# Domain and range, functions of 2 variables

1. Oct 18, 2007

### hotcommodity

I'm having a bit of trouble grasping the domain and range of functions of 2 variables. Does anyone know of any helpful tutorials that will help me get the hang of this concept? Any help appreciated.

2. Oct 18, 2007

### ice109

tutorial starts here and now
$$f(x,y)=\frac{1}{\sqrt{x^2+y^2}}$$
what is the domain and range of this function?

Last edited: Oct 18, 2007
3. Oct 18, 2007

### hotcommodity

Its domain is x^2 + y^2 must be greater than or equal to zero, and its range is from minus to plus infinity?...

4. Oct 18, 2007

### ice109

do you know what domain means? you're like 1/3 of the way there. part of your answer is wrong and you didn't answer the question about the domain. but you do have something...

the range is actually discontinuous but again you're close

edit

the domain is actually discontinuous

can the function ever be negative? did you just guess?

Last edited: Oct 18, 2007
5. Oct 18, 2007

### d_leet

How is the range of that function discontinuous?

6. Oct 18, 2007

### ice109

bah i meant domain, and i just realized the grievous error he made in the range as well.

7. Oct 18, 2007

### ice109

you've probably confused the hell out of him

8. Oct 19, 2007

### hotcommodity

ice's post kind of inspired me to take the bull by the horns, so I opened up a calc book of mine, different from my class text, and I think I've got a handle on it. The domain would be D = {(x,y)|x^2 + y^2 >0}, and the range would be R = { z | z = f(x,y), x^2 + y^2 >0}. Is that right?

9. Oct 19, 2007

### CRGreathouse

Yes, but you should probably simplify those. Are you working over the reals or the complex numbers? Oh, of course you are working over the reals -- you used a >.

So for which (x, y) in R^2 is it true that x^2 + y^2 > 0? The strict quadrants, right... $x\neq0\neq y$. What is the range simplified the same way?

10. Oct 19, 2007

### hotcommodity

I'm not quite sure. I know the range is in R^1, and the range is every value that f(x, y) can take on... What are the "strict quadrants?"

Edit: Also, since the function takes an pair (x, y) from R^2, and turns it into R^1, is that what they call a mapping?

Last edited: Oct 19, 2007
11. Oct 19, 2007

### ice109

yes a function is a mapping. let me do this one for you and then give you another one

the domain for this function is all values for x and all values of except the point (0,0), i think you know why. and the range is $(-\infty,0)$ and $(0,\infty)$ notice unbounded below and open notation, the parenthesis instead of brackets, and the open and unbounded above, again the parenthesis instead of brackets. do you know why the range doesn't include zero?

try this now

$$f(x,y)=ln|x-y|$$

what is the domain and range of this function, describe it in words, set builder notation, and what it would look like on a graph

12. Oct 20, 2007

### hotcommodity

First of all I appreciate the help.

A few questions, are you saying that the range of the function does not include zero, or that it does? And shouldn't it be for all z since z = f(x,y) ?

Ok, for $$f(x,y)=ln|x-y|$$

The domain of this function is the set of all x and y pairs such that x - y is greater than zero. The range of the function is the set z such that z = f(x,y), and z is greater than zero, and also x and y must fulfill the constraint for the domain. So D = {(x, y) | x > y}, and R = {z | z = f(x, y), x > y}. And graphically, I think it would exist in the first octant, and have a positive slope.

13. Oct 20, 2007

### ice109

yes the range of the first function doesn't contain zero. is the function zero anywhere?

look how i wrote the range of the previous function. it is pointless to write the range implicitly as you have. give me in interval notation. you're essentially saying that the range of the function f(x,y) is all values of f(x,y).

and about the domain graphically you're wrong