Question

What is the inverse function of `f(x) = x^5+x`?

Answer 1

The inverse function is expressible in terms of the Bring Radical, as follows:

`f^(-1)(y) = BR(-y)`

Explanation:

Let `y = f(x) = x^5 + x`.

Then `x^5 + x - y = 0`

This is a quintic equation in Bring-Jerrard normal form.

It is not solvable using normal radicals.

The Bring Radical of a number `a` is a root of `x^5+x+a = 0`, chosen so that the Bring Radical of a Real number is itself Real.

This is obviously well defined for Real numbers since this quintic equation has exactly one Real root if `a` is Real. You can see this because `f'(x) = 5x^4+1 > 0` for all `x in RR`, so `f` is strictly monotonically increasing.

In our case, `a = -y`, so `f^(-1)(y) = BR(-y)`

For more info on the Bring Radical see https://en.wikipedia.org/wiki/Bring_radical