Thread: 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?

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 .

What about and

Originally Posted by Ackbeet

What about and

How can you have x take on both those values at the same time?

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.

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:



Another thing is that in this problem , the second case represents the empty set. That is the reason of the Ackbeet's remark.

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!