The domain of definition of `f (x) = sin ^(-1) {log_(2)(x^(2) + 3x + 4)}`, is

Options

A. `[-4, -1//4] `
B. ` [-3, -1//3]`
C. `[-2, -1]`
D. none of these

Explanation

To find the domain of the function \( f(x) = \sin^{-1}(\log_2(x^2 + 3x + 4)) \), we need to ensure that the expression inside the sine inverse function is valid. This involves two main conditions:

1. The argument of the logarithm must be positive.
2. The output of the logarithm must lie within the range of the sine inverse function, which is \([-1, 1]\).

### Step 1: Ensure the argument of the logarithm is positive
We need to check when \( x^2 + 3x + 4 > 0 \).

The quadratic \( x^2 + 3x + 4 \) can be analyzed using the discriminant \( D \):
\[ D = b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot 4 = 9 - 16 = -7 \]
Since the discriminant is negative, the quadratic does not have real roots and opens upwards (as the coefficient of \( x^2 \) is positive). Therefore, \( x^2 + 3x + 4 > 0 \) for all \( x \in \mathbb{R} \).

### Step 2: Ensure the output of the logarithm lies within the range of sine inverse
Next, we need to satisfy:
\[ -1 \leq \log_2(x^2 + 3x + 4) \leq 1 \]

#### Sub-step 2.1: Solve \( \log_2(x^2 + 3x + 4) \leq 1 \)
This can be rewritten using the properties of logarithms:
\[ x^2 + 3x + 4 \leq 2^1 = 2 \]
Rearranging gives:
\[ x^2 + 3x + 2 \leq 0 \]
Factoring the quadratic:
\[ (x + 2)(x + 1) \leq 0 \]
To find the intervals where this inequality holds, we find the roots:
- The roots are \( x = -2 \) and \( x = -1 \).

Next, we test intervals around these points:
- For \( x < -2 \) (e.g., \( x = -3 \)): \( (-3 + 2)(-3 + 1) = (-1)(-2) > 0 \)
- For \( -2 < x < -1 \) (e.g., \( x = -1.5 \)): \( (-1.5 + 2)(-1.5 + 1) = (0.5)(-0.5) < 0 \)
- For \( x > -1 \) (e.g., \( x = 0 \)): \( (0 + 2)(0 + 1) > 0 \)

Thus, the solution to \( (x + 2)(x + 1) \leq 0 \) is:
\[ -2 \leq x \leq -1 \]

#### Sub-step 2.2: Solve \( \log_2(x^2 + 3x + 4) \geq -1 \)
This can be rewritten as:
\[ x^2 + 3x + 4 \geq 2^{-1} = \frac{1}{2} \]
Rearranging gives:
\[ x^2 + 3x + \frac{7}{2} \geq 0 \]
Calculating the discriminant:
\[ D = 3^2 - 4 \cdot 1 \cdot \frac{7}{2} = 9 - 14 = -5 \]
Since the discriminant is negative, the quadratic \( x^2 + 3x + \frac{7}{2} \) is always positive for all \( x \in \mathbb{R} \).

### Final Step: Combine the results
The only restriction on \( x \) comes from the first inequality \( -2 \leq x \leq -1 \). Therefore, the domain of the function \( f(x) \) is:
\[ \boxed{[-2, -1]} \]