could someone help me determine an inverse?

Question

what's inverse of f(x)= x^(5) + 2x^(3) + x -1?

Answer 1

let

y = f'(x)

f(y) = x

y^5 + 2y^3 + y -1 = x

y(y^4 + 2y^2 + 1) - 1 = x
y(y^2 + 1)^2 - 1 = x
(y^4 + y^2) = sqrt(x + 1)

let u = y^2

(u^2 + u + (1/2)^2 - (1/2)^2)
(u + 1/2)^2 - 1/4 = sqrt(x + 1)
(u + 1/2)^2 = sqrt(x + 1) + 1/4
u = sqrt(sqrt(x + 1) + 1/4) - 1/2
u = y^2

y = sqrt(sqrt(sqrt(x + 1) + 1/4) - 1/2)

f'(x) = sqrt(sqrt(sqrt(x + 1) + 1/4) - 1/2)

this is the inverse.

Answer 2

The usual method is to swap x and y and then solve for y. You'd be solving

x = y^5 + 2y³ + y - 1 for y. Good luck with that. Graph shows it satisfies the horizontal line test, so it DOES have an inverse, but.....? The polynomial has no obvious factors, not that they'd help.

My ti-89 solves it as y = -(2x³ - 1)^(1/5). I've been replaced by a machine.