Question

What is the domain and range of `y = (2x^2)/( x^2 - 1)`?

Answer 1

The domain is `x in (-oo,-1) uu (-1,1) uu (1,+oo)`
The range is `y in (-oo,0] uu (2,+oo)`

Explanation:

The function is

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

We factorise the denominator

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

Therefore,

`x!=1` and `x!=-1`

The domain of y is `x in (-oo,-1) uu (-1,1) uu (1,+oo)`

Let's rearrage the function

`y(x^2-1)=2x^2`

`yx^2-y=2x^2`

`yx^2-2x^2=y`

`x^2=y/(y-2)`

`x=sqrt(y/(y-2))`

For `x` to a solution, `y/(y-2)>=0`

Let `f(y)=y/(y-2)`

We need a sign chart

`color(white)(aaaa)` `y` `color(white)(aaaa)` `-oo` `color(white)(aaaaaa)` `0` `color(white)(aaaaaaa)` `2` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `y` `color(white)(aaaaaaaa)` `-` `color(white)(aaa)` `0` `color(white)(aaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `y-2` `color(white)(aaaaa)` `-` `color(white)(aaa)` `color(white)(aaa)` `-` `color(white)(aa)` `||` `color(white)(aa)` `+`

`color(white)(aaaa)` `f(y)` `color(white)(aaaaaa)` `+` `color(white)(aaa)` `0` `color(white)(aa)` `-` `color(white)(aa)` `||` `color(white)(aa)` `+`

Therefore,

`f(y)>=0` when `y in (-oo,0] uu (2,+oo)`

graph{2(x^2)/(x^2-1) [-16.02, 16.02, -8.01, 8.01]}