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

1 Answer
Apr 5, 2016

Answer:

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)#.