To find the inverse of the function $f(x) = 2xe^x + 1$, we need to follow these steps:

Step 1: Replace $f(x)$ with $y$

$y = 2xe^x + 1$ We aim to solve for $x$ in terms of $y$.

Step 2: Isolate $2xe^x$

Subtract 1 from both sides: $y - 1 = 2xe^x$

Step 3: Attempt to solve for $x$

This equation involves both $x$ and $e^x$, which is a transcendental equation. It cannot be solved for $x$ algebraically using elementary functions. However, such equations can sometimes be solved using the Lambert W function, which is the inverse function of $x e^x$.

To express $x$ in terms of $y$, we first try to get the equation into a form that involves the Lambert W function.

Step 4: Rewrite the equation to use the Lambert W function

Divide both sides by 2 to simplify: $\frac{y - 1}{2} = xe^x$

Now, let’s apply the substitution $u = x e^x$, so: $u = \frac{y - 1}{2}$

Thus, solving for $x$, we apply the Lambert W function, denoted as $W(z)$: $x = W\left(\frac{y - 1}{2}\right)$

Final Answer:

The inverse function is: $f^{-1}(y) = W\left(\frac{y - 1}{2}\right)$

The Lambert W function is a special function that generally requires numerical methods or software for evaluation.

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Would you like further details on how the Lambert W function works or how to compute specific values for the inverse function?

Here are 5 related questions to explore further:

1. How can you approximate values of the Lambert W function?
2. What types of equations can be solved using the Lambert W function?
3. What is the significance of transcendental equations?
4. Can the inverse of every function be expressed using elementary functions?
5. How does the domain of a function affect the existence of an inverse?

Tip: The Lambert W function is widely used to solve equations where a variable appears both as a base and an exponent (e.g., $x e^x$).