What is the natural domain and range for f(x,y) = 1/sqrt(x^2-y)?

[Derill03]

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 I've approached this correctly?

 

[Dick]

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.

 

[HallsofIvy]

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

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

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

can someone tell me if I've approached this correctly?

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