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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