Question

How do you find the domain and range of `g(x)= (3+x^2)/(4-x^2)`?

Answer 1

The domain is `x in (-oo,-2)uu(-2,2)uu(2,+oo)`
The range is `g(x) in (-oo,-1)uu [3/4,+oo)`

Explanation:

As you cannot divide by `0`, the denominator is `!=0`

`4-x^2!=0`

`(2+x)(2-x)!=0`

`=>`, `x!=-2` and `x!=2`

Therefore,

The domain is `x in (-oo,-2)uu(-2,2)uu(2,+oo)`

Let

`y=(3+x^2)/(4-x^2)`

`y(4-x^2)=3+x^2`

`4y-yx^2=3+x^2`

`x^2(1+y)=4y-3`

`x^2=(4y-3)/(1+y)`

`x=sqrt((4y-3)/(1+y))`

Therefore,

`(4y-3)/(1+y)>=0`

`=>`, `y!=-1`

Let `g(y)=(4y-3)/(1+y)`

We build a sign chart

`color(white)(aaaa)` `y` `color(white)(aaaa)` `-oo` `color(white)(aaaaaa)` `-1` `color(white)(aaaaaaa)` `3/4` `color(white)(aaaaaaa)` `+oo`

`color(white)(aaaa)` `1+y` `color(white)(aaaaaa)` `-` `color(white)(aaa)` `||` `color(white)(aaaa)` `+` `color(white)(aaaaaaa)` `+`

`color(white)(aaaa)` `4y-3` `color(white)(aaaaa)` `-` `color(white)(aaa)` `||` `color(white)(aaaa)` `-` `color(white)(aa)` `0` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `g(y)` `color(white)(aaaaaaa)` `+` `color(white)(aaa)` `||` `color(white)(aaaa)` `-` `color(white)(aa)` `0` `color(white)(aaaa)` `+`

Therefore,

`g(y)>=0` when `y in (-oo,-1)uu [3/4,+oo)`

graph{(3+x^2)/(4-x^2) [-12.66, 12.65, -6.33, 6.33]}