the domain of the function
`f(x)=log_(3+x)(x^(2)-1)` is

Options

A. `(-3,-1)cup(1,infty)`
B. `(-3,-2)cup(-2,-1)cup(1,infty)`
C. `[-3,-1),cup[1,infty)`
D. `[-3,-2)cup(-2,-1)cup[1,infty)`

Explanation

To find the domain of the function \( f(x) = \log_{(3+x)}(x^2 - 1) \), we need to ensure that both the base of the logarithm and the argument of the logarithm meet certain conditions.

### Step 1: Conditions for the base of the logarithm
The base \( 3 + x \) must be greater than 0 and cannot be equal to 1.

1. **Condition 1:** \( 3 + x > 0 \)
- This simplifies to \( x > -3 \).

2. **Condition 2:** \( 3 + x \neq 1 \)
- This simplifies to \( x \neq -2 \).

### Step 2: Conditions for the argument of the logarithm
The argument \( x^2 - 1 \) must be greater than 0.

1. **Condition 3:** \( x^2 - 1 > 0 \)
- This factors to \( (x - 1)(x + 1) > 0 \).
- The critical points are \( x = -1 \) and \( x = 1 \). We can analyze the intervals:
- For \( x < -1 \): both factors are negative, so the product is positive.
- For \( -1 < x < 1 \): one factor is negative and the other is positive, so the product is negative.
- For \( x > 1 \): both factors are positive, so the product is positive.
- Thus, the solution to \( (x - 1)(x + 1) > 0 \) is \( x < -1 \) or \( x > 1 \).

### Step 3: Combine the conditions
Now we combine all the conditions derived:

- From Condition 1: \( x > -3 \)
- From Condition 2: \( x \neq -2 \)
- From Condition 3: \( x < -1 \) or \( x > 1 \)

### Step 4: Determine the valid intervals
1. For \( x < -1 \):
- The interval \( (-3, -1) \) is valid, but we must exclude \( -2 \).
- Thus, this interval becomes \( (-3, -2) \cup (-2, -1) \).

2. For \( x > 1 \):
- The interval \( (1, \infty) \) is valid.

### Final Domain
Combining these results, the domain of the function is:
\[ (-3, -2) \cup (-2, -1) \cup (1, \infty) \]