Determining the domain and range of multi-variable function

  • #1
15
0

Homework Statement


f(x,y) = 1/y^2-x

find the domain of f.

Given c ∈ R \ {0} find (x, y) ∈ R 2 such that f(x, y) = c. Finally determine the range of f.

Homework Equations


I know that the domain of the function is anywhere that the function is defined.

The Attempt at a Solution


in the case of this question i can see that the function is going to be undefined where y^2-x =0
since anything over 0 is undefined.

I believe that for the range part of the question it is essentially saying, define the range such that x and y are real numbers but please correct me if i'm wrong.

When i used to be dealing with functions such as f(x) = x+1, in that case i knew that the domain was where the function was defined over the x-axis and i knew that the range was where the function was defined over the y-axis.

But since the domain now is including X and Y i have no idea by what they mean by finding the range.

So if someone could help me define the range of this function and show me the steps to defining the range of any other multi-variable like this one i would greatly appreciate it!
This question is purely for study purposes for exam in start of January so feel free to go into as much detail as you want if you feel it would aid understanding. Thanks again!
 
Last edited:

Answers and Replies

  • #2
35,138
6,892

Homework Statement


f(x,y) = 1/y^2-x
You need to write the right side as 1/(y2 - x).
What you wrote means ##\frac 1 {y^2} - x##, which isn't what you meant, based on your work below.
cathal84 said:
find the domain of f.

Given c ∈ R \ {0} find (x, y) ∈ R 2 such that f(x, y) = c. Finally determine the range of f.

Homework Equations


I know that the domain of the function is anywhere that the function is defined.

The Attempt at a Solution


in the case of this question i can see that the function is going to be undefined where y^2-x =0
since anything over 0 is undefined.

I believe that for the range part of the question it is essentially saying, define the range such that x and y are real numbers but please correct me if i'm wrong.
The range is the set of real numbers z such that ##z = \frac 1 {y^2 - x}##. Obviously 0 is not in the range of this function. Are there any other values that aren't in the range?
cathal84 said:
When i used to be dealing with functions such as f(x) = x+1, in that case i knew that the domain was where the function was defined over the x-axis and i knew that the range was where the function was defined over the y-axis.

But since the domain now is including X and Y i have no idea by what they mean by finding the range.

So if someone could help me define the range of this function and show me the steps to defining the range of any other multi-variable like this one i would greatly appreciate it!
This question is purely for study purposes for exam in start of January so feel free to go into as much detail as you want if you feel it would aid understanding. Thanks again!
 
  • #3
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
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Homework Statement


f(x,y) = 1/y^2-x

find the domain of f.

Given c ∈ R \ {0} find (x, y) ∈ R 2 such that f(x, y) = c. Finally determine the range of f.

You wrote
$$f(x,y) = \frac{1}{y^2} - x,$$
at least when we read your expression using standard parsing rules. If you mean something else you need to use parentheses.
If you mean
$$f(x,y) = \frac{1}{y^2-x},$$
then write it in plain text as f(x,y) = 1/(y^2-x).
 
  • #4
15
0
You need to write the right side as 1/(y2 - x).
What you wrote means ##\frac 1 {y^2} - x##, which isn't what you meant, based on your work below.
The range is the set of real numbers z such that ##z = \frac 1 {y^2 - x}##. Obviously 0 is not in the range of this function. Are there any other values that aren't in the range?
You wrote
$$f(x,y) = \frac{1}{y^2} - x,$$
at least when we read your expression using standard parsing rules. If you mean something else you need to use parentheses.
If you mean
$$f(x,y) = \frac{1}{y^2-x},$$
then write it in plain text as f(x,y) = 1/(y^2-x).
You need to write the right side as 1/(y2 - x).
What you wrote means ##\frac 1 {y^2} - x##, which isn't what you meant, based on your work below.
The range is the set of real numbers z such that ##z = \frac 1 {y^2 - x}##. Obviously 0 is not in the range of this function. Are there any other values that aren't in the range?
ah yes sorry ill know for next time to put the brackets in thanks for the advice.
I don't see why any other numbers would not be in the range besides 0 but then again I'm just looking at where the function is undefined but thats how you define the domain, so i am not sure, think i am just confusing myself. what exactly am i looking to do differently when defining the range?
 

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