The domain of the function `f(x)=sqrt("In"_((|x|-1))(x^(2)+4x+4))` is

Options

A. `[-3,-1] cup [1,2]`
B. `(-2, -1) cup [2,oo)`
C. `(-oo,-3] cup (-2,-1) cup (2,oo)`
D. None of these

Explanation

To find the domain of the function \( f(x) = \sqrt{\log_{|x|-1}(x^2 + 4x + 4)} \), we need to ensure that the expression inside the square root is non-negative and that the logarithm is defined.

### Step 1: Conditions for the logarithm
The logarithm \( \log_{|x|-1}(x^2 + 4x + 4) \) is defined under the following conditions:
1. The base \( |x| - 1 \) must be greater than 0.
2. The base \( |x| - 1 \) must not equal 1.
3. The argument \( x^2 + 4x + 4 \) must be greater than 0.

### Step 2: Analyze the base \( |x| - 1 \)
1. **Condition 1**: \( |x| - 1 > 0 \)
- This implies \( |x| > 1 \).
- Therefore, \( x < -1 \) or \( x > 1 \).

2. **Condition 2**: \( |x| - 1 \neq 1 \)
- This implies \( |x| \neq 2 \).
- Therefore, \( x \neq -2 \) and \( x \neq 2 \).

### Step 3: Analyze the argument \( x^2 + 4x + 4 \)
The expression \( x^2 + 4x + 4 \) can be factored as:
\[ x^2 + 4x + 4 = (x + 2)^2 \]
This expression is always non-negative since it is a perfect square. It is equal to 0 when \( x = -2 \). However, since \( x \neq -2 \) from the previous condition, we can conclude that:
\[ x^2 + 4x + 4 > 0 \quad \text{for all } x \neq -2. \]

### Step 4: Combine the conditions
From the conditions derived:
1. \( x < -1 \) or \( x > 1 \)
2. \( x \neq -2 \) and \( x \neq 2 \)

Now we can summarize the valid intervals:
- For \( x < -1 \): The interval is \( (-\infty, -2) \cup (-2, -1) \).
- For \( x > 1 \): The interval is \( (1, 2) \cup (2, \infty) \).

### Step 5: Final domain
Combining all the intervals, we get the domain of the function:
\[ \text{Domain of } f(x) = (-\infty, -2) \cup (-2, -1) \cup (1, 2) \cup (2, \infty). \]