The domain of `f(x)=(log_(2)(x+3))/(x^(2)+3x+2)` is

Options

A. `R-{-1,-2}`
B. `(-2,oo)`
C. `R-{-1,-2,-3}`
D. `(-3,oo)-{-1,-2}`

Explanation

To find the domain of the function \( f(x) = \frac{\log_2(x + 3)}{x^2 + 3x + 2} \), we need to ensure that both the logarithm and the denominator are defined.

### Step 1: Determine the condition for the logarithm
The logarithmic function \( \log_2(x + 3) \) is defined when its argument is greater than zero:
\[ x + 3 > 0 \]
Solving this inequality:
\[ x > -3 \]

### Step 2: Determine the condition for the denominator
The denominator \( x^2 + 3x + 2 \) must not be equal to zero:
\[ x^2 + 3x + 2 \neq 0 \]
We can factor the quadratic:
\[ (x + 1)(x + 2) \neq 0 \]
This gives us the conditions:
\[ x + 1 \neq 0 \quad \text{and} \quad x + 2 \neq 0 \]
Which leads to:
\[ x \neq -1 \quad \text{and} \quad x \neq -2 \]

### Step 3: Combine the conditions
From Step 1, we have \( x > -3 \). From Step 2, we have \( x \neq -1 \) and \( x \neq -2 \).

Now, we can summarize the conditions:
- \( x > -3 \)
- \( x \neq -1 \)
- \( x \neq -2 \)

### Step 4: Express the domain
The domain of \( f(x) \) can be expressed in interval notation. Since \( x \) must be greater than \(-3\) but cannot equal \(-1\) or \(-2\), we can write the domain as:
\[ (-3, -2) \cup (-2, -1) \cup (-1, \infty) \]

### Final Answer:
The domain of \( f(x) \) is:
\[ (-3, -2) \cup (-2, -1) \cup (-1, \infty) \]