# Inverse function of $ax + bx^3$

I am trying to find the inverse of the function $$y = f(x) = ax + bx^3$$, i.e. $$x = f^{-1}(y)$$.

(The equation arises in the modeling of a certain type of transmission used in robots)

Looking at the curve it seems that there should be some straightforward expression, but I am at a loss how to come up with it.

Can anyone help?

• The expression is not pretty. Ask WA. – lhf Apr 11 at 12:12

By Cardano's formula for the roots of a cubic equation, if $$b\neq0$$ the inverse is given by $$\begin{eqnarray*} x&=&\sqrt[3]{\frac{y}{2b}+\sqrt{\frac{y^2}{4b^2}+\frac{a^3}{27b^3}}} +\sqrt[3]{\frac{y}{2b}-\sqrt{\frac{y^2}{4b^2}+\frac{a^3}{27b^3}}}\\ &=&\frac{1}{\sqrt[3]{b}}\left(\sqrt[3]{\frac{y}{2}+\sqrt{\frac{y^2}{4}+\frac{a^3}{27b}}} +\sqrt[3]{\frac{y}{2}-\sqrt{\frac{y^2}{4}+\frac{a^3}{27b}}}\right). \end{eqnarray*}$$ Note that $$\tfrac{y^2}{4}+\tfrac{a^3}{27b}\geq0$$ for all $$y\in\Bbb{R}$$ if and only if $$ab\geq0$$, and in this case the inner square roots are real numbers, and so both cube roots correspond to a unique real number.
On the other hand, if $$ab<0$$ then the square roots are imaginary, and each cube root correponds to three complex numbers. In this case you must choose the cube roots in such a way that their product is $$-\tfrac{a}{3b}$$ to get the appropriate value of $$x$$.
• Thanks, this is really helpful. In our problem $a$ and $b$ are both positive and non-zero, so this will do the trick. – guero64 Apr 11 at 13:04