https://file.aiquickdraw.com/mathbot/file/1726248215871bej3o3wq.jpg

Solve the equation in details and step bay step?

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.

Let f(x) = (2^x - 2^(-x)) / (2^x + 2^(-x)), then determine the inverse of the function f(x).

Learn how to find the inverse of the function f(x) = (2^x - 2^(-x)) / (2^x + 2^(-x)) with a detailed step-by-step solution. This problem covers key mathematical concepts such as inverse functions, exponential functions, and logarithms. Suitable for students in Grades 10-12.

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