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

Question

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!

Answer 1

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

Answer 2

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 $-1$.

Answer 3

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.

Answer 4

I just revisited this answer due to a comment, then I came across this question. I had worked out the solution to the cubic and quartic equations in high school. I reproduce the solution to the cubic equation here.

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Cubic equation solution

Since $(ax+b)-(x+c)$ is a factor of $(ax+b)^3-(x+c)^3$, I expanded $$ \begin{align} 0 &=(ax+b)^3-(x+c)^3\\ &=\left(a^3-1\right)x^3+3\left(a^2b-c\right)x^2+3\left(ab^2-c^2\right)x+\left(b^3-c^3\right)\tag1 \end{align} $$ which has a root at $x=\frac{c-b}{a-1}$. However, trying to solve for $a,b,c$ from the coefficients of $(1)$ is not easy.

If we get rid of the quadratic term by setting $c=a^2b$, things get significantly simpler: $$ \begin{align} 0 &=(ax+b)^3-(x+a^2b)^3\\ &=\left(a^3-1\right)x^3+3ab^2\left(1-a^3\right)x+b^3\left(1-a^6\right)\\ &=\left(a^3-1\right)\left(x^3-3ab^2x-b^3\left(1+a^3\right)\right)\tag2 \end{align} $$ which has a root at $x=b(a+1)$. It is much easier to solve for $a,b$ from the coefficients of $(2)$.

First, note that $$ \left(b^3\left(1+a^3\right)\right)^2-\frac4{27}\left(3ab^2\right)^3=\left(b^3\left(1-a^3\right)\right)^2\tag3 $$ Using $(3)$ we get $$ b^3=\frac12b^3\left(1+a^3\right)+\left[\frac14\left(b^3\left(1+a^3\right)\right)^2-\frac1{27}\left(3ab^2\right)^3\right]^{1/2}\tag4 $$ and $$ a^3b^3=\frac12b^3\left(1+a^3\right)-\left[\frac14\left(b^3\left(1+a^3\right)\right)^2-\frac1{27}\left(3ab^2\right)^3\right]^{1/2}\tag5 $$ Setting $c=3ab^2$ and $d=b^3\left(1+a^3\right)$, we get that $$ x^3-cx-d=0\tag6 $$ has a root at $$ \begin{align} x &=b+ab\\ &=\overbrace{\left[\frac12d+\left[\frac14d^2-\frac1{27}c^3\right]^{1/2}\right]^{1/3}}^\text{$b$ from $(4)$}+\overbrace{\left[\frac12d-\left[\frac14d^2-\frac1{27}c^3\right]^{1/2}\right]^{1/3}}^\text{$ab$ from $(5)$}\\ &=\left[\frac12d+\left[\frac14d^2-\frac1{27}c^3\right]^{1/2}\right]^{1/3}+\frac{\frac13c}{\left[\frac12d+\left[\frac14d^2-\frac1{27}c^3\right]^{1/2}\right]^{1/3}}\tag7 \end{align} $$ Note that $$ x^3+ax^2+bx+c=u^3-\left(\frac{a^2}3-b\right)u-\left(\frac{ab}3-\frac{2a^3}{27}-c\right)\tag8 $$ where $x=u-\frac13a$. That is, $(8)$ converts the general cubic polynomial into the form of $(6)$ so that we can apply $(7)$.

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Answer to the Question

We can apply $(7)$ directly to the question with $c=-1$ and $d=y$: $$ x^3+x=y\implies x=\left[\frac12y+\left[\frac14y^2+\frac1{27}\right]^{1/2}\right]^{1/3}-\frac{\frac13}{\left[\frac12y+\left[\frac14y^2+\frac1{27}\right]^{1/2}\right]^{1/3}}\tag9 $$

Answer 5

Its inverse function is $ x= y^3-y $

It is obtained by interchanging $x$ and $y$ or by mirroring graph with respect to line through origin $ x=y. $