Question

How do you find the inverse of `f(x) = root5((3 x - 5) / (x - 5))`?

Answer 1

Let `y = f(x)` and solve for `x` to find:

`f^(-1)(y) = 10/(y^5-3) + 5`

Explanation:

Let `y = f(x) = root(5)((3x-5)/(x-5)`

Then:

`y^5 = (3x-5)/(x-5) = (3x-15+10)/(x-5) = 3+10/(x-5)`

Subtract `3` from both ends to get:

`y^5 - 3 = 10/(x-5)`

Multiply both sides by `(x-5)/(y^5 - 3)` to get:

`x - 5 = 10/(y^5-3)`

Add `5` to both sides to get:

`x = 10/(y^5-3) + 5`

So:

`f^(-1)(y) = 10/(y^5-3) + 5`

Answer 2

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

Explanation:

Write as

`y=root5((3x-5)/(x-5))`

Switch `x` and `y` and solve for `y`.

`x=root5((3y-5)/(y-5))`

`x^5=(3y-5)/(y-5)`

`x^5(y-5)=3y-5`

`x^5y-5x^5=3y-5`

`x^5y-3y=5x^5-5`

`y(x^5-3)=5(x^5-1)`

`y=(5(x^5-1))/(x^5-3)`