Prove that $y = x^3 + {5 \over 9} x $ is invertible

Question

Prove that $y = x^3 + {5 \over 9} x $ is invertible

How do I even express x in terms of y for cubic equations? For even powers I could use for example substitution, but no idea comes to my mind for odd ones.

I've asked W|A to solve for $x$ and it suggests that i should substitute $x$ with some $t+{ \lambda \over t}$, but that eventually brings an equation with even higher order and additionally there appears some constant $\lambda$ to be determined which seems to be way too complicated. I believe there exists a simpler solution.

UPD:

I'm not supposed to use derivative or other methods from "calculus".

So eventually this comes to a couple of questions:

1. Is there a simpler way of expressing $x$ in terms of $y$ than the above?
2. Is it possible to generalize the solution for the polynomials of the following kind: $y = x^n + px + q$ where $n$ is odd; $n, p>0$ and $q \in \mathbb R$?

Answer 1

Let $f(x) = x^3+\frac{5}{9} x$. Then, for reals $a\neq b$, $$f(a)=f(b) \iff a^3-b^3 +\frac{5}{9}(a-b)=0 \iff (a-b)\left(a^2+ab+b^2+\frac{5}{9}\right)=0.$$ But $$\left(a^2+ab+b^2+\frac{5}{9}\right) = \left(a+\frac{b}{2}\right)^2 + \frac{3b^2}{4}+\frac{5}{9}>0.$$ Consequently, $a=b$, a contradiction.

Answer 2

The derivative of function $x\mapsto x^3+\frac59x$ is positive for every $x$, so the function is monotonically increasing. Further it is continuous with $\lim_{x\to+\infty} f(x)=+\infty$ and $\lim_{x\to-\infty} f(x)=-\infty$.

Then with intermediate value theorem it can be proved that the function is surjective, and the fact that is monotically increasing tells us directly that it is injective as well.

This together proves that the function is bijective, or equivalently has an inverse.