How do you find the inverse of #f(x) = x^5+x#?

1 Answer
Jul 18, 2015

#f^(-1)(y) = BR(-y)# where #BR# is the Bring radical.

https://en.wikipedia.org/wiki/Bring_radical

Explanation:

We can prove that #f(x)# has an inverse easily enough, in that

#d/(dx)f(x) = 5x^4+1 >= 1# for all #x in RR#

so #f(x)# is strictly monotonically increasing.

The problem is that a quintic of the form #x^5+x+a = 0# does not generally have a solution expressible using normal arithmetic operations and radicals. This is where the Bring radical comes in.

The Bring radical of a number #a# is a root of #x^5+x+a = 0#.

If #a# is real, then it is the unique real root.

So if #y = f(x) = x^5 + x#, then #x^5 + x + (-y) = 0#

and #BR(-y)# is the real root of #x^5 + x + (-y) = 0#