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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?

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    $\begingroup$ The expression is not pretty. Ask WA. $\endgroup$ – lhf Apr 11 at 12:12
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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 $x$.

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    $\begingroup$ Thanks, this is really helpful. In our problem $a$ and $b$ are both positive and non-zero, so this will do the trick. $\endgroup$ – guero64 Apr 11 at 13:04

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