Domain and range, functions of 2 variables

In summary: And graphically, I think it would exist in the first octant, and have a positive slope.In summary, the conversation revolved around understanding the domain and range of functions with two variables. The participants discussed helpful tutorials and clarified concepts like domain, range, and mapping. They also worked through examples of functions and their corresponding domains and ranges, such as f(x,y)=\frac{1}{\sqrt{x^2+y^2}} and f(x,y)=ln|x-y|.
  • #1
hotcommodity
436
0
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.
 
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  • #2
tutorial starts here and now
[tex]f(x,y)=\frac{1}{\sqrt{x^2+y^2}}[/tex]
what is the domain and range of this function?
 
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  • #3
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
hotcommodity said:
Its domain is x^2 + y^2 must be greater than or equal to zero, and its range is from minus to plus infinity?...

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?
 
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  • #5
ice109 said:
the range is actually discontinuous but again you're close

How is the range of that function discontinuous?
 
  • #6
d_leet said:
How is the range of that function discontinuous?

bah i meant domain, and i just realized the grievous error he made in the range as well.
 
  • #7
you've probably confused the hell out of him
 
  • #8
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
hotcommodity said:
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?

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... [itex]x\neq0\neq y[/itex]. What is the range simplified the same way?
 
  • #10
CRGreathouse said:
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... [itex]x\neq0\neq y[/itex]. What is the range simplified the same way?

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?
 
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  • #11
hotcommodity said:
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?

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 [itex](-\infty,0)[/itex] and [itex] (0,\infty)[/itex] 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

[tex] f(x,y)=ln|x-y|[/tex]

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
ice109 said:
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 [itex](-\infty,0)[/itex] and [itex] (0,\infty)[/itex] 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

[tex] f(x,y)=ln|x-y|[/tex]

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

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 [tex] f(x,y)=ln|x-y|[/tex]

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
hotcommodity said:
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 [tex] f(x,y)=ln|x-y|[/tex]

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.

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
 

1. What is the definition of a function of 2 variables?

A function of 2 variables is a mathematical relationship between two quantities, where the value of one variable is determined by the value of the other variable. In other words, each input (x) is associated with a unique output (y).

2. How is the domain of a function of 2 variables determined?

The domain of a function of 2 variables is the set of all possible input values (x) that can be used to evaluate the function. It is typically determined by identifying any restrictions on the variables, such as the presence of square roots or division by zero, and finding the set of values that satisfy those restrictions.

3. What is the range of a function of 2 variables?

The range of a function of 2 variables is the set of all possible output values (y) that can be obtained by evaluating the function for all possible input values. It is determined by plugging in different values for the input variables and observing the corresponding output values.

4. How do you graph a function of 2 variables?

To graph a function of 2 variables, you can create a table of values by choosing different input values and calculating the corresponding output values. Then, plot these points on a coordinate plane and connect them to create a curve. Alternatively, you can use a graphing calculator or software to plot the function.

5. What is the difference between a function of 2 variables and a relation?

A function of 2 variables is a special type of relation where each input (x) is associated with exactly one output (y). In a relation, one input can be associated with multiple outputs, meaning that it does not pass the vertical line test. Additionally, a function of 2 variables can be represented by an equation or a graph, while a relation can be represented by a set of ordered pairs.

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