The natural domain of the function `f(x)=sqrt(sin^(-1)(2x)+(pi)/(3))` is

Options

A. `[-(1)/(2),(1)/(2)]`
B. `[-(sqrt3)/(4),(1)/(2)]`
C. `[(13)/(4),(1)/(2)]`
D. `[-(sqrt3)/(2),1]`

Answer

To find the natural domain of the function \( f(x) = \sqrt{\sin^{-1}(2x) + \frac{\pi}{3}} \), we need to ensure that the expression inside the square root is non-negative. This leads us to the following steps:

### Step 1: Set up the inequality
We require:
\[ \sin^{-1}(2x) + \frac{\pi}{3} \geq 0 \]
This can be rearranged to:
\[ \sin^{-1}(2x) \geq -\frac{\pi}{3} \]

### Step 2: Understand the range of \(\sin^{-1}\)
The function \(\sin^{-1}(y)\) is defined for \(y \in [-1, 1]\) and its range is \([- \frac{\pi}{2}, \frac{\pi}{2}]\). Therefore, for \(\sin^{-1}(2x)\) to be defined, we must have:
\[ -1 \leq 2x \leq 1 \]

### Step 3: Solve the inequalities for \(x\)
From the inequality \(2x \geq -1\):
\[ x \geq -\frac{1}{2} \]
From the inequality \(2x \leq 1\):
\[ x \leq \frac{1}{2} \]

### Step 4: Combine the results
Combining both inequalities, we have:
\[ -\frac{1}{2} \leq x \leq \frac{1}{2} \]

### Step 5: Check the condition for \(\sin^{-1}(2x) \geq -\frac{\pi}{3}\)
Next, we need to find when \(\sin^{-1}(2x) \geq -\frac{\pi}{3}\). This means:
\[ 2x \geq \sin\left(-\frac{\pi}{3}\right) \]
Since \(\sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}\), we have:
\[ 2x \geq -\frac{\sqrt{3}}{2} \]
Thus:
\[ x \geq -\frac{\sqrt{3}}{4} \]

### Step 6: Determine the final domain
Now we combine this with the previous results:
\[ -\frac{1}{2} \leq x \leq \frac{1}{2} \quad \text{and} \quad x \geq -\frac{\sqrt{3}}{4} \]

The value \(-\frac{\sqrt{3}}{4} \approx -0.433\) is greater than \(-\frac{1}{2}\). Therefore, the effective lower bound for \(x\) is \(-\frac{\sqrt{3}}{4}\).

### Final Answer
The natural domain of the function \( f(x) \) is:
\[ \left[-\frac{\sqrt{3}}{4}, \frac{1}{2}\right] \]