Need help with the following: if `f(x) = 2x-3;` `g(x) = 1/(x+1),` find `(f@g)^(-1)(x).`

Hello!

First found the composition:

`(f@g)(x)=f(g(x))=f(1/(x+1))=2*1/(x+1)-3=-(3x+1)/(x+1).`

This function is defined everywhere except `x=-1.`

To find the inverse function, we have to solve for `x` the equation `(f@g)(x)=y:`

`-(3x+1)/(x+1)=y,`  or `(3x+1)/(x+1)=-y.`

Multiply both sides by `(x+1)` and obtain

`3x+1=-xy-y.`

Move the terms with `x` to the left and without `x` to the right:

`x(y+3)=-y-1=-(y+1),`

so  `x=-(y+1)/(y+3)`  (of course `y!=-3` ).

Thus the answer is: the function `(f@g)^(-1)(y)` exists for all `y!=-3` and is equal to `-(y+1)/(y+3).`