Inverse of $y= x^{3} +x+1$

Question

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!

Answer 1

Why won't Cardano work?

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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

Answer 2

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.