Question

How do you find the domain and range of `g(x)=sqrt((x+3)/(x-2))`?

Answer 1

Range: `" " x in (-oo, -3] uu (2, + oo)`

Domain: `" " g(x) in [0,1 ) uu (1, + oo)`

Explanation:

In order to find the domain of this function, you need to find all the values that `x` can take for which `g(x)` is defined.

Right from the start, you know that the denominator of the fraction cannot be equal to `2` because that would make the function undefined.

`x - 2 != 0 implies x != 2`

Now, you know that when working with real numbers, you can only take the square root of a positive number.

This implies that you must have

`(x+3)/(x-2) >= 0`

You know that when `x = -3`, you have

`(-3 + 3)/(-3 - 2) = 0/(-5) >= 0`

so you can say that `x = {-3}` will be included in the domain of the function.

Now, in order to have

`(x+3)/(x-2) > 0`

you must look at two possible situations

`color(white)(a)`

• `x + 3 >0" " ul(and) " " x -2 > 0`

In this case, you must have

`x + 3 > 0 implies x >= -3`

and

`x - 2 > 0 implies x > 2`

This implies that the solution interval will be

`(-3, + oo) nn (2, + oo) = (2, + oo)`

This tells you that any value of `x` that is greater than `2` will satisfy the inequality `(x+3)/(x-2) > 0`.

`color(white)(a)`

• `x + 3 <0" "ul(and)" " x - 2 < 0`

In this case, you must have

`x + 3 < 0 implies x < -3`

and

`x - 2 < 0 implies x < 2`

This implies that the solution interval will be

`(- oo, -3) nn (-oo, 2) = (-oo, -3)`

This tells you that any value of `x` that is less than `-3` will also satisfy the inequality `(x + 3)/(x - 2) > 0`.

Therefore, the domain of the function will be--remember that `x = -3` is also included in the domain!

`"domain: " color(darkgreen)(ul(color(black)(x in (-oo, -3] uu (2, oo))))`

This tells you that any value of `x` that is less than or equal to `-3` or greater than `2` will get you

`(x+3)/(x-2) >= 0`

Now, to find the range of the function, you must determine the values that `g(x)` can take for any value of `x` that is part of its domain.

Since you're working with real numbers, you can say that taking the square root of a positive number will always return a positive number.

`g(x) >= 0`

You know that when `x = -3`, you have

`g(-3) = sqrt( (-3 + 3)/(-3 - 2)) = 0`

Now, it's important to realize that the range of the function will not include `1` because you can never have

`color(red)(cancel(color(black)(x))) + 3 = color(red)(cancel(color(black)(x))) -2`

`3 != - 2`

This means that you don't have a value of `x` for which `(x+3)/(x-2) = 1`.

Therefore, the range of the function will be

`"range: " color(darkgreen)(ul(color(black)(g(x) in [0, 1) uu (1, + oo))))`

graph{sqrt( (x+3)/(x-2)) [-16.02, 16.01, -8.01, 8]}