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Domain and range f(x,y)

  1. Mar 9, 2009 #1
    define the natural domain and range for f(x,y) = 1/sqrt(x^2-y)

    I get D= {(x,y)|x^2-y>0} same as saying y can not equal x^2

    R= {(x,Y)|f>0}

    can someone tell me if ive approached this correctly?
     
  2. jcsd
  3. Mar 9, 2009 #2

    Dick

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    Suppose x=1 and y=2. That's a problem point for defining f too. And your range should be a set of real numbers, not ordered pairs.
     
  4. Mar 9, 2009 #3

    HallsofIvy

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    No, it's not. That would be [itex]x^2- y\ne 0[/itex]. You are requiring that y be less than x2.

    And, as Dick said, the range of f is a set of numbers, not of pairs of numbers.
     
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