Question

How do you verify that `f(x)=(3-x)/x; g(x)=3/(x+1)` are inverses?

Answer 1

If `f(x)` and `g(x)` are inverses, then both of the following must be true:

`f(g(x)) = x` and `g(f(x)) = x`

Explanation:

Check `f(g(x))`:

Start with `f(x)`:

`f(x)=(3-x)/x`

Substitute `g(x)` for for every x in `f(x)`:

`f(g(x))=(3-g(x))/g(x)`

Substitute the right of `g(x)=3/(x+1)` into every `g(x)` on the right side of the above equation:

`f(g(x))=(3-3/(x+1))/(3/(x+1))`

Multiply the right side by 1 in form of `(x+1)/(x+1)`:

`f(g(x))=(x+1)/(x+1)(3-3/(x+1))/(3/(x+1))`

This makes the embedded denominators disappear:

`f(g(x))=(3(x+1)-3)/3`

There is a common factor of `3/3` that becomes 1:

`f(g(x))=x+1-1`

`f(g(x))=x`

Check `g(f(x))`

Start with `g(x)`:

`g(x)=3/(x+1)`

Substitute `f(x)` for every x:

`g(f(x))=3/(f(x)+1)`

Substitute the right side of `f(x)=(3-x)/x` for every `f(x)` on the right side of the above equation:

`g(f(x))=3/((3-x)/x+1)`

Multiply by 1 in the form of `x/x`:

`g(f(x))=x/x 3/((3-x)/x+1)`

This makes the embedded denominator disappear:

`g(f(x))= (3x)/(3-x+x)`

`-x + x` becomes 0:

`g(f(x))= (3x)/3`

`g(f(x))= x`

Both are true, therefore, they are inverses.