To find the inverse of the function

$y = \frac{2^x}{2^x + 1},$

we will follow these steps:

1. Replace $y$ with $x$:

$x = \frac{2^y}{2^y + 1}.$

2. Solve for $y$:

Multiply both sides by $(2^y + 1)$ to eliminate the denominator: $x(2^y + 1) = 2^y.$

Expand: $2^y \cdot x + x = 2^y.$

Rearrange terms to isolate $2^y$: $2^y \cdot x - 2^y = -x.$

Factor out $2^y$: $2^y(x - 1) = -x.$

Solve for $2^y$: $2^y = \frac{-x}{x - 1}.$

3. Apply the logarithm:

Take the base-2 logarithm of both sides to solve for $y$: $y = \log_2\left(\frac{-x}{x - 1}\right).$

Final Answer:

The inverse function is: $y^{-1}(x) = \log_2\left(\frac{-x}{x - 1}\right).$

Let me know if you'd like a detailed explanation of logarithmic properties or clarification on any step!

Explore Further:

1. What are the domain and range of the original and inverse functions?
2. Can the inverse function be graphed easily?
3. What is the behavior of the original function as $x \to \pm\infty$?
4. How does the base-2 logarithm affect the inverse function's growth rate?
5. Could the function's inverse be expressed in another logarithmic base?

Tip: Always verify inverses by substituting one function into the other to confirm you get back to $x$.