Find the domain and range of `f(x)=[log(sin^(-1)sqrt(x^2+3x+2))]`.

Answer

To find the domain and range of the function \( f(x) = \log(\sin^{-1}(\sqrt{x^2 + 3x + 2})) \), we will follow these steps:

### Step 1: Determine the expression inside the logarithm
The function involves a logarithm, which is defined only for positive values. Therefore, we need to ensure that:
\[ \sin^{-1}(\sqrt{x^2 + 3x + 2}) > 0 \]
This implies:
\[ \sqrt{x^2 + 3x + 2} > 0 \]

### Step 2: Analyze the square root
The expression inside the square root is:
\[ x^2 + 3x + 2 \]
We can factor this quadratic:
\[ x^2 + 3x + 2 = (x + 1)(x + 2) \]
To find where this expression is greater than zero, we need to find the roots:
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]
\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]
Next, we analyze the intervals determined by these roots: \( (-\infty, -2) \), \( (-2, -1) \), and \( (-1, \infty) \).

### Step 3: Test intervals for positivity
1. For \( x < -2 \) (e.g., \( x = -3 \)):
\[ (-3 + 1)(-3 + 2) = (-2)(-1) = 2 > 0 \]
2. For \( -2 < x < -1 \) (e.g., \( x = -1.5 \)):
\[ (-1.5 + 1)(-1.5 + 2) = (-0.5)(0.5) = -0.25 < 0 \]
3. For \( x > -1 \) (e.g., \( x = 0 \)):
\[ (0 + 1)(0 + 2) = (1)(2) = 2 > 0 \]

Thus, \( x^2 + 3x + 2 > 0 \) for \( x \in (-\infty, -2) \cup (-1, \infty) \).

### Step 4: Ensure the argument of sine inverse is valid
Next, we need to ensure that:
\[ \sqrt{x^2 + 3x + 2} \leq 1 \]
Squaring both sides gives:
\[ x^2 + 3x + 2 \leq 1 \]
This simplifies to:
\[ x^2 + 3x + 1 \leq 0 \]
Now, we find the roots of the quadratic \( x^2 + 3x + 1 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-3 \pm \sqrt{5}}{2} \]
Let \( \alpha = \frac{-3 - \sqrt{5}}{2} \) and \( \beta = \frac{-3 + \sqrt{5}}{2} \).

### Step 5: Analyze the intervals for the quadratic inequality
The quadratic \( x^2 + 3x + 1 \) opens upwards, so it is less than or equal to zero between its roots:
\[ \alpha \leq x \leq \beta \]

### Step 6: Find the intersection of intervals
The domain of \( f(x) \) is the intersection of the intervals:
1. From \( x^2 + 3x + 2 > 0 \): \( (-\infty, -2) \cup (-1, \infty) \)
2. From \( x^2 + 3x + 1 \leq 0 \): \( \left[\frac{-3 - \sqrt{5}}{2}, \frac{-3 + \sqrt{5}}{2}\right] \)

The common portion of these intervals gives us the domain:
\[ \text{Domain: } \left[\frac{-3 - \sqrt{5}}{2}, -2\right) \cup (-1, \frac{-3 + \sqrt{5}}{2}] \]

### Step 7: Determine the range of \( f(x) \)
Since \( \sin^{-1}(y) \) is defined for \( 0 < y \leq 1 \), and \( \log(z) \) is defined for \( z > 0 \), we have:
\[ 0 < \sin^{-1}(\sqrt{x^2 + 3x + 2}) \leq \frac{\pi}{2} \]
Taking the logarithm gives:
\[ -\infty < \log(\sin^{-1}(\sqrt{x^2 + 3x + 2})) \leq 0 \]
Thus, the range of \( f(x) \) is:
\[ \text{Range: } (-\infty, 0] \]

### Final Answer
- **Domain:** \( \left[\frac{-3 - \sqrt{5}}{2}, -2\right) \cup (-1, \frac{-3 + \sqrt{5}}{2}] \)
- **Range:** \( (-\infty, 0] \)