Find the domain the range of each of the following function: `f(x)=1/(sqrt(4+3sinx))`

Explanation

To find the domain and range of the function \( f(x) = \frac{1}{\sqrt{4 + 3 \sin x}} \), we will follow these steps:

### Step 1: Determine the conditions for the function to be defined
The function \( f(x) \) has a square root in the denominator, which means that the expression under the square root must be positive:
\[ 4 + 3 \sin x > 0 \]
This ensures that the denominator is not zero and the square root is defined.

### Step 2: Solve the inequality
To solve the inequality \( 4 + 3 \sin x > 0 \):
\[ 3 \sin x > -4 \]
\[ \sin x > -\frac{4}{3} \]
Since the sine function \( \sin x \) has a range of \([-1, 1]\), the condition \( \sin x > -\frac{4}{3} \) is always satisfied for all \( x \) because \(-\frac{4}{3}\) is less than \(-1\). Thus, the domain of \( f(x) \) is:
\[ \text{Domain: } (-\infty, \infty) \]

### Step 3: Determine the range of the function
Next, we need to find the range of \( f(x) \). We know that:
\[ -1 \leq \sin x \leq 1 \]
Multiplying the entire inequality by 3 gives:
\[ -3 \leq 3 \sin x \leq 3 \]
Adding 4 to all parts of the inequality results in:
\[ 1 \leq 4 + 3 \sin x \leq 7 \]
Now, taking the square root of the entire inequality:
\[ \sqrt{1} \leq \sqrt{4 + 3 \sin x} \leq \sqrt{7} \]
This simplifies to:
\[ 1 \leq \sqrt{4 + 3 \sin x} \leq \sqrt{7} \]

### Step 4: Invert the inequality to find the range of \( f(x) \)
Since \( f(x) = \frac{1}{\sqrt{4 + 3 \sin x}} \), we need to take the reciprocal of the values in the inequality. When we take the reciprocal, we must reverse the inequality signs:
\[ \frac{1}{\sqrt{7}} \leq f(x) \leq 1 \]

### Conclusion
Thus, the range of \( f(x) \) is:
\[ \text{Range: } \left[\frac{1}{\sqrt{7}}, 1\right] \]

### Final Answer
- **Domain:** \( (-\infty, \infty) \)
- **Range:** \( \left[\frac{1}{\sqrt{7}}, 1\right] \)