Question

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

Answer 1

The domain is `(-oo, 2) uu (2, oo)`, i.e. `RR "\" { 2 }`

The range is `(-oo, 1) uu (1, oo)`, i.e. `RR "\" { 1 }`

Explanation:

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

When `x = 2`, the denominator of `f(x)` is `0`, but the numerator is non-zero. So `f(x)` is undefined and has a vertical asymptote at `x = 2`.

If we let `y = f(x) = 1+1/(x-2)`, then:

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

So:

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

So:

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

So:

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

This is well defined for all `y in RR` except `y=1`, where the inverse function `f^(-1)(y)` has a vertical asymptote - so the original function has a horizontal asymptote `y=1`.

Since `y=1` is not in the domain of the inverse function, it is not in the range of the original function.

To summarise:

The domain of `(x-1)/(x-2)` is `RR "\" { 2 } = (-oo, 2) uu (2, oo)`

The range of `(x-1)/(x-2)` is `RR "\" { 1 } = (-oo, 1) uu (1, oo)`

graph{(x-1)/(x-2) [-8.13, 11.87, -3.88, 6.12]}