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?

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.