# Domain and range f(x,y)

1. Mar 9, 2009

### 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 ive approached this correctly?

2. Mar 9, 2009

### 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.

3. Mar 9, 2009

### HallsofIvy

Staff Emeritus
No, it's not. That would be $x^2- y\ne 0$. 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.