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The function given is:

\[
f(x) = \left( \ln \frac{x + 1}{x - 1} \right) \cdot \sqrt{(x + 1)(x + 2)}
\]

To find the domain of this function, we need to analyze two aspects:

1. **The logarithmic part:**
   \[
   \ln \frac{x + 1}{x - 1}
   \]
   The argument of the logarithm, \(\frac{x + 1}{x - 1}\), must be **positive**. This gives us the inequality:
   \[
   \frac{x + 1}{x - 1} > 0
   \]
   We can solve this inequality by finding the critical points and using sign analysis:
   - Critical points occur when \(x + 1 = 0 \Rightarrow x = -1\) and \(x - 1 = 0 \Rightarrow x = 1\).
   
   To solve \(\frac{x + 1}{x - 1} > 0\), we will divide the number line into intervals based on these critical points and check the sign of the expression in each interval.

2. **The square root part:**
   \[
   \sqrt{(x + 1)(x + 2)}
   \]
   The expression inside the square root, \((x + 1)(x + 2)\), must be **non-negative**. This leads to the inequality:
   \[
   (x + 1)(x + 2) \geq 0
   \]
   Again, we find the critical points by solving \(x + 1 = 0 \Rightarrow x = -1\) and \(x + 2 = 0 \Rightarrow x = -2\), and use a sign diagram to determine where the product is non-negative.

---

### Step 1: Solve \(\frac{x + 1}{x - 1} > 0\)
Critical points: \(x = -1\) and \(x = 1\).

We analyze the sign of \(\frac{x + 1}{x - 1}\) in the intervals determined by these points:

- For \(x < -1\) (e.g., \(x = -2\)): \(\frac{-2 + 1}{-2 - 1} = \frac{-1}{-3} > 0\)
- For \(-1 < x < 1\) (e.g., \(x = 0\)): \(\frac{0 + 1}{0 - 1} = \frac{1}{-1} < 0\)
- For \(x > 1\) (e.g., \(x = 2\)): \(\frac{2 + 1}{2 - 1} = \frac{3}{1} > 0\)

Thus, \(\frac{x + 1}{x - 1} > 0\) in the intervals \(x < -1\) and \(x > 1\).

### Step 2: Solve \((x + 1)(x + 2) \geq 0\)
Critical points: \(x = -1\) and \(x = -2\).

We analyze the sign of \((x + 1)(x + 2)\) in the intervals:

- For \(x < -2\) (e.g., \(x = -3\)): \((-3 + 1)(-3 + 2) = (-2)(-1) > 0\)
- For \(-2 < x < -1\) (e.g., \(x = -1.5\)): \((-1.5 + 1)(-1.5 + 2) = (-0.5)(0.5) < 0\)
- For \(x > -1\) (e.g., \(x = 0\)): \((0 + 1)(0 + 2) = (1)(2) > 0\)

Thus, \((x + 1)(x + 2) \geq 0\) in the intervals \(x \leq -2\) and \(x \geq -1\).

### Step 3: Combine the results
- From the logarithmic part: \(x < -1\) or \(x > 1\).
- From the square root part: \(x \leq -2\) or \(x \geq -1\).

The domain of the function \(f(x)\) is the intersection of these two conditions. The valid intervals are:
- \(x \leq -2\) (from the square root part)
- \(x > 1\) (from both parts)

### Final Domain:
The domain of \(f(x)\) is \( (-\infty, -2] \cup (1, \infty) \).

Would you like further details or a graph of the sign diagrams?

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Here are some follow-up questions:

1. What happens to the function as \(x\) approaches \( -1 \) or \(1\)?
2. How does the logarithmic part affect the overall behavior of the function for positive and negative \(x\)?
3. Can the function be extended beyond this domain?
4. What is the behavior of \(f(x)\) at \(x = 1\) and \(x = -2\)?
5. How would the domain change if the square root function were removed?

**Tip:** Always consider the critical points of each part of a composite function separately when finding the domain.

Find the domain of f(x) = (ln((x+1)/(x-1))) * sqrt((x+1)(x+2)) using sign diagrams.

Find the domain of the function f(x) = (ln((x+1)/(x-1))) * sqrt((x+1)(x+2)). This problem requires knowledge of logarithmic and square root functions, and involves solving inequalities using sign diagrams. Suitable for advanced high school students in Grades 11-12.

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