Harder-Than-It-Seems: Finding inverse of $f(x)=x^3-x-12$?

Question

This may seem simple, but I have had long days of frustration with finding the inverse of the function

$$f(x)=x^3-x-12.$$

I got this on some homework and it did not ask for the inverse. However I wanted to try anyways. So this is as far as I got:

$$x=y^3-y-12$$ $$x+12=y^3-y$$ $$y(y^2-1)=x+12$$ $$y(y+1)(y-1)=x+12.$$

And that is absolutely as far as I go. In the end, I have no idea how this turns out to be the answer:

A step-by-step walk-through in finding the inverse is gratefully appreciated!

Answer 1

Finding the inverse of function f(x) is effectively solving the equation f(x)= y. There exist a "cubic formula" which can be used to solve such an equation but it is very complicated as you can see from the "inverse" you were given.

Answer 2

There are two classic ways of solving cubics.

• A magical Tartaglia way:

  Let $y = u + v$. Then the equation becomes $u^3 + 3uv(u+v) + v^3 - (u + v) = x+12$. Since the choice of $u$ and $v$ is so far arbitrary, we may let $3uv=1$, which cancels two of the terms leaving $u^3 + v^3 = x+12$, and because we let $3uv=1$ it also results in $u^3 + \frac{1}{(3u)^3} = x+12$.

  The latter is a quadratic equation for $u^3$.

• A "modern" Vieta way is based on the fact that $$\cos 3x = 4\cos^3x - 3 \cos{x} = 3(\frac{4}{3}\cos^3x - \cos{x})$$

  Scale $y = \sqrt{\frac{4}{3}} z$, so that $y^3 - y = \sqrt{\frac{4}{3}}(\frac{4}{3}z^3 - z)$ and let $z = \cos \theta$. It gives $$\cos{3\theta} = 3\sqrt{\frac{3}{4}}(x + 12)$$

Answer 3

This answer is a remark regarding your question. To make the answer seem more related to the question, I will remark that the answer to your question for the general cubic equation is here.

I'll note here, and this may be of some interest to you, that there is a similar equation for roots of quartic equations (equations of degree $4$), see here. However, there is no solution for polynomials of degree $\geq 5$. This is because the symmetric group $S_n$ is not solvable if $n\geq 5$; this result is called Abel's theorem. This does not state, however, that some polynomials of degree $\geq 5$ have no solution (this would contradict the fundamental theorem of algebra which states that the field $\mathbf{C}$ is algebraically closed). The proof of this result can be done using Galois theory.

Answer 4

The inverse function $y(x)$ satisfies $x=y^3-y-12$, or the cubic equation

$$y^3-y-(x+12)=0$$ Use Cardano’s formula to obtain

$$y(x)=\sqrt[3]{\frac {x+12}2+\sqrt{\frac{(x+12)^2}4-\frac1{27}}} +\sqrt[3]{\frac {x+12}2-\sqrt{\frac{(x+12)^2}4-\frac1{27}}} $$