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

1 Answer
Jim H
Apr 5, 2016

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).

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