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Let's break down the problem step by step.

We are given the function:

\[
f(x) = \left( \frac{2 - e^{-x}}{e^{-x} - 1} \right)^{1/2}
\]

The problem asks us to:
- (a) Find the domain of \( f(x) \)
- (b) Find the inverse \( f^{-1} \)
- (c) Find the domain of the inverse \( f^{-1} \)

### Part (a): Find the domain of \( f(x) \)

The domain of a function refers to the set of values for which the function is defined. For this function, there are two main considerations:

1. The **fraction** inside the square root must be defined, which means the denominator cannot be zero:
   \[
   e^{-x} - 1 \neq 0
   \]
   Solving this:
   \[
   e^{-x} \neq 1 \quad \Rightarrow \quad -x \neq 0 \quad \Rightarrow \quad x \neq 0
   \]
   So, \( x \neq 0 \).

2. The **expression inside the square root** must be non-negative (since we're dealing with real numbers):
   \[
   \frac{2 - e^{-x}}{e^{-x} - 1} \geq 0
   \]
   Let's examine when this fraction is non-negative:
   - The numerator, \( 2 - e^{-x} \), is positive when \( e^{-x} < 2 \), which is true for all \( x \) since \( e^{-x} \) is always less than 2.
   - The denominator, \( e^{-x} - 1 \), is positive when \( e^{-x} > 1 \), i.e., \( x < 0 \), and negative when \( e^{-x} < 1 \), i.e., \( x > 0 \).

   Thus, the fraction is non-negative for \( x < 0 \) and negative for \( x > 0 \).

**Conclusion:**
The domain of \( f(x) \) is \( x < 0 \), or \( (-\infty, 0) \).

### Part (b): Find the inverse \( f^{-1}(x) \)

To find the inverse, we first rewrite the equation for \( f(x) \):

\[
y = \left( \frac{2 - e^{-x}}{e^{-x} - 1} \right)^{1/2}
\]

Now square both sides to get rid of the square root:

\[
y^2 = \frac{2 - e^{-x}}{e^{-x} - 1}
\]

Now, solve for \( x \):
1. Multiply both sides by \( e^{-x} - 1 \):
   \[
   y^2(e^{-x} - 1) = 2 - e^{-x}
   \]
2. Distribute \( y^2 \) on the left-hand side:
   \[
   y^2 e^{-x} - y^2 = 2 - e^{-x}
   \]
3. Bring all terms involving \( e^{-x} \) to one side:
   \[
   y^2 e^{-x} + e^{-x} = 2 + y^2
   \]
4. Factor out \( e^{-x} \) on the left-hand side:
   \[
   e^{-x}(y^2 + 1) = 2 + y^2
   \]
5. Solve for \( e^{-x} \):
   \[
   e^{-x} = \frac{2 + y^2}{y^2 + 1}
   \]
6. Take the natural logarithm of both sides:
   \[
   -x = \ln \left( \frac{2 + y^2}{y^2 + 1} \right)
   \]
7. Solve for \( x \):
   \[
   x = -\ln \left( \frac{2 + y^2}{y^2 + 1} \right)
   \]

Thus, the inverse function is:

\[
f^{-1}(y) = -\ln \left( \frac{2 + y^2}{y^2 + 1} \right)
\]

### Part (c): Find the domain of the inverse \( f^{-1}(x) \)

The domain of the inverse is the range of the original function. From the form of \( f(x) \), we see that:
- The square root ensures that \( f(x) \geq 0 \).
- As \( x \to -\infty \), \( f(x) \to 0 \).
- As \( x \to 0^- \), \( f(x) \to \infty \).

Thus, the range of \( f(x) \) is \( [0, \infty) \), meaning the domain of the inverse function \( f^{-1}(x) \) is \( [0, \infty) \).

---

Would you like more detailed steps or further clarification? Here are some related questions for deeper understanding:

1. How do you identify when a fraction is non-negative?
2. What does the range of a function tell you about its inverse?
3. How can we find the domain of more complex expressions like \( f(x) \)?
4. Why do we take the logarithm when solving for the inverse?
5. Can the function \( f(x) \) ever be negative?

**Tip:** Always check for restrictions on the function (like square roots or denominators) when determining the domain.

Let f(x) = (2 - e^(-x) / (e^(-x) - 1))^(1/2). Find (a) the domain of f(x), (b) the inverse of f(x), and (c) the domain of the inverse.

Learn how to find the domain of f(x) = (2 - e^(-x) / (e^(-x) - 1))^(1/2), its inverse, and the domain of the inverse. This covers key calculus concepts, including solving for the domain and using logarithms to find the inverse. Ideal for undergraduate math courses.

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