Question

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

Answer 1

The only critical number is `e^(-1/2)`

Explanation:

`f(x) = x^(x^2) = e^ln(x^(x^2)) = e^(x^2lnx)`.

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^2lnx)`.

`f'(x) = e^(x^2lnx)[d/dx(x^2lnx)]`

`= e^(x^2lnx)[(2x)lnx+x^2 (1/x)]`

`= e^(x^2lnx)(2xlnx+x)`

`= x^(x^2)x(2lnx+1)`.

`f'(x)` 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

`2lnx+1 = 0`

So, `lnx=-1/2`

And, `x=e^(-1/2)`.