Question

What is the inverse function of `g(x)= (x + 2) / (x - 3)`?

Answer 1

`g^(-1)(x) = (3x+2)/(x-1), color(white)("XXX")x!=1`

Explanation:

Replacing `g(x)` with `y`
`color(white)("XXX")y=(x-3)/(x+2)`

`rArrcolor(white)("XXX")y*(x-3) = x+2`

`rArrcolor(white)("XXX")xy-3y=x+2`

`rArrcolor(white)("XXX")xy-x=3y+2`

`rArrcolor(white)("XXX")x(y-1)=3y+2`

and provided `x!=1`
`rArrcolor(white)("XXX")x = (3y+2)/(y-1)`

We could write this as a function in terms of `y` as

`color(white)("XXX")f(y) = (3y+2)/(y-1)`
or in terms of `x` as
`color(white)("XXX")f(x) = (3x+2)/(x-1)`
with
`color(white)("XXX")f(x)=g^(-1)(x)`