Question

If f(x) = 2/3x - 6, find f^-1. Justify your answer using the composition of functions f(f^-1(x))=f^-1(f(x))

Answer 1

`f^(-1)(x)= (3)/(2)(x+6)`

`f(f^(-1)(x)) = f^(-1)(f(x))= x`

Explanation:

Given

`f(x) = (2)/(3)x - 6`

Required

Determine
`f^(-1)(x)`

`f(x) = (2)/(3)x - 6`

Represent f(x) as y

`y=(2)/(3)x - 6`

Swap the positions of x and y

`x=(2)/(3)y - 6`

Add 6 to both sides

`x+6=(2)/(3)y - 6+6`

`x+6=(2)/(3)y`

Multiply both sides by 3

`3(x+6)=(2)/(3)y * 3`

`3(x+6)=2y`

Divide both sides by 2

`(3)/(2)(x+6)=(2y)/(2)`

`(3)/(2)(x+6)=y`

`y= (3)/(2)(x+6)`

Replace y with
`f^(-1)(x)`

`f^(-1)(x)= (3)/(2)(x+6)`

Justify the result:

First, we solve for
`f(f^(-1)(x))`

`f(x) = (2)/(3)x - 6`

`f(x) = (2)/(3)x - 6` becomes

`f(f^(-1)(x)) = (2)/(3)x - 6`

Substitute
`(3)/(2)(x + 6)` for x

`f(f^(-1)(x)) = (2)/(3)((3)/(2)(x + 6)) - 6`

`f(f^(-1)(x)) = (2*3)/(3*2)(x + 6)- 6`

`f(f^(-1)(x)) = (6)/(6)(x + 6)- 6`

`f(f^(-1)(x)) = 1*(x + 6)- 6`

`f(f^(-1)(x)) = x + 6- 6`

`f(f^(-1)(x)) = x`

Next, we solve
`f^(-1)(f(x))`

`f^(-1)(x)= (3)/(2)(x+6)`

`f^(-1)(x)= (3)/(2)(x+6)` becomes

`f^(-1)(f(x))= (3)/(2)(x+6)`

Substitute
`(2)/(3)x - 6` for x

`f^(-1)(f(x))= (3)/(2)((2)/(3)x) - 6+6`

`f^(-1)(f(x))= (3)/(2)((2)/(3)x)`

`f^(-1)(f(x))= (2*3)/(3*2)x`

`f^(-1)(f(x))= (6)/(6)x`

`f^(-1)(f(x))= 1*x`

`f^(-1)(f(x))= x`

Hence:

`f(f^(-1)(x)) = f^(-1)(f(x))= x`