If `0lexle(pi)/(2)`, what is the maximum value of the function `f(x)="sin"(1)/(3)x`?

Options

A. 0
B. `(1)/(3)`
C. `(1)/(2)`
D. `(sqrt(3))/(2)`

Explanation

To find the maximum value of the function \( f(x) = \sin\left(\frac{1}{3}x\right) \) for \( 0 \leq x \leq \frac{\pi}{2} \), we can follow these steps:

### Step 1: Determine the range of \( \frac{1}{3}x \)
Given that \( x \) lies between \( 0 \) and \( \frac{\pi}{2} \), we can find the corresponding range for \( \frac{1}{3}x \).

\[ 0 \leq \frac{1}{3}x \leq \frac{1}{3} \cdot \frac{\pi}{2} \]

Calculating \( \frac{1}{3} \cdot \frac{\pi}{2} \):

\[ \frac{1}{3} \cdot \frac{\pi}{2} = \frac{\pi}{6} \]

So, the range of \( \frac{1}{3}x \) is:

\[ 0 \leq \frac{1}{3}x \leq \frac{\pi}{6} \]

### Step 2: Analyze the sine function
The function \( \sin\left(\frac{1}{3}x\right) \) is increasing in the interval \( [0, \frac{\pi}{6}] \) because the sine function increases from \( 0 \) to \( \frac{\pi}{2} \).

### Step 3: Find the maximum value
Since \( \sin\left(\frac{1}{3}x\right) \) is increasing, it will achieve its maximum value at the upper limit of its range, which is \( \frac{\pi}{6} \).

Thus, we need to evaluate:

\[ f\left(\frac{\pi}{2}\right) = \sin\left(\frac{1}{3} \cdot \frac{\pi}{2}\right) = \sin\left(\frac{\pi}{6}\right) \]

### Step 4: Calculate \( \sin\left(\frac{\pi}{6}\right) \)
The value of \( \sin\left(\frac{\pi}{6}\right) \) is:

\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]

### Conclusion
Therefore, the maximum value of the function \( f(x) = \sin\left(\frac{1}{3}x\right) \) for \( 0 \leq x \leq \frac{\pi}{2} \) is:

\[ \boxed{\frac{1}{2}} \]