# Find Domain of f(x,y)=sqrt(5xy-2y^2)

• KUphysstudent
In summary: ## is always satisfied by a real number x such that 5xy≥2y2, but the minus sign still appears in the inequality.

## Homework Statement

Find the domain of f(x,y)=sqrt(5xy-2y^2)

## The Attempt at a Solution

As far as I understand it's all about solving inequalities when you have this kind of problem.
The problem doesn't state it is in ℝ2 but I'm pretty sure it is.
So no negative numbers under the root.
Therefore 5xy ≥ -2y^2
x ≥ -2y/5 or -y ≥ 5x/2
But how exactly is this useful information?
So I looked up the topic on Youtube, Multivariable Calculus: Finding and Sketching the Domain.
It sounds perfect and the guy has two examples and one even has a root in it.
Sadly I still don't get it :/

KUphysstudent said:
Therefore 5xy ≥ -2y^2
x ≥ -2y/5 or -y ≥ 5x/2
Be careful with divisions in inequalities. The sign matters.
KUphysstudent said:
But how exactly is this useful information?
That tells you where the function has a chance to be defined.

Yea signs in inequalities shift direction if you multiply or divide with a negative number, I knew there was something so I googled it before hand.
If x ≥ -2y/5 or -y ≥ 5x/2 tells me where the function is defined, well then that is its domain. What format is commonly used to denote it?
I think we use {(x,y) ∈ ℝ2 ; 5xy ≥ -2y2}
so (x,y) are a plane in ℝ where 5xy ≥ -2y2. Does that make sense?

Edit: My quoting skills lack as much as my math.

KUphysstudent said:
The problem doesn't state it is in ℝ2 but I'm pretty sure it is.
Yes. The function takes as input pairs of real numbers, and outputs a real number, so it's a map from some subset of ##\mathbb R^2## to ##\mathbb R##.
KUphysstudent said:
Yea signs in inequalities shift direction if you multiply or divide with a negative number, I knew there was something so I googled it before hand.
If x ≥ -2y/5 or -y ≥ 5x/2 tells me where the function is defined, well then that is its domain.
mfb's point was that this is not valid. When you divide by a variable, you don't know what its sign is, so you can't tell whether the inequality switches direction or not.

The square root is defined, in this context, only when ##5xy - 2y^2 \ge 0##. Factor that expression to see where the product of the two factors will be nonnegative.
KUphysstudent said:
What format is commonly used to denote it?
I think we use {(x,y) ∈ ℝ2 ; 5xy ≥ -2y2}
so (x,y) are a plane in ℝ where 5xy ≥ -2y2. Does that make sense?
No, in part because the domain is not a plane. The domain of this function is some subset of the real plane.

I second @Mark44 suggestion, factor ##5xy-2y^2 = y(5x-2y)##. You want the region where both factors are positive or both are negative, or zero. I like to use plots like this, where I have shaded where each factor is positive. Here you should be able to describe where both factors are positive or neither are.

#### Attachments

• areas.jpg
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LCKurtz said:
I second @Mark44 suggestion, factor ##5xy-2y^2 = y(5x-2y)##. You want the region where both factors are positive or both are negative, or zero. I like to use plots like this, where I have shaded where each factor is positive. Here you should be able to describe where both factors are positive or neither are.
View attachment 231726

No matter how I solve the inequality there will be a negative sign, so I get the complex part of the function if I try would sketch it. So wouldn't sketching it and taking everything the outside the complex part be a thing you could do? Just a thought I'm going to look at the factoring part :)

KUphysstudent said:
No matter how I solve the inequality there will be a negative sign, so I get the complex part of the function if I try would sketch it. So wouldn't sketching it and taking everything the outside the complex part be a thing you could do? Just a thought I'm going to look at the factoring part :)

I mean, don't we already assume that the number is positive since it under the radical? meaning you can basically throw the negative symbol away to begin with so we have 5xy≥2y2

KUphysstudent said:
No matter how I solve the inequality there will be a negative sign, so I get the complex part of the function if I try would sketch it. So wouldn't sketching it and taking everything the outside the complex part be a thing you could do? Just a thought I'm going to look at the factoring part :)
Huh? You can consider two cases separately, e.g. y<0 and y>0.
KUphysstudent said:
I mean, don't we already assume that the number is positive since it under the radical?
The overall expression yes, but not the individual variables.

KUphysstudent said:
I mean, don't we already assume that the number is positive since it under the radical? meaning you can basically throw the negative symbol away to begin with so we have 5xy≥2y2
You're not "throwing away" the minus sign. It's true that the inequality ##5xy - 2y^2 \ge 0## is equivalent to ##5xy \ge 2y^2##, but this step doesn't do you any good. In your work in post #1, you then divided both sides by y, which is really an invalid step, since y could be negative, positive, of zero, so you can't tell which inequality symbol should be used.

Again, what you need to do is to factor ##5xy - 2y^2## into ##y(5x - 2y)##, and investigate the x and y intervals that make the product nonnegative.

mfb said:
Huh? You can consider two cases separately, e.g. y<0 and y>0.The overall expression yes, but not the individual variables.

Yea okay that makes sense.

Okay I'm still very confused, there must be something fundamentally wrong with my results.
So if I start out with 5xy-2y2 ≥ 0 → 5xy ≤ -2y2 → 5x ≤ -2y → -5x/2 ≥ y , so this is wrong but what is the problem? I cannot remember ever learning about inequalities in school. And reading online tells me it's just algebra and flipping the sign whenever you multiply or divide by something negative.
Holy Moly I just realized that I for some reason keep moving -2y2 like I have to add it... oh god. -2y2*-2y2 = 0 *(-2y2). I don't know what is wrong with me hehe.
ok so with that out of the way I get to add and not multiply. so we have 5xy-2y2 ≥ 0 → 5xy ≥ 2y2 → 5x ≥ 2y → 5x/2 ≥ y.
I cannot believe I got into uni…
y=5x/2 gives me a line, like the one mark has in his figure. and if I only want the positive or negative x,y-values at once then I can see the domain. But how exactly did you determine that it was only positives or negatives?

Edit:
Holy Moly I just realized that I for some reason keep moving -2y2 like I have to add it
Should have been:
Holy Moly I just realized that I for some reason keep moving -2y2 like I have to multiply it

Last edited:
KUphysstudent said:
Yea okay that makes sense.

Okay I'm still very confused, there must be something fundamentally wrong with my results.
So if I start out with 5xy-2y2 ≥ 0 → 5xy ≤ -2y2
No.
##5xy - 2y^2 \ge 0 \Leftrightarrow 5xy \ge 2y^2##
I added ##2y^2## to both sides of the inequality.
You're playing fast and loose with the signs and your algebra.
KUphysstudent said:
→ 5x ≤ -2y → -5x/2 ≥ y , so this is wrong but what is the problem? I cannot remember ever learning about inequalities in school. And reading online tells me it's just algebra and flipping the sign whenever you multiply or divide by something negative.
Holy Moly I just realized that I for some reason keep moving -2y2 like I have to add it... oh god. -2y2*-2y2 = 0 *(-2y2). I don't know what is wrong with me hehe.
ok so with that out of the way I get to add and not multiply. so we have 5xy-2y2 ≥ 0 → 5xy ≥ 2y2 → 5x ≥ 2y → 5x/2 ≥ y.
I cannot believe I got into uni…
y=5x/2 gives me a line, like the one mark has in his figure. and if I only want the positive or negative x,y-values at once then I can see the domain. But how exactly did you determine that it was only positives or negatives?

KUphysstudent said:
5xy ≤ -2y2 → 5x ≤ -2y
Consider x=1, y=-1. Clearly 5xy=-5 is smaller than y2=1, but 5x=5 is not smaller than -2y=-2.
If you divide an inequality by a negative number (here: y) you have to swap the direction of the inequality.

##a<b \Leftrightarrow -a > -b##

@KUphysstudent Did you ever understand which regions answer your question by looking at the graphic I gave you in post #5?

Mark44 said:
No.
##5xy - 2y^2 \ge 0 \Leftrightarrow 5xy \ge 2y^2##
I added ##2y^2## to both sides of the inequality.
You're playing fast and loose with the signs and your algebra.

Yea I get the algebra part, I just kept making the same mistake.
LCKurtz said:
@KUphysstudent Did you ever understand which regions answer your question by looking at the graphic I gave you in post #5?

I think I do, the bold line is the one you get from the inequality and the green area is to depict the positive x- and y-values. so you put in an x-value and get a y-value and in the first quadrant it is everything below the line. In the third quadrant it is everything above the line.
I cannot explain this very well, but here is how I think of it. if you have 5x/2=y and set x=2 you will get y=5, so if you instead of using 5x/2=y but then use 2y/5=x and plugin y-values you get all the points below the graph.
I think but it is likely wrong :)

mfb said:
Consider x=1, y=-1. Clearly 5xy=-5 is smaller than y2=1, but 5x=5 is not smaller than -2y=-2.
If you divide an inequality by a negative number (here: y) you have to swap the direction of the inequality.

##a<b \Leftrightarrow -a > -b##
That makes sense. so you can basically check every time you have manipulated the equation to see if the equality still is correct, if not you did it wrong.

LCKurtz said:
@KUphysstudent Did you ever understand which regions answer your question by looking at the graphic I gave you in post #5?
KUphysstudent said:
I think I do, the bold line is the one you get from the inequality and the green area is to depict the positive x- and y-values. so you put in an x-value and get a y-value and in the first quadrant it is everything below the line. In the third quadrant it is everything above the line.
I cannot explain this very well, but here is how I think of it. if you have 5x/2=y and set x=2 you will get y=5, so if you instead of using 5x/2=y but then use 2y/5=x and plugin y-values you get all the points below the graph.
I think but it is likely wrong :)
Yes, but it is easier just to say the green area and the white area. That's where the factors are either both positive or both negative. If you have trouble figuring out the areas, try this. The points on the line ##5x-2y = 0## are the the only places in the plane where ##5x-2y=0##. Everyplace else it is either ##>0## or ##<0##. You can tell which side is which by simply plugging in a point not on the line. For example, at ##(1,0)## you get ##5\cdot 1 -2\cdot 0 = 5 > 0## so that is the side where ##5x-2y>0##, which I colored yellow. It's easy to visualize the regions that way.

## 1. What is the domain of the function f(x,y) = sqrt(5xy-2y^2)?

The domain of a function refers to all the possible values of the independent variables (x and y) for which the function is defined. In this case, since we have a square root function, the expression inside the square root must be greater than or equal to 0 for the function to be defined. Therefore, the domain of f(x,y) = sqrt(5xy-2y^2) is all the values of x and y that make 5xy-2y^2 greater than or equal to 0.

## 2. How do you find the domain of a function with two variables?

To find the domain of a function with two variables, you need to consider the restrictions on the values of the variables that make the function defined. In this case, we have a square root function, so the expression inside the square root must be greater than or equal to 0. Thus, we need to solve the inequality 5xy-2y^2 ≥ 0 for x and y to determine the domain.

## 3. Can the domain of a function with two variables be expressed as an interval?

Yes, the domain of a function with two variables can be expressed as an interval. In this case, since we have an inequality to solve, the domain will be expressed as a set of values that satisfy the inequality, which can be written as an interval on a number line.

## 4. How can you graph the domain of a function with two variables?

To graph the domain of a function with two variables, you can plot the points that satisfy the inequality and shade the region on the coordinate plane. In this case, we can plot the points (0,0) and (0,5) on the x-axis and y-axis, respectively, and then draw a line connecting these points. The region above this line will be shaded, representing the domain of the function.

## 5. Are there any values of x and y that would make the function undefined?

Yes, there are values of x and y that would make the function undefined. In this case, if the expression inside the square root is negative, the function would be undefined. Therefore, any values of x and y that make 5xy-2y^2 less than 0 would make the function undefined.