Question

How do you find the domain and range of `f(x) = (x-2)/(x+1)`?

Answer 1

The domain is restricted as the denominator cannot be zero.

i.e. if x + 1 = 0, then x = -1

Hence, the domain can take all real values except x = -1

`f(x) = (x - 2)/(x + 1)`

i.e. `f(x) = (x + 1)/(x + 1) - 3/(x + 1)`

or, `f(x) = 1 - 3/(x + 1)`

as x gets very small or very big, 3/(x + 1) tends towards zero

i.e. f(x) approaches a limit of 1

Hence, the range of f(x) takes all real values except 3

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Answer 2

The domain of `f(x)` is `D_f(x)=RR-{-1}`
The range of `f(x)`, `R_f(x)=RR-{1}`

Explanation:

As you cannot divide by `0`, `x!=-1`

The domain of `f(x)` is `D_f(x)=RR-{-1}`

To find the range, we need `f^-1(x)`

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

Interchange `y` and `x`

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

We rewrite `y` as function of `x`

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

`xy+x=y-2`

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

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

So,

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

The domain of `f^-1(x)` is `D_(f^-1)(x)=RR-{1}`

This is the range of `f(x)`, `R_f(x)=RR-{1}`