# help with multivariable domain

• Oct 21st 2011, 04:51 PM
Kuma
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?
• Oct 21st 2011, 05:26 PM
SammyS
Re: help with multivariable domain
Quote:

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 .
• Oct 21st 2011, 06:09 PM
Ackbeet
Re: help with multivariable domain
What about \$\displaystyle x > 2\$ and \$\displaystyle x < -y^{2}?\$
• Oct 21st 2011, 09:13 PM
Kuma
Re: help with multivariable domain
Quote:

Originally Posted by Ackbeet
What about \$\displaystyle x > 2\$ and \$\displaystyle x < -y^{2}?\$

How can you have x take on both those values at the same time?
• Oct 21st 2011, 10:39 PM
SammyS
Re: help with multivariable domain
Quote:

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.
• Oct 21st 2011, 10:52 PM
FernandoRevilla
Re: help with multivariable domain
Quote:

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:

\$\displaystyle \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 \$\displaystyle (x>2)\;\wedge\; (x<-y^2)\$ represents the empty set. That is the reason of the Ackbeet's remark.
• Oct 22nd 2011, 01:34 AM
Ackbeet
Re: help with multivariable domain
Quote:

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!