1+ln(x))=0$$
$$x^x≠0, (1+ln(x))=0$$
$$ln(x)=-1$$
$$x=\frac{1}{e}$$
$$x = 0.36787944117144233$$
That's why.
