TITLE: Why does the minimum value of $x^x$ equal $1/e$?
QUESTION [11 upvotes]: The graph of $y=x^x$ looks like this:

As we can see, the graph has a minimum value at a turning point. According to WolframAlpha, this point is at $x=1/e$.
I know that $e$ is the number for exponential growth and $\frac{d}{dx}e^x=e^x$, but these ideas seem unrelated to the fact that the mimum value of $x^x$ is $1/e$. Is this just pure coincidence, or could someone provide an intuitive explanation (i.e. more than just a proof) of why this is?

REPLY [2 votes]: Try to minimize logarithm of  $ y= x^x,$ i.e., $y=x \, \log x$ 
Its derivative is
$$ 1 + \log(x) $$
When equated to zero, it solves to that the minimum of $y(x)$ ocuurs at 
$$ x= \dfrac{1}{e}= 0.36788$$
and the minimum value is
$$ y_{min}= \dfrac{1}{{e}^{\frac{1}{e}}} \approx 0.6922$$
The tiny red dot shown in your graph for minimum is:
$$ (x,y) = (0.36788, 0.6922) $$
Note that the minimum value is not $\dfrac{1}{e}\, ! \, $ ... but is the reciprocal of $e^{th} $root of $e.$