The problem involves determining the inverse of the function $f(x) = x^3 + x$. Let us carefully solve this step by step:

Step 1: Express $y = f(x)$

We write: $y = x^3 + x$

Step 2: Swap $x$ and $y$ to find $f^{-1}(x)$

$x = y^3 + y$

Step 3: Solve for $y$ in terms of $x$

This equation is not trivial to solve algebraically due to the cubic nature of $y^3 + y - x = 0$. A general solution involves using numerical methods or the Cardano's formula for solving cubic equations.

Using Cardano’s formula:

The general cubic equation $y^3 + py + q = 0$ has the solution: [ y = \sqrt[3]{-\frac{q}{2} + \sqrt{\left(\frac{q}{2}\right)^2 + \left(\frac{p}{3}\right)^3}} + \sqrt[3]{-\frac{q}{2} - \sqrt{\left(\frac{q}{2}\right)^2 + \left(\frac{p}{3}\right)^3}} ]

In this case: $y^3 + y - x = 0$ Here, $p = 1$ and $q = -x$.

Substitute $p$ and $q$ into Cardano’s formula: [ y = \sqrt[3]{\frac{x}{2} + \sqrt{\left(\frac{x}{2}\right)^2 + \left(\frac{1}{3}\right)^3}} + \sqrt[3]{\frac{x}{2} - \sqrt{\left(\frac{x}{2}\right)^2 + \left(\frac{1}{3}\right)^3}} ]

Step 4: Final Answer

The inverse function $f^{-1}(x)$ is: [ f^{-1}(x) = \sqrt[3]{\frac{x}{2} + \sqrt{\left(\frac{x}{2}\right)^2 + \left(\frac{1}{3}\right)^3}} + \sqrt[3]{\frac{x}{2} - \sqrt{\left(\frac{x}{2}\right)^2 + \left(\frac{1}{3}\right)^3}} ]

This solution is exact but can also be simplified for numerical evaluation if needed.

Would you like me to elaborate on the derivation, explore numerical solutions, or explain this process in more detail?

Here are 5 related questions to expand your understanding:

1. How does the Cardano's formula work for solving general cubic equations?
2. Can we approximate the solution to $f^{-1}(x)$ numerically for specific values of $x$?
3. How do we verify that the calculated $f^{-1}(x)$ satisfies $f(f^{-1}(x)) = x$?
4. What are the implications of the cubic equation having more than one real root?
5. How can we determine the derivative of the inverse function $f^{-1}(x)$?

Tip:

For functions like $f(x) = x^3 + x$, numerical methods (such as Newton-Raphson) can provide approximate values of the inverse function efficiently when an exact symbolic solution is unnecessary.