Question

How do you find the inverse of `f( x ) = \frac { 5x - 2} { x + 5}`?

Answer 1

Substitute `x = f^-1(x)` for every x; this will allow you to make the substitution `f(f^-1(x)) = x`
Solve the resulting equation for `f^-1(x)`
Verify that `f(f^-1(x)) = x` and `f^-1(f(x)) = x`

Explanation:

Given: `f(x) = (5x-2)/(x+5)`

Substitute `x = f^-1(x)` for every x:

`f(f^-1(x)) = (5(f^-1(x))-2)/((f^-1(x))+5)`

Substitute the property `f(f^-1(x)) = x` into the left side:

`x = (5f^-1(x)-2)/(f^-1(x)+5)`

Solve the equation for `f^-1(x)`:

`x(f^-1(x)+5) = 5(f^-1(x))-2`

`xf^-1(x)+5x = 5(f^-1(x))-2`

`xf^-1(x))-5(f^-1(x)) = -5x-2`

`(x-5)f^-1(x)= -5x-2`

`f^-1(x)= (5x+2)/(5-x)`

Verify that `f(f^-1(x)) = x` and `f^-1(f(x)) = x`

Start with:

`f(x) = (5x-2)/(x+5)`

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

`f(f^-1(x)) = (5((5x+2)/(5-x))-2)/(((5x+2)/(5-x))+5)`

`f(f^-1(x)) = (5(5x+2)-2(5-x))/(5x+2+5(5-x))`

`f(f^-1(x)) = (25x+10-10+2x)/(5x+2+25-5x)`

`f(f^-1(x)) = (27x)/27`

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

Start with:

`f^-1(x)= (5x+2)/(5-x)`

Substitute `x=f(x)`

`f^-1(f(x))= (5((5x-2)/(x+5))+2)/(5-((5x-2)/(x+5)))`

`f^-1(f(x))= (5(5x-2)+2(x+5))/(5(x+5)-5x+2)`

`f^-1(f(x))= (25x-10+2x+10)/(5x+25-5x+2)`

`f^-1(f(x))= (27x)/27`

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

Both are verified `f^-1(x)= (5x+2)/(5-x)`