Simplifying inverses of polynomials to real functions

Question

For example, I have a polynomial: $$f(x)=-\frac{1}{4}x^3 + \frac{3}{4}x + \frac{1}{2}$$ And one of its inverses is: $$f^{-1}(x) = \frac{-(1 + i \sqrt{3}) (-2 x + 2 \sqrt{(x - 1) x} + 1)^{2/3} + i \sqrt{3} - 1}{2 \sqrt[3]{-2 x + 2 \sqrt{(x - 1) x} + 1}}$$ In the real domain 0<x<1 it has a real codomain -1<y<1.

Is it possible to simplify inverses of polynomials to "real functions"? By which I mean I want the new function to have same values in a given domain as the original one and I want to be able to find any value in that domain by doing real number calculations only.

Answer 1

Consider the cubic equation $$-\frac{1}{4}x^3 + \frac{3}{4}x + \frac{1}{2}-y=0$$

I shall follow the steps given here.

The discriminant $$\Delta=-\frac{27}{16} (y-1) y$$ is negative for any $y <0$ and $y >1$. So, one real root except for $0 \leq y \leq 1$.

On the other side, we have $p=-3$ and $q=4 y-2$

$$y<0 \implies x=2 \cosh \left(\frac{1}{3} \cosh ^{-1}(1-2 y)\right)$$ $$y>1 \implies x=-2 \cosh \left(\frac{1}{3} \cosh ^{-1}(2 y-1)\right)$$

Now, the problem is that, for $0 \leq x \leq 1$, $\frac 12 \leq y \leq 1$. So $\Delta >0$ and three real roots. Using the trigonomatric method $$x_k=2 \cos \left(\frac{2 \pi k}{3}-\frac{1}{3} \cos ^{-1}(1-2 y)\right)\qquad \text{with} \qquad k=0,1,2$$