Taking the inverse of a one-to-one polynomial

Question

I'm trying to take the inverse of:

$$f(x)=\frac{4x^3}{x^2+1}$$

When looking at the graph, it seems to be fully inversible (it is one-to-one), so I should be able to end up with another equation that is mirrored in the $x=y$ axis. However, I cannot get the formula in its nicest form, a form where the $y$ term is only on the left.

My process: exchange x and y, and try to solve for y:

$$x=\frac{4y^3}{y^2+1}$$ $$x(y^2+1)=4y^3$$ $$xy^2+x=4y^3$$ $$4y^3-xy^2=x$$ $$y^2(4y-x)=x$$ $$y^2=\frac{x}{4y-x}$$ $$f^-1(x)=y=\sqrt{\frac{x}{4y-x}}$$

This does seem to be correct, but the equation has a stray $y$ on the right side which I cant get rid of. Also, the inverse function has lost half of its graph because of the root.

So is $f(x)$ fully inversible, and if so how do I achieve this form?

Answer 1

The real solution of the cubic equation $4 x^3 = (x^2 + 1) y$ (for $y \ne 0$) is

$$ x = \frac{1}{12} \left(\sqrt [3]{216\,y+{y}^{3}+12\,\sqrt {3\,{y}^{4}+324\,{y}^{2}}}+ {\frac {{y}^{2}}{\sqrt [3]{216\,y+{y}^{3}+12\,\sqrt {3\,{y}^{4}+ 324\,{y}^{2}}}}}+y\right) $$

Answer 2

The cubic equation being $$4x^3-y x^2-y=0$$ we have $$\Delta=-4 y^2 \left(y^2+108\right) \quad < 0 \quad \forall y$$ So, only one real root. Using the hyperbolic method, we then have

$$x=\frac{1}{12} y \left(1+2 \cosh \left(\frac{1}{3} \cosh ^{-1}\left(1+\frac{216}{y^2}\right)\right)\right)$$