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We were going over inverse functions in class and the method for mathematically finding an inverse function (i.e. replace y with x in the function and solve implicitly for y). I noticed, however, this method seemed impossible for a function such as $\mathbf {f(x)=x^3+x}$ After trying for a while, I used the solve() function on my calculator to find y implicitly. The result was $$y=\frac{6^\frac 13\left(2^\frac 13\sqrt {3(27x^2+4)}+9x\right)^\frac 23}{6\left(\sqrt{3\left(27x^2+4\right)}+9x\right)^\frac 13}$$

I have graph this and it is indeed the inverse function. My question is: How was this conclusion reached? (no this question was not for homework, yes I am aware of and okay with the fact that this may be well beyond my level; I'd like to know anyways)

EDIT: Someone else asked the exact same question on this forum. I had actually read the responses to that particular post, they did not answer my specific question and, to the best of my knowledge, gave methods to solve for the graph the inverse function. Also while the question titles are the same, the nature of the question really is not at the heart of it.

Theseus
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  • If you read the full context of both questions (as well as the responses on the first questions) you will see they are not duplicates. Furthermore, the answers didn't really answer my question. If the problem is title duplication I can reword the question title. – Theseus Aug 24 '16 at 11:10
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    Solving for the function implicitly is pretty much all you can do. Most functions are of course not invertible. There is a theorem that gives conditions for when a function in locally invertible. It's called the inverse function theorem. Generally even if a function is invertible, finding the inverse function might be difficult. It's just that the examples and exercises one gets are always easy. – Adomas Baliuka Aug 24 '16 at 11:22
  • @AdomasBaliuka To solve implicitly is to get the the inverse function assuming you use the method I describe above. The confusion lies in the plethora of numbers that the solution incurs when the greatest coefficient of the original expression is 1. Since my calculator was able to find the inverse function, it must be mathematically possible. My question is where to start as simple algebra seems to get me nowhere. – Theseus Aug 24 '16 at 12:27
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    First of all, I think the accepted answer http://math.stackexchange.com/a/60913/10756 actually answers your question to explicitly find the inverse function, although it leaves the last details to the reader. Secondly, @AdomasBaliuka's comment that a global inverse cannot be found is not true. In the case of functions $f:\mathbb{R} \to \mathbb{R}$ a global inverse exists if $f$ is strictly monotonically increasing or decreasing. – Jaap Eldering Aug 24 '16 at 23:52
  • The formula for solving a cubic equation is nasty, but at least one exists. Good thing you didn't specify $x^5$ instead of $x^3$. –  Aug 25 '16 at 00:05
  • @eldering I did not claim the function in the question is not globally invertible, rather my comment was a remark on what could happen in the general case. On another note, it may further be worth noting the implications of Bungo's comment: Even if a global inverse exists, there might not be a way to specify it in a way that doesn't involve Limits. For example the inverse of a polinomial may not be expressible by nested addition, multiplikation and roots. – Adomas Baliuka Aug 25 '16 at 07:59
  • If you are wondering about higher degree inverses, it is notable that anything with $x^5$ or higher has no closed form solution. And the messiness of the inverse of an $x^4$ polynomial to the $x^3$ polynomial is like the messiness of the inverse of the $x^3$ polynomial to the quadratic equation. – Simply Beautiful Art Aug 27 '16 at 00:39

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The real solution of the cubic $$x^3+x-y=0$$ is given by

$$x=\frac{2 }{\sqrt{3}}\sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{3 \sqrt{3}}{2}y\right)\right)$$ using the so simple hyperbolic soluion for cubic equations.

Does it look nicer than $$x=\frac{\sqrt[3]{\sqrt{3} \sqrt{27 y^2+4}+9 y}}{\sqrt[3]{2} 3^{2/3}}-\frac{\sqrt[3]{\frac{2}{3}} }{\sqrt[3]{\sqrt{3} \sqrt{27y^2+4}+9 y}}$$ given using Cardano formaula ?

Claude Leibovici
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