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Domain of f(x,y)=x^y

  1. Nov 10, 2012 #1
    1. The problem statement, all variables and given/known data
    What is the domain of f(x,y)=x^y




    3. The attempt at a solution
    I thought it would be all real pairs except (0,0)
    but it is x>0 and y all real numbers?
     
  2. jcsd
  3. Nov 10, 2012 #2
    If ##z=x^y##, then ##ln(z)=yln(x)##

    Do you see why x cannot be less than zero?
     
  4. Nov 10, 2012 #3

    Dick

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    How do you define (-1)^(1/2) or even worse (-1)^sqrt(2)?
     
  5. Nov 10, 2012 #4

    Zondrina

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    Look at this. Now where is ln(x) defined? Is it discontinuous anywhere? Are there any points where the function isn't defined...
     
  6. Nov 10, 2012 #5
    Sorry, I don't think I fully understand
    Would it not be defined when y is something like 2 or whatever, even if x is negative?
     
  7. Nov 11, 2012 #6

    SammyS

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    How about when x < 0, if y = 1/2 or y = √(2). What then?
     
  8. Nov 11, 2012 #7

    loy

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    see this If z=x^y, then ln(z)=yln(x) , x cannot be less than zero since x is not defined if it is less than 0 . if you dont believe, you can press your calculator to see whether it has a result when you put in x a value which is less than 0.
     
  9. Nov 11, 2012 #8

    loy

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    but as long as x>0 , y can be defined in any value in R.
    given x>0,
    IF y>0 , f(x,y)=x^y .
    IF y<0 , f(x,y)=x^(-y) = 1/(x^y)
    IF y=0 , f(x,y)=1.
    BUT when x<0 , the function is not defined , meaning , you can't get any answer, the function is discontinuous in x<0.
    the domain of x is (0,∞) ,while y is (-∞,∞)
     
  10. Nov 11, 2012 #9

    vela

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    It depends on your definition of exponentiation. If you were to restrict y to the integers, then yes, you could consistently define xy for negative values of x. On the other hand, for real numbers x and y, xy is often defined by xy=ey log x, which isn't defined if x<0.
     
  11. Nov 11, 2012 #10
    I see
    This is the first time I'm hearing of such a definition, I'll have to learn more about it
    But thank you, I think I understand now
     
  12. Nov 11, 2012 #11

    SammyS

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    Your line,
    IF y<0 , f(x,y)=x^(-y) = 1/(x^y)​
    is incorrect.

    You can say
    [itex]\displaystyle \text{If }\ y<0\,,\ \text{ then }f(x,y)=x^{-|y|} = \frac{1}{x^{|y|}}\ .[/itex]​
     
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