To find the inverse of the function $f(x) = \frac{8}{1 + x}$, follow these steps:

Step 1: Write the function in terms of $y$

Let $y = f(x)$, so: $y = \frac{8}{1 + x}$

Step 2: Solve for $x$

Now, solve for $x$ in terms of $y$:

1. Multiply both sides by $1 + x$ to eliminate the denominator: $y(1 + x) = 8$

2. Expand the left side: $y + yx = 8$

3. Isolate $yx$ by subtracting $y$ from both sides: $yx = 8 - y$

4. Solve for $x$ by dividing both sides by $y$: $x = \frac{8 - y}{y}$

Step 3: Write the inverse function

Now, replace $y$ with $x$ to represent the inverse function: $f^{-1}(x) = \frac{8 - x}{x}$

So, the inverse function is: $f^{-1}(x) = \frac{8 - x}{x}$

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Would you like more details, or do you have any questions?

Here are some related questions you might find useful:

1. What is the domain of $f^{-1}(x)$?
2. How do you verify that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$?
3. How does the behavior of $f(x)$ compare to its inverse?
4. What is the range of the original function $f(x)$?
5. Can we determine any asymptotes for both $f(x)$ and $f^{-1}(x)$?

Tip: When finding an inverse, always swap $x$ and $y$, then solve for the new $y$. This represents the "undoing" of the original function.