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

Options

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

Answer

To find the domain of the function \( f(x) = \frac{\log_2(x + 3)}{x^2 + 3x + 2} \), we need to consider two main conditions:

1. The argument of the logarithm must be positive.
2. The denominator must not be zero.

### Step 1: Determine the condition for the logarithm

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

\[ x + 3 > 0 \]

Solving this inequality:

\[ x > -3 \]

### Step 2: Factor the denominator

Next, we need to factor the denominator \( x^2 + 3x + 2 \):

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

### Step 3: Set the denominator not equal to zero

The function will be undefined when the denominator is zero. Therefore, we set the factored form equal to zero:

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

This gives us the values:

\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]
\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]

### Step 4: Combine the conditions

Now, we combine the conditions from Steps 1 and 3. We know:

- \( x > -3 \) (from the logarithm)
- \( x \neq -1 \) and \( x \neq -2 \) (from the denominator)

### Step 5: Write the final domain

The domain of the function can be expressed in interval notation. Since \( x \) must be greater than \(-3\) but cannot be \(-1\) or \(-2\), we can write:

\[ \text{Domain of } f(x) = (-3, -2) \cup (-2, -1) \cup (-1, \infty) \]

### Final Answer

The domain of the function \( f(x) = \frac{\log_2(x + 3)}{x^2 + 3x + 2} \) is:

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