Question

How do you find the inverse of `f(x) = x^3 + 2x` and is it a function?

Answer 1

`f^(-1)(y) = root(3)((9y+sqrt(81y^2+96))/18) + root(3)((9y-sqrt(81y^2+96))/18)`

which is a function.

Explanation:

Let `y = f(x) = x^3+2x`

We want to solve:

`x^3+2x-y = 0`

for `x` (i.e. treating `y` as a constant)

Note that `x^3+2x = x(x^2+2)` is strictly monotonic increasing. So there will be exactly one Real root for any Real value of `y`.

graph{x^3+2x [-10, 10, -5, 5]}

Use Cardano's method...

Let `x = u+v`. Then:

`u^3+v^3+(3uv+2)(u+v)-y=0`

Add the constraint `v=-2/(3u)` to eliminate the `(u+v)` term.

Then our equation becomes:

`u^3-8/(27u^3)-y = 0`

Multiply through by `27u^3` and rearrange slightly to get:

`27(u^3)^2-27y(u^3)-8 = 0`

Using the quadratic formula, we get:

`u^3 = (27y+-sqrt((-27y)^2-4(27)(-8)))/(2*27)`

`=(27y+-sqrt(729y^2+864))/54`

`=(27y+-3sqrt(81y^2+96))/54`

`=(9y+-sqrt(81y^2+96))/18`

Since these roots are Real and the derivation is symmetric in `u` and `v`, we can use one of these roots for `u^3` and the other for `v^3` to find the Real root of the cubic:

`x = root(3)((9y+sqrt(81y^2+96))/18) + root(3)((9y-sqrt(81y^2+96))/18)`

The other two roots are non-Real complex roots and need not concern us.

This Real zero gives us an inverse function for the original cubic function `f(x)`

`f^(-1)(y) = root(3)((9y+sqrt(81y^2+96))/18) + root(3)((9y-sqrt(81y^2+96))/18)`