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

1 Answer
George C.
Sep 12, 2015

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

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