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Can you help me find the inverse function for $y= x^3 +x+1$? This question proposed by one of my friends and I don't know the real source of the problem. We can't use Cardano's method for solve this problem.

Thanks!

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closed as off-topic by Shailesh, abiessu, Rohan, Stefan4024, Namaste Dec 24 '17 at 14:26

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  • $\begingroup$ Cardano works, and is the only thing that will work... $x^3 + x$ is always increasing, so you never get casus irreducibils en.wikipedia.org/wiki/Cubic_function#Cardano%27s_method $\endgroup$ – Will Jagy Dec 24 '17 at 3:14
  • $\begingroup$ Possible duplicate of The Inverse of $y=x^5-x^3+x$ in terms $y=$ $\endgroup$ – Gil-Galad Dec 24 '17 at 3:29
  • $\begingroup$ @WillJagy: it is not the only thing. Lagrange inversion theorem applies, too. $$x=-\sum_{k\geq 0}\binom{3k}{k}\frac{(-1)^k (1-y)^{2k+1}}{2k+1}.$$ $\endgroup$ – Jack D'Aurizio Dec 24 '17 at 9:35
  • $\begingroup$ @JackD'Aurizio I don't know that one. Maybe I should have said only easy, closed form, finite number of terms, and so on. $\endgroup$ – Will Jagy Dec 24 '17 at 17:48
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Why won't Cardano work?


I'll tackle this problem the algebraic way because it's the most straightforward, despite the intimidating algebra.

Replacing $x$ with $y$, we get$$y^3+y+1=x\quad\implies\quad y^3+y+1-x=0$$Now recall Cardano's formula for the cubic $x^3+qx+r=0$ who has the real root at$$x_1=\left\{-\frac r2+\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}+\left\{-\frac r2-\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}$$Substituting $q=1$ and $r=1-x$ gives$$\begin{align*}y=\sqrt[3]{-\frac {1-x}2+\sqrt{\frac {27x^2-54x+31}{108}}}-\sqrt[3]{\frac {1-x}2+\sqrt{\frac {27x^2-54x+31}{108}}}\end{align*}$$And that's your inverse! Here's a visual diagram in Desmos

Inverse

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  • $\begingroup$ I'm pretty sure "can't" meant "not allowed to" in the question. $\endgroup$ – Will Jagy Dec 24 '17 at 3:40
  • $\begingroup$ Notice that the square roots always come out real, as $27 x^2 - 54x + 31 >0$ $\endgroup$ – Will Jagy Dec 24 '17 at 3:46
  • $\begingroup$ "Can" means.. bending the numerical solution to an analytical one. $\endgroup$ – Narasimham Dec 24 '17 at 3:59
  • $\begingroup$ @WillJagy Oh... I see. In that case, I'll have to rework the answer. $\endgroup$ – Crescendo Dec 24 '17 at 14:43
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    $\begingroup$ Crescendo, I wouldn't worry about it. You gave a good answer. At the moment you have seven upvotes, so you have helped some people. My impression when an OP says a friend gave a question is that there is a minor contest somewhere that is being hidden. Add in an unrealistic restriction on techniques, for me that confirms it. You did a good job. $\endgroup$ – Will Jagy Dec 24 '17 at 17:41
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Given

$$ y = x^3+x+1 $$

Interchange $x,y$ to see inverse function graph.

$$ x= y^3+y+1 $$

Using Cardano it is quite possible to find the solution $f(x)$ in terms of $y$ analytically. There is one real root at $x=1$ coinciding at its inflection point.

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  • $\begingroup$ Unfortunately your respond is not correct. $\endgroup$ – BarzanHayati Dec 28 '17 at 18:17

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