Question

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

Answer 1

Most important limitation is that the denominator may not be `0`

Explanation:

So the domain is limited by `x!=2`

As `x->2` the fraction gets larger and larger, whether you go to `2` from above or below. In "the language":

`lim_(x->2^-) f(x)=-oo` and `lim_(x->2^+) f(x)=+oo`

If `x` gets larger the fraction will become smaller, but never quite reaching `0`, so the range is `f(x)!=0` or:

`lim_(x->-oo) f(x)=0` and `lim_(x->+oo) f(x)=0`
graph{1/(x-2 [-10, 10, -5, 5]}

`x=2andf(x)=0` are called asymptotes .

Answer 2

The domain of an inverse function is the range of the function and vice versa.

Explanation:

For `f(x) = 1/(x-2)`, the domain of `f` is all reals except `2`, so the range of `f^-1` is all reals except `2`.

The range of `f` is all reals except `0`, so the domain of `f^-1` is all reals except `0`.

Notice that is we solve `y = 1/(x-2)` for `x`, we get:

`y(x-2)=1`

`xy-2y=1`

`xy=2y+1`

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

We can see from this that for the original function, `f`, we can get every number for `y` except `0`.
That is the range of `f` and the domain of `f^-1`