The equation in the image is as follows:

$f(x) = \frac{2^x - 2^{-x}}{2^x + 2^{-x}}$

We are tasked with finding the inverse of this function, $f^{-1}(x)$. Let's go through the steps to solve for the inverse.

Step-by-step Solution:

1. Express the function in terms of $y$: $y = \frac{2^x - 2^{-x}}{2^x + 2^{-x}}$ Now, we will solve for $x$ in terms of $y$.

2. Simplify the expression using exponent rules: Notice that $2^{-x} = \frac{1}{2^x}$. So we rewrite the function: $y = \frac{2^x - \frac{1}{2^x}}{2^x + \frac{1}{2^x}}$

   Multiply the numerator and denominator by $2^x$ to clear the fractions: $y = \frac{(2^x)^2 - 1}{(2^x)^2 + 1}$ Let $z = 2^x$, so the equation becomes: $y = \frac{z^2 - 1}{z^2 + 1}$

3. Solve for $z$: Multiply both sides of the equation by $z^2 + 1$: $y(z^2 + 1) = z^2 - 1$ Expand both sides: $yz^2 + y = z^2 - 1$ Rearrange to collect all terms involving $z^2$ on one side: $z^2 - yz^2 = y + 1$ Factor out $z^2$ on the left-hand side: $z^2(1 - y) = y + 1$ Solve for $z^2$: $z^2 = \frac{y + 1}{1 - y}$ Take the square root of both sides to solve for $z$: $z = \sqrt{\frac{y + 1}{1 - y}}$ Recall that $z = 2^x$, so we now have: $2^x = \sqrt{\frac{y + 1}{1 - y}}$

4. Solve for $x$: Take the logarithm base 2 of both sides: $x = \log_2 \left( \sqrt{\frac{y + 1}{1 - y}} \right)$ Simplify the logarithmic expression: $x = \frac{1}{2} \log_2 \left( \frac{y + 1}{1 - y} \right)$

Thus, the inverse function is:

$f^{-1}(x) = \frac{1}{2} \log_2 \left( \frac{x + 1}{1 - x} \right)$

This is the inverse of the given function $f(x)$.

Let me know if you need any clarification or further details.

Here are some related questions you might consider:

1. How does the inverse function behave for different values of $x$?
2. What are the domain and range of the original function $f(x)$ and its inverse?
3. How would the process change if the base of the exponentials was different?
4. Can you derive the derivative of the original function $f(x)$?
5. How does the function $f(x)$ compare to other similar hyperbolic functions?

Tip: Always verify the domain and range of both the original function and the inverse to ensure there are no contradictions.