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?
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.
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!