inverse of $f:(0,4/9) \to \mathbb{R}$ such that $f(x) = x - x^{3/2}$

Question

I'm trying to find the inverse of $f(x)=x-x^{3/2}$ in the interval from zero to its maximum (at $x=4/9$). WolframAlpha gives me an expression involving complex terms, but this is no good, as the formula will be part of a moodle calculated question that only accepts reals. Any thoughts?

Answer 1

Found a solution. After changing variables so that $u = \sqrt{x}$, one gets $f(x) = x-x^{3/2} = u^2-u^3$. Thus, all one has to do is solve the cubic $u^3 - u^2 + y = 0$. This can be accomplished be means of a trigonometric identity (Tschirnhaus-Vieta approach).

Within the required range of $x$, which translates to $ u \in (0, 2/3) $, the solution is: $$ u = \frac{1}{3} + \frac{2}{3} \cos \left[ \frac{1}{3} \arccos\left( 1- \frac{27}{2} y\right) - \frac{2\pi}{3} \right] $$