# Math Help - help with multivariable domain

1. ## help with multivariable domain

i want to find and sketch the domain of

f(x,y) = 1/[(x+y²)(2-x)]^½

so what I did

2-x > 0, thus x <2

x+y² > 0, so x > -y²

that right?

2. ## Re: help with multivariable domain

Originally Posted by Kuma
i want to find and sketch the domain of

f(x,y) = 1/[(x+y²)(2-x)]^½

so what I did

2-x > 0, thus x <2

x+y² > 0, so x > -y²

that right?
That is correct, x < 2, and x > -y² .

3. ## Re: help with multivariable domain

What about $x > 2$ and $x < -y^{2}?$

4. ## Re: help with multivariable domain

Originally Posted by Ackbeet
What about $x > 2$ and $x < -y^{2}?$
How can you have x take on both those values at the same time?

5. ## Re: help with multivariable domain

Originally Posted by Kuma
How can you have x take on both those values at the same time?
It's impossible. y² is ≥ 0 , so -y² ≤ 0 . Clearly 2 > 0 .

I don't know what Ackbeet was thinking.

6. ## Re: help with multivariable domain

Originally Posted by Kuma
How can you have x take on both those values at the same time?
It is necessary to express all the possible cases:

$\exists f(x,y)\Leftrightarrow [(x<2)\;\wedge \;(x>-y^2)]\;\vee\;[(x>2)\;\wedge\; (x<-y^2)]$

Another thing is that in this problem , the second case $(x>2)\;\wedge\; (x<-y^2)$ represents the empty set. That is the reason of the Ackbeet's remark.

7. ## Re: help with multivariable domain

Originally Posted by SammyS
It's impossible. y² is ≥ 0 , so -y² ≤ 0 . Clearly 2 > 0 .

I don't know what Ackbeet was thinking.
Yeah, all I wanted to do was point out that the product of two numbers can be positive if both of the multiplicands are negative. I hadn't gotten so far as to determine if there actually were any numbers in the set or not. So the answer is, I wasn't thinking as deeply as was necessary!