The domain of a multivariable function

[DottZakapa]

TL;DR

domain of multivariable function

hey there
I'm struggling on finding the domain of the following function

log (xy²)+x²y)

I then do

xy(y+x)>0

but then i don't know what to do with xy

one attempt

\begin{cases}
y+x>0\\
x>0\\
y>0
\end{cases} union
\begin{cases}
y+x<0\\
x<0\\
y<0 \end{cases}

but this doesn't lead to the correct solution

 

[member 587159]

Indeed, you have to describe the set

$$V=\{(x,y) \in \mathbb{R}^2 \mid xy(x+y) > 0\}$$

When is a product $abc > 0$. Exactly in the following cases:

$a,b,c > 0$
$a,b < 0, c > 0$
$a,c < 0, b > 0$
$b,c < 0, a > 0$

so you have to consider 4 cases (by symmetry, actually only 3).

Note that the second case you gave is wrong. Then the product will be $<0$ and you need $> 0$.

 

[DottZakapa]

so there is a total of 4 systems to be solved right? in fact it works
thanks

 

[member 587159]

so there is a total of 4 systems to be solved right? in fact it works
thanks

Yes, and then you have to take the union of the four solution sets. Maybe you can write the union in the end a little nicer. I did not try it myself though.

 

[DottZakapa]

what if you have
ab>0 ?
is
a,b>0
a>0,b<0
a<0, b<0

 

[mfb]

If a>0, b<0, how could ab>0 be true?

 

[HallsofIvy]

First the function in your first post, log(xy^2)+ x^2y), has two right parentheses and only one left parenthesis so is ambiguous. Do you mean log(xy^2)+ x^2y or log(xy^2+ x^2y)? The domain for the first is "x&gt; 0, y\ne 0". The domain for the second is the set of all x, y such that xy^2+ x^2y= xy(x+ y)&gt; 0 The product of three numbers is positive if all three numbers are positive or if one is positive and the other two negative. Of course if x and y are both positive so is x+ y so the other possibilities are
(1) x is positive, y and x+y negative. That is, y< -x.
(2) y is positive, x and x+y negative. That is, x< -y.

In terms of sets, "x and y positive" is the first quadrant.

y= -x, with x> 0, is the "half-line" in the fourth quadrant from the origin at 45 degrees to the x-axis. y< -x is the region in the fourth quadrant below that line.

x= -y, with y> 0, is the "half-line" in the second quadrant from the origin at 45 degrees to the x-axis. x< -y is the region in the second quadrant below that line.