# Inverse of $y = x^3 + x$?

Can you help me find the inverse function for $y = x^3 + x$? This question was posed at the beginning of AP Calculus, so we can't use any math beyond precalc.

Thanks!

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You'll have to use the cubic formula, I reckon. –  Guess who it is. Aug 31 '11 at 8:29
And since the formula for solving cubic equations is probably not on the syllabus, there is probably no way for you to do the problem. So either the problem is a mistake, or you aren't quoting it correctly. –  Gerry Myerson Aug 31 '11 at 9:13
Are you sure you're really asked to find the inverse? I would guess that the problem asks you to show that the function has an inverse (which is much easier). –  Hans Lundmark Aug 31 '11 at 9:56
Until badatmath returns and answers Hans's question, there is no point in going on with Cardano's formulas. –  GEdgar Aug 31 '11 at 13:14
Oh, well I don't know. I never read the direct statement of the problem so it's possible it was something else. Thanks everyone, it's interesting to learn about Cardano's formulas anyway. –  badatmath Aug 31 '11 at 16:53

Write $x^{3}+x-y=0$ and make $x=u+v$. You get $$u^{3}+v^{3}-y+\left( u+v\right) \left( 3uv+1\right) =0.$$ Then solve the system

$$\left\{ \begin{array}{c} u^{3}+v^{3}=y \\ u^{3}v^{3}=-\frac{1}{27}, \end{array} \right.$$

which is equivalent to solving a quadratic equation in $u^3$ or $v^3$, because you know the sum of the two numbers $u^3,v^3$ and their product, e.g. the equation

$$\left( u^{3}\right) ^{2}-yu^{3}-\frac{1}{27}=0.$$

This technique is known as Cardano's method.

Reference: Sebastião e Silva and Silva Paulo, Compêndio de Álgebra, VII ano, pp.215-216, 1963

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Good, though I'd be surprised if AP Calculus students would be expected to come up with this. –  Gerry Myerson Aug 31 '11 at 9:15
I was about to write this very derivation out... (which I picked up from Uspensky's book); nice! –  Guess who it is. Aug 31 '11 at 9:17
@Gerry Myerson: I don't know what AP Calculus is. I learned Cardano's method in an historical note in my text book after the discussion of the quadratic equation, in the 11th grade. –  Américo Tavares Aug 31 '11 at 9:28
I'd be (pleasantly) surprised if any American 11th grade textbook presented Cardano's method. –  Gerry Myerson Aug 31 '11 at 13:52
@Gerry Myerson: Nowadays in Portugal 11th grade textbooks do not present it either, as far as I know. –  Américo Tavares Aug 31 '11 at 14:18

I seriously doubt that you were expected to find the inverse function explicitly, as is probably clear from the earlier answers. But it is quite instructive to convince yourself that there is an inverse function. It may even be that the intention of the question was to illustrate that while it can be easy to see that a function has an inverse, it can be extremely difficult, or impossible, to write the inverse down explcitly. In this case, while there is an explicit formula for the inverse, it is complicated. Sometimes, there just is no way to write down the function explicitly in terms of a formula.

But it is already instructive to ask why the function has an inverse. A function has an inverse if and only if it is one to one and onto (that is, a bijection). The function $f: \mathbb{R} \to \mathbb{R}$ is onto (surjective), because $f(x) \to -\infty$ as $x \to -\infty$ and $f(x) \to +\infty$ as $x \to +\infty$ (and, because it is continuous everywhere, it takes all intermediate values, though I'm not sure how you would prove even that at a precalculus level, you need some sort of continuity). The function $f$ is one-to-one (injective) because if $x^3 + x = z^3 + z$ and $x \neq z,$ we easily obtain $x^2 + xz + z^2 = -1$, so that $(x+\frac{z}{2})^2 + \frac{3z^2}{4} = -1,$ a contradiction.

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For what it is worth, the real solution looks something like this$$x = \sqrt[3]{\frac{y + \sqrt{\tfrac{4}{27} + y^2}}{2}} - \sqrt[3]{\frac{\tfrac{2}{27}}{y + \sqrt{\tfrac{4}{27} + y^2}}}$$ and there are also two complex solutions related to the cube roots of $-1$.

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