# What are the critical points of f(x) =x^(x^2)?

Apr 5, 2016

The only critical number is ${e}^{- \frac{1}{2}}$

#### Explanation:

$f \left(x\right) = {x}^{{x}^{2}} = {e}^{\ln} \left({x}^{{x}^{2}}\right) = {e}^{{x}^{2} \ln x}$.

The domain of ${x}^{{x}^{2}}$ is the set of positive real numbers together wit the negative integers.
The negative integers are discrete, so we cannot find a (real) derivative of $f$ at those elements of the domain.
For the calculus, we can restrict the domain to the positive real numbers, which is the domain of ${e}^{{x}^{2} \ln x}$.

$f ' \left(x\right) = {e}^{{x}^{2} \ln x} \left[\frac{d}{\mathrm{dx}} \left({x}^{2} \ln x\right)\right]$

$= {e}^{{x}^{2} \ln x} \left[\left(2 x\right) \ln x + {x}^{2} \left(\frac{1}{x}\right)\right]$

$= {e}^{{x}^{2} \ln x} \left(2 x \ln x + x\right)$

$= {x}^{{x}^{2}} x \left(2 \ln x + 1\right)$.

$f ' \left(x\right)$ is never undefined on the domain of $f$.

Since $0$ is not in the domain of $f$, the only critical number is the solution to

$2 \ln x + 1 = 0$

So, $\ln x = - \frac{1}{2}$

And, $x = {e}^{- \frac{1}{2}}$.