Question

How do you use composition of functions to show that `f(x)=(2+x)/x` and `f^-1(x) = 2/(x-1)` are inverses?

Answer 1

Step by step explanation is given below.

Explanation:

To prove `f(x)` and `f^-1(x)` we need to show
`(fcirc f^-1)(x)=x` and `(f^-1circf)(x)=x`

`f(x)=(2+x)/x` and `f^-1(x) = 2/(x-1)`

`(fcircf^-1)(x)`

`= (2+2/(x-1))/(2/(x-1))`

`=((2(x-1))/(x-1)+2/(x-1))/(2/(x-1))`

`=(2(x-1)+2)/(x-1) xx (x-1)/2`

`=(2x-2+2)/(x-1)xx(x-1)/2`

`=(2x)/(x-1)xx(x-1)/2`

`=(cancel(2)x)/cancel(x-1)xxcancel(x-1)/cancel(2)`

`=x`

We got that `(fcirc f^-1)(x)=x`

Similarly, we can prove `(f^-1circf)(x)=x`
`(f^-1circf)(x)`

`=2/((2+x)/x - 1)`

`=2/((2+x)/x- x/x)`

`=2/((2+x-x)/x)`

`=2/((2+cancel(x)-cancel(x))/x)`

`=2/(2/x)`

`=2/1 xx x/2`

`=cancel(2)/1xxx/cancel(2)`

`=x`

Therefore, `(f^-1circf)(x)=x`

Thus were are able to show that the functions are inverses of each other.