The value of the function `f(x)=3sinsqrt((pi^(2))/(16)-x^(2))` is defined what `x in `

Options

A. `[-(pi)/(4),(pi)/(4)]`
B. `[0,(3)/(sqrt(2))]`
C. `(-3,3)`
D. none of these

Answer

To find the domain of the function \( f(x) = 3 \sin \sqrt{\frac{\pi^2}{16} - x^2} \), we need to ensure that the expression under the square root is non-negative. This is because the square root function is only defined for non-negative values.

### Step-by-Step Solution:

1. **Identify the expression under the square root**:
\[ \frac{\pi^2}{16} - x^2 \]

2. **Set up the inequality**:
To find the values of \( x \) for which the function is defined, we need:
\[ \frac{\pi^2}{16} - x^2 \geq 0 \]

3. **Rearrange the inequality**:
This can be rearranged to:
\[ \frac{\pi^2}{16} \geq x^2 \]

4. **Take the square root of both sides**:
Taking the square root gives:
\[ \sqrt{\frac{\pi^2}{16}} \geq |x| \]
Simplifying the left side:
\[ \frac{\pi}{4} \geq |x| \]

5. **Express the absolute value inequality**:
The inequality \( |x| \leq \frac{\pi}{4} \) can be expressed as:
\[ -\frac{\pi}{4} \leq x \leq \frac{\pi}{4} \]

6. **Write the domain in interval notation**:
Therefore, the domain of the function \( f(x) \) is:
\[ x \in \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \]

### Conclusion:
The function \( f(x) = 3 \sin \sqrt{\frac{\pi^2}{16} - x^2} \) is defined for \( x \) in the interval:
\[ \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \]