What is the inverse of $f(x)=x^3-3x$?

Question

I found this problem in the Spivak calculus book, and have spent probably 5 hours with no progress being made. I found online that Cardano's method can be used, but I doubt this book would require that knowledge, and the problem is only labelled as "pretty hard," not "extremely hard." (with asterisks).

The only thing I figured out was that it's gonna be 3 separate functions. I have tried dividing $x^3-3x-c$ by a linear factor, I have tried guessing the type of functions in the inverse by looking at the graph, I have tried assuming x would be some sum $p+q$ and solving for each variable's expression. I'm about 90% sure there is a much simpler solution that I can no longer see from falling deeper and deeper into these wrong ideas. What am I missing?

Answer 1

We have $x^3-3x-y=0.$

By Cardano's method that you suggested,

$$x=f^{-1}(y)\\=\sqrt[3]{\frac y2+\sqrt{\frac{y^2}4-1}}+ \sqrt[3]{\frac y2-\sqrt{\frac{y^2}4-1}}$$ for $x>1.$

$$x=f^{-1}(y)\\=w^2\sqrt[3]{\frac y2+\sqrt{\frac{y^2}4-1}}+w \sqrt[3]{\frac y2-\sqrt{\frac{y^2}4-1}}$$ for $-1<x<1.$

$$x=f^{-1}(y)=w\sqrt[3]{\frac y2+\sqrt{\frac{y^2}4-1}}+w^2 \sqrt[3]{\frac y2-\sqrt{\frac{y^2}4-1}}$$ for $x<-1.$ Here, $w=e^{\frac{2\pi i}3}$.

Answer 2

If you consider the cubic equation$$x^3-3x-y=0$$ its discriminant is $$\Delta=-27 \left(y^2-4\right)$$ which is negative if $|y|>2$. For this case, only one real root.

Following the steps given here, the solution is simply $$x=2\frac{| y| }{y}\,\cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{| y| }{2}\right)\right)$$