# Thread: Finding the range of a relation

1. ## Finding the range of a relation

I have troubles in working out the range of the following relations:

y=x-a, x<0, a>0

y=(a/x)+a, a>0

May you please list out the steps because I really want to understand the way~

Also:

y=square root x, x>_0 ---> Shouldn't the range be "all real numbers"? I saw the answer is [0,infinity[

(Because for example, assuming that x=81, so y=square root 81, which equals either (-9) or (9), meaning that negative numbers are also accepted, so I think it is "all real numbers") Am I right?

2. ## Re: Finding the range of a relation

As for the square root one, there is a major misunderstanding.

If you had $\displaystyle y^2 = x$, then yes, your range of y values would be all reals.

But you have the $\displaystyle y = \sqrt x$, which is understood to be THE positive square root (singular) of x.

So you will never get a negative output.

The following is also true, and might help:

If $\displaystyle a^2 = 4$ then a = +/- 2

If $\displaystyle b = \sqrt 9$, then b = 3.

3. ## Re: Finding the range of a relation

hi kevinlam2490

you have thewrong idea.square root of 81 is just 9,because its value is always positive.

it is much different than solving the equation x^2=81.