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

Options

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

Explanation

To find the domain of the function \( f(x) = \log_{(3+x)}(x^2 - 1) \), we need to ensure that the logarithm is defined. This involves checking two conditions:

1. The base of the logarithm, \( 3 + x \), must be greater than 0 and not equal to 1.
2. The argument of the logarithm, \( x^2 - 1 \), must be greater than 0.

Let's solve these conditions step by step.

### Step 1: Ensure the base \( 3 + x > 0 \)
To find when the base is positive:
\[ 3 + x > 0 \]
Subtracting 3 from both sides gives:
\[ x > -3 \]

### Step 2: Ensure the base is not equal to 1
Next, we need to ensure that the base is not equal to 1:
\[ 3 + x \neq 1 \]
Subtracting 3 from both sides gives:
\[ x \neq -2 \]

### Step 3: Ensure the argument \( x^2 - 1 > 0 \)
Now, we check when the argument is positive:
\[ x^2 - 1 > 0 \]
Factoring gives:
\[ (x - 1)(x + 1) > 0 \]
To solve this inequality, we find the critical points, which are \( x = -1 \) and \( x = 1 \). We will test the intervals defined by these points: \( (-\infty, -1) \), \( (-1, 1) \), and \( (1, \infty) \).

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

- For \( -1 < x < 1 \) (e.g., \( x = 0 \)):
\((x - 1)(x + 1) = (-1)(1) = -1 < 0\) (False)

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

Thus, the solution to \( (x - 1)(x + 1) > 0 \) is:
\[ x \in (-\infty, -1) \cup (1, \infty) \]

### Step 4: Combine the conditions
Now we combine the conditions:
1. From \( 3 + x > 0 \), we have \( x > -3 \).
2. From \( 3 + x \neq 1 \), we have \( x \neq -2 \).
3. From \( x^2 - 1 > 0 \), we have \( x \in (-\infty, -1) \cup (1, \infty) \).

The intersection of these conditions gives:
- From \( x > -3 \) and \( x \in (-\infty, -1) \), we have \( -3 < x < -1 \) but excluding \( -2 \).
- The interval \( (1, \infty) \) remains valid.

Thus, the final domain of \( f(x) \) is:
\[ \text{Domain: } (-3, -2) \cup (-2, -1) \cup (1, \infty) \]