The range of the function `f(x) = tan sqrt((pi^(2))/(9)-x^(2))`, is

Options

A. `[0, sqrt3]`
B. `(0, sqrt3)`
C. `[0, sqrt3)`
D. `(0, sqrt3]`

Explanation

To find the range of the function \( f(x) = \tan\left(\sqrt{\frac{\pi^2}{9} - x^2}\right) \), we will follow these steps:

### Step 1: Determine the domain of the function
The expression inside the square root must be non-negative for \( f(x) \) to be defined. Therefore, we need to solve the inequality:

\[ \frac{\pi^2}{9} - x^2 \geq 0 \]

This simplifies to:

\[ \frac{\pi^2}{9} \geq x^2 \]

Taking the square root of both sides, we find:

\[ -\frac{\pi}{3} \leq x \leq \frac{\pi}{3} \]

### Step 2: Analyze the square root function
The function \( \sqrt{\frac{\pi^2}{9} - x^2} \) will take values from:

\[ \sqrt{0} \text{ (when } x = \pm \frac{\pi}{3}\text{) to } \sqrt{\frac{\pi^2}{9}} \text{ (when } x = 0\text{)} \]

Calculating \( \sqrt{\frac{\pi^2}{9}} \):

\[ \sqrt{\frac{\pi^2}{9}} = \frac{\pi}{3} \]

Thus, \( \sqrt{\frac{\pi^2}{9} - x^2} \) varies from \( 0 \) to \( \frac{\pi}{3} \).

### Step 3: Find the range of the tangent function
Next, we need to find the range of \( f(x) = \tan\left(\sqrt{\frac{\pi^2}{9} - x^2}\right) \) as \( \sqrt{\frac{\pi^2}{9} - x^2} \) varies from \( 0 \) to \( \frac{\pi}{3} \).

- At \( \sqrt{\frac{\pi^2}{9} - x^2} = 0 \):

\[ f(x) = \tan(0) = 0 \]

- At \( \sqrt{\frac{\pi^2}{9} - x^2} = \frac{\pi}{3} \):

\[ f(x) = \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]

### Step 4: Conclusion about the range
Since \( \tan(x) \) is an increasing function in the interval \( \left(0, \frac{\pi}{3}\right) \), the range of \( f(x) \) will be:

\[ [0, \sqrt{3}] \]

Thus, the range of the function \( f(x) \) is:

\[ \boxed{[0, \sqrt{3}]} \]