Question

What is the inverse function of `h(x)= 3-(x+4)/(x-7)`?

Answer 1

`h^-1(y) = 7 - 11/(y-2)` with restriction `y != 2`

Explanation:

First re-factor the equation defining `h(x)` to have just one occurrence of `x`, then move the operations away from the term in `x` to leave it isolated.

`y = h(x) = 3 - (x+4)/(x-7)`

`=3-((x-7)+11)/(x-7)`

`=3-cancel(x-7)/cancel(x-7)-11/(x-7)`

`=3 - 1 - 11/(x-7)`

`=2-11/(x-7)`

Subtract `2` from both ends to get:

`y - 2 = -11/(x-7)`

Multiply both sides by `(x-7)` to get:

`(x-7)(y - 2) = -11`

Divide both sides by `(y-2)` to get:

`x -7 = -11/(y - 2)`

Add `7` to both sides to get:

`x = 7-11/(y-2)`

This defines `x` in terms of `y = h(x)`, so:

`h^-1(y) = 7 - 11/(y-2)`

with restriction `y != 2`