Domain and range in two coordinates let $f(x,y) = \sqrt{(x-1)(y-1)}$; what is the domain and the range?
I wanted to know if I was doing the domain part right and needed help with the range.
For the domain I got $(x-1)(y-1)$ are good as long as they are any real number. is that right?
 A: We need for $f(x, y) = \sqrt{(x - 1)(y - 1)}$ to be defined. So we must have $(x - 1)(y - 1) \geq 0$. Why?  
Then recall: The domain of a function are those values on which the function is defined. So we know that $f(x, y)$ is defined for all $(x, y)$ in the plane for which $(x - 1)(y - 1)\geq 0.\;$ We can pick, say, any $\,x\,$ we want. This will then determine the values of $\,y\,$ on which the function $\;f(x, y)\,$ is defined. We can determine $\,y\,$ as follows: $$(x-1) \lt 0 \implies (y - 1) \leq 0, \;\; (x-1) = 0 \implies y\in \mathbb R,\;\;\; (x -1) \gt 0 \implies (y - 1) \geq 0$$
All points $(x, y)$ satisfying the above inequalities are points in the domain of $f(x, y)$, which will points in $\mathbb R^2$, the real plane. And it will follow that the range will necessarily be points in $\mathbb R$ (the real line) such greater than or equal to zero: the set of all non-negative reals.
A: HINT: Assuming that you are working in the real numbers, the domain of $f$ is the set of $\langle x,y\rangle\in\Bbb R^2$ such that $(x-1)(y-1)\ge 0$: those are precisely the pairs for which the square root exists in $\Bbb R$. This means, for instance, that $\langle 2,0\rangle$ is not in the domain of $f$, since $(2-1)(0-1)=-1$ is negative. Thus, to describe the domain of $f$ in a nice way you will have to solve the inequality
$$(x-1)(y-1)\ge 0\;.$$
A product of two real numbers is non-negative precisely when one of the following holds:


*

*both factors are positive;  

*both factors are negative;  

*at least one of the factors is zero.


Can you describe the regions in the plane where:


*

*$x-1>0$ and $y-1>0$;  

*$x-1<0$ and $y-1<0$; and  

*$x-1=0$ or $y-1=0$?


The range is the set of possible values of $f(x,y)$. Clearly all of these are non-negative. Suppose that $z$ is any non-negative real number; can you find values of $x$ and $y$ (in terms of $z$) such that $f(x,y)=z$? If can do so for all $z\ge 0$, you’ll have shown that the range of $f$ is the set of all non-negative real numbers.
A: The square root is not defined (in the reals) if the thing under it is negative, so you need $x-1$ and $y-1$ to be of the same sign.  For any $x$ there are some $y$ that are acceptable, but the domain is a region of the plane.
