Question

How do I find the inverse of `f(x)=(x+3)/(x-2)`?

Answer 1

Please see the explanation. `f^-1(x) = 5/(x - 1) + 2`

Explanation:

let `x = f^-1(x)` and then substitute everywhere there is an x:

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

Let left side becomes x by definition:

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

Add zero to the numerator in the form -2 + 2:

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

Regroup:

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

Break into two fractions:

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

Substitute 1 for `(f^-1(x) -2)/(f^-1(x) - 2)` and 5 for `3 + 2`

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

Subtract 1 from both sides:

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

Multiply both sides by `(f^-1(x) - 2)/(x - 1)`

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

Add 2 to both sides:

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

To prove that it is an inverse, one must show that `f(f^-1(x)) = x and f^-1(f(x)) = x`.

`f(f^-1(x)) = ((5/(x - 1) + 2) + 3)/((5/(x - 1) + 2) - 2)`

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

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

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

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

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

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

Half of the proof is done.

Now for `f^-1(f(x))`:

`f^-1(f(x)) = 5/((x + 3)/(x - 2) - 1) + 2`

`f^-1(f(x)) = 5/((x + 3)/(x - 2) - 1)(x - 2)/(x - 2) + 2`

`f^-1(f(x)) = (5(x - 2))/(x + 3 - (x - 2)) + 2`

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

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

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

Q.E.D.