Does $x^x$ function have a name?

$f(x) = x^x = e^{x \ln x}$

Let's differentiate this function!

The first derivative:

$f'(x) = e^{x \ln x} ( 1 + \ln x )$

The stationary point:

$e^{x \ln x} ( 1 + \ln x ) = 0$

Since the exponential function is always positive, this can only stand, if the bracketed expression is 0.

$1 + \ln x = 0$

$\ln x = -1$

$x = \frac {1}{ e }$

The second derivative:

$f'' (x) = e^ { x \ln x } ( 1 + \ln x )^2 + \frac { e^{ x \ln x } } {x}$

$f''( \frac {1}{ e }) = 0 + \frac { e ^ { \frac {1}{ e } × \ln \frac {1}{ e } } } { \frac {1}{ e } } = { e } ^ {1 - \frac {1}{ e } } > 0$

$\text {Hence, } x = \frac {1}{ e } \text { is a minimum point. }$

The minimum value:

$( \frac {1}{ e } ) ^ { \frac {1}{ e } } = \boxed { 0.692 \text { (3 d. p.) } }$

$x^x = e^{x \ln x}$ . We know that $e^x$ is an increasing function of x.

You can find the minimum of $x^x$ by finding the minimum of $e^{x \ln x}$ .

Tthis can be done by setting its derivative $(1+ \ln x)$ to zero, yielding $x = \frac{-1}{e}$ . The second derivative $\frac{1}{x}$ is positive for $x > 0$ , so the function is everywhere convex, and the unique extremum is indeed a global minimum.

The minimum value of $x^x = \large e^ {\frac{-1}{e} } = 0.692$