Prove that there is a differentiable function $f$ such that $[f(x)]^5 + f(x) + x = 0$. Spivak Ch. 12, Q. 14. Prove that there is a differentiable function $f$ such that $[f(x)]^5 + f(x) + x = 0$. (My textbook offers the following hint: Show that $f$ can be expressed as an inverse function.)
My Progress
$f(x) = -[f(x)]^5 - x$
$x = -[f \ (f^{-1} (x) )]^5 - f^{-1}(x)$
$-x^5 - x = f^{-1}(x)$
However, I do not know where to go next. The only thing that comes to mind is that $f$ is one-one since $f^{-1}$ is certainly a function. 
Any hints please?

I think the answer of Alex R. is on track, but I want to make things a bit clearer.
Let $g(y) = y^5 + y.$ Argue from one variable results that $g$ is a bijection of $\mathbb R$ onto $\mathbb R,$ with a differentiable inverse $g^{-1}.$ Define $f(x) = g^{-1}(-x).$ Then $f$ is differentiable on $\mathbb R,$ and we have $g(f(x)) = -x.$ This is the desired result.
