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 two conditions are satisfied:

1. The logarithm must be defined.
2. The denominator must not be zero.

### Step 1: Determine when the logarithm is defined

The logarithm \( \log_2 (x+3) \) is defined when its argument is greater than zero:

\[ x + 3 > 0 \]

Solving this inequality:

\[ x > -3 \]

### Step 2: Determine when the denominator is not zero

Next, we need to ensure that the denominator \( x^2 + 3x + 2 \) is not equal to zero. We can factor the quadratic:

\[ x^2 + 3x + 2 = (x + 1)(x + 2) \]

Setting the denominator equal to zero gives us:

\[ (x + 1)(x + 2) = 0 \]

This implies:

\[ x + 1 = 0 \quad \text{or} \quad x + 2 = 0 \]

Thus, the roots are:

\[ x = -1 \quad \text{and} \quad x = -2 \]

So, the function is undefined at \( x = -1 \) and \( x = -2 \).

### Step 3: Combine the conditions

From Step 1, we have \( x > -3 \). From Step 2, we know \( x \) cannot be \( -1 \) or \( -2 \).

Thus, the domain of \( f(x) \) can be expressed in interval notation as:

\[ (-3, -2) \cup (-2, -1) \cup (-1, \infty) \]

### Final Domain

The final domain of the function \( f(x) \) is:

\[ (-3, -2) \cup (-2, -1) \cup (-1, \infty) \]