What is the Domain of the f(x)^g(x) Function?

  • Thread starter Thread starter karkas
  • Start date Start date
  • Tags Tags
    Domain Function
AI Thread Summary
The discussion focuses on determining the domain of the function h(x) = (x^2 - 4)^x. It is established that for the base x^2 - 4 to be positive, x must be greater than 2 or less than -2. Additionally, the exponent x introduces further restrictions, particularly when x is negative, as negative bases raised to rational exponents can lead to undefined values. The consensus suggests that the domain of the function x^x is typically considered to be (0, ∞), with complications arising for negative values. Overall, the domain of h(x) is primarily constrained by the conditions on both the base and the exponent.
karkas
Messages
131
Reaction score
1

Homework Statement


We have to find the domain range (hope I'm using this right) of some functions, one of them being h(x)=f(x)^g(x), say h(x)=(x^2-4)^(x). I've been looking around and couldn't find the domain range of the x^x function, so I am kinda stuck on this one.


Homework Equations



Find the domain of h(x)=(x^2 -4)

The Attempt at a Solution


Well there's a somewhat obvious requirement (I think), that x^2-4>0 . So we get x>2 or x<-2. Other than that though I don't see any other requirement that stems from the exponent. Maybe that isn't even the case , however. Any suggestions? Thanks!
 
Physics news on Phys.org
xx does pose a problem.

When x is a negative number, xq, where q is a rational number, is only defined if the numerator of q is even (and it is applied before applying the denominator), or if the denominator of q is odd.

xr is undefined if r is irrational.

To my knowledge, the usual practice is to consider the domain of xx to be (0, ∞). While you can make a case for including a subset of the negative rational numbers, to do so puts a lot of holes in the domain for x < 0 .
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top