4

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?

Brian M. Scott
  • 616,228
jain smit
  • 333

3 Answers3

4

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.

amWhy
  • 209,954
1

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:

  1. both factors are positive;
  2. both factors are negative;
  3. at least one of the factors is zero.

Can you describe the regions in the plane where:

  1. $x-1>0$ and $y-1>0$;
  2. $x-1<0$ and $y-1<0$; and
  3. $x-1=0$ 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.

Brian M. Scott
  • 616,228
0

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.

Ross Millikan
  • 374,822
  • so pretty much any x and y values(real numbers) work as long as they have the same signs? – jain smit Jun 07 '13 at 23:55
  • @bobadam: No, you need $[x \ge 1$ and $y \ge 1]$ or $[x \le 1$ and $y \le 1]$. It would be as you say if the quantity under the radical were $xy$ – Ross Millikan Jun 08 '13 at 02:02