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How can one find the inverse of functions like $f(x) = x^3 + x$?

I know how to do it for explicit quadratic functions; how do I express $x$ as a function of $y$ here?

Travis Willse
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raj
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    where is the implicit function ? – KonKan May 03 '16 at 17:16
  • As the function is strictly monotonous you can write down the inverse on the whole domain. You have to use cardanos formula for the cubic equation and solve $x=y^3+y$. – MrYouMath May 03 '16 at 17:23

2 Answers2

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In general, the inverse of a invertible closed-form function need not itself have a closed form. Simple examples of this include $x \mapsto x^5 + x$ and a suitable restriction of $x \mapsto x^x$.

Since the function $f$ is a cubic polynomial function, we can find an explicit inverse of the given function, $f(x) := x^3 + x$, by writing $y = x^3 + x$ using Cardano's Formula or the equivalent to solve for $x$ and hence inverting the given function. The (somewhat unpleasant) result is: $$\color{#bf0000}{\boxed{f^{-1}(y) = \frac{1}{6} \left(108 y+12 \sqrt{81 y^2+12}\right)^{1/3} - \frac{2}{\left(108 y+12 \sqrt{81 y^2+12}\right)^{1/3}}}} .$$

See Jack D'Aurizio's nice answer for a more compact way to write this function, using $\sinh$ and $\operatorname{arsinh}$.

Like all quadratic polynomial functions, some cubic polynomial functions are not one-to-one and hence do not admit a (global) inverse. A cubic polynomial $a x^3 + b x^2 + c x + d$, $a \neq 0$, is invertible iff its derivative does not have two real roots (which would correspond to local extrema of the cubic polynomial), and this is the case iff the discriminant of the derivative is nonpositive, or equivalently, iff $$b^2 \leq 3 a c .$$

Travis Willse
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    It would be interesting to mention what the Lagrange inversion formula (https://en.wikipedia.org/wiki/Lagrange_inversion_theorem) has to say about such inverse function. – Jack D'Aurizio May 19 '16 at 18:19
  • @JackD'Aurizio I haven't worked it out, but perhaps the inversion formula leads to a nice form for the general term of the power series of the inverse---is this the sort of feature you have in mind? – Travis Willse May 19 '16 at 18:25
  • Exactly! :D $\phantom{}$ – Jack D'Aurizio May 19 '16 at 18:39
  • Travis, I'd appreciate a link to "Cardano's Formula or the equivalent". If you're referring to this method which can be used to solve a depressed cubic, I don't understand how you can use the same method to find the inverse of a function. – Chewers Jingoist May 20 '16 at 00:37
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I would like to write an addendum to Travis' answer.

By Lagrange's inversion formula, the solution $x$ of $x^3-x+z=0$ can be written as: $$ x = \sum_{k\geq 0}\binom{3k}{k}\frac{z^{2k+1}}{2k+1} $$ under the assumption $|z|< \frac{2}{\sqrt{27}} $ that makes such a series convergent.

It follows that the inverse function $g(y)$ of $f(x)=x^3+x$ can be represented through: $$ g(y) = \sum_{k\geq 0}\binom{3k}{k}\frac{(-1)^k y^{2k+1}}{2k+1} \tag{1}$$ for any $y\in\left(-\frac{2}{\sqrt{27}},\frac{2}{\sqrt{27}}\right)$.

The connection with the cubic formula (and the classical method for solving third-degree equations through the sine/cosine triplication formula) lies here:

$$ g(y) = \frac{2}{\sqrt{3}}\,\sinh\left[\frac{1}{3}\,\text{arcsinh}\left(\frac{3\sqrt{3}y}{2}\right)\right]\tag{2} $$

and the interesting fact is that both $(1)$ and $(2)$ give more compact formulas compared to Cardano's one involving $\sqrt[3]{\cdot}$.

Jack D'Aurizio
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