Question

How do you find the domain and range of `sqrt((13x)/((x^2)-1)`?

Answer 1

The domain is `x in (-1,0] uu (+1,+oo)`
The range is `y in [0,+oo)`

Explanation:

What's uner the `sqrt` sign is `>=0`

So,

`(13x)/(x^2-1)>=0`

`(13x)/((x+1)(x-1))>=0`

Let `f(x)=(13x)/((x+1)(x-1))`

We can build a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaa)` `-1` `color(white)(aaaaa)` `0` `color(white)(aaaaaaa)` `+1` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+1` `color(white)(aaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `||` `color(white)(aa)` `+`

`color(white)(aaaa)` `x` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `||` `color(white)(aa)` `+`

`color(white)(aaaa)` `x-1` `color(white)(aaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aa)` `+`

Therefore,

`f(x)>=0` when `x in (-1,0] uu (+1,+oo)`

The domain is `x in (-1,0] uu (+1,+oo)`

Let,

`y=sqrt((13x)/(x^2-1))`

When `x=-1`, `=>`, `y=+oo`

When `x=0`, `=>`, `y=0`

The range is `y in [0,+oo)`