*If `f(x)=(2x)/(x-5)` find `f^(-1)(x)` and the domain of `f^(-1)(x)` .*

One method is to let `y=(2x)/(x-5)` , then exchange x and y, then solve for y. Thus:

`x=(2y)/(y-5)` exchange x and y

`x(y-5)=2y`

`xy - 5x=2y`

`xy-2y=5x`

`y(x-2)=5x`

`y=(5x)/(x-2)` This is a new function which is the inverse

of the given function.

We exchange x and y because the input of the inverse function is the output of the original function and vice versa. In other words, the domain of the inverse function is the range of the original function.

**Thus `f(x)=(2x)/(x-5) => f^(-1)(x)=(5x)/(x-2)` **. **The domain of `f^(-1)(x)` is `x!= 2` .**

** Note that the range of the original function is `y != 2` as there is a horizontal asymptote at y=2. The domain of the original function is `x != 5` , and the range of the inverse function is `y != 5` **