Question

How do you find the inverse of `f(x)=(2x+7)/(3x-1)`?

Answer 1

Let `y=f(x)`.

Eliminate all but one of the `x`'s then rearrange to isolate and find a formula for `x` in terms of `y`. This is the inverse function.

`f^(-1)(y) = (y+7)/(3y-2)`

Explanation:

Let `y = f(x) = (2x+7)/(3x-1)`

`y = (2x+7)/(3x-1)`

I like to reduce the number of occurrences of `x` to one. Then it is clear how to isolate `x` on one side of the equation.

The `(2x ...) / (3x ...)` is going to make things a little messy -

unless we multiply through by `3` first, so let's do that:

Multiply both sides by `3` to get:

`3y = (3(2x+7))/(3x-1)`

`=(6x+21)/(3x-1)`

`=(6x-2+23)/(3x-1)`

`=(2(3x-1)+23)/(3x-1)`

`=2+23/(3x-1)`

Subtract `2` from both sides to get:

`3y - 2 = 23/(3x-1)`

Multiply both sides by `(3x-1)` to get:

`(3y-2)(3x-1) = 23`

Divide both sides by `(3y-2)` to get:

`3x-1 = 23/(3y-2)`

Add `1` to both sides to get:

`3x = 23/(3y-2)+1`

`=23/(3y-2)+(3y-2)/(3y-2)`

`=(23+(3y-2))/(3y-2)`

`=(3y+21)/(3y-2)`

`=(3(y+7))/(3y-2)`

Divide both sides by `3` to get:

`x = (y+7)/(3y-2)`

So `f^(-1)(y) = (y+7)/(3y-2)`

Answer 2

`f^-1(x)=(x+7)/(3x-2)`

Explanation:

Replace `f(x)` with `y`.

`y=(2x+7)/(3x-1)`

Switch all occurrences of `x` with `y` and the `y` with `x`.

`x=(2y+7)/(3y-1)`

Solve for `y`. This may seem daunting, but just start by trying to eliminate fractions by cross-multiplying.

`x(3y-1)=2y+7`

Distribute on the left.

`3xy-x=2y+7`

Get all terms with `y` on one side of the equation. Move anything not including a `y` term to the other side of the equation.

`3xy-2y=x+7`

Factor a `y` term on the left.

`y(3x-2)=x+7`

Divide both sides by `(3x-2)` to solve for `y`.

`y=(x+7)/(3x-2)`

Since this is the inverse function, we can write is with `f^-1(x)` instead of `y`. All `f^-1(x)` means is the inverse of the function `f(x)`.

`f^-1(x)=(x+7)/(3x-2)`