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?

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- Oct 21st 2011, 04:51 PMKumahelp 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 PMSammySRe: help with multivariable domain
- Oct 21st 2011, 06:09 PMAckbeetRe: help with multivariable domain
What about $\displaystyle x > 2$ and $\displaystyle x < -y^{2}?$

- Oct 21st 2011, 09:13 PMKumaRe: help with multivariable domain
- Oct 21st 2011, 10:39 PMSammySRe: help with multivariable domain
- Oct 21st 2011, 10:52 PMFernandoRevillaRe: help with multivariable domain
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 AMAckbeetRe: help with multivariable domain
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!