Find the domain and range of the real function `f(x)=sqrt(9-x^(2))`

Answer

To find the domain and range of the function \( f(x) = \sqrt{9 - x^2} \), we will follow these steps:

### Step 1: Determine the Domain

1. **Identify the condition for the square root**: The expression inside the square root must be non-negative. Therefore, we need:
\[ 9 - x^2 \geq 0 \]

2. **Rearranging the inequality**: This can be rearranged to:
\[ x^2 \leq 9 \]

3. **Taking square roots**: Taking the square root of both sides gives:
\[ -3 \leq x \leq 3 \]

4. **Conclusion about the domain**: Thus, the domain of \( f(x) \) is:
\[ \text{Domain} = [-3, 3] \]

### Step 2: Determine the Range

1. **Express the function**: We can rewrite the function as:
\[ y = \sqrt{9 - x^2} \]

2. **Identify the minimum and maximum values of \( y \)**: Since \( 9 - x^2 \) is non-negative within the domain, we need to find the maximum value of \( y \). The maximum occurs when \( x^2 \) is minimized, which is at \( x = 0 \):
\[ y = \sqrt{9 - 0^2} = \sqrt{9} = 3 \]

3. **Identify the minimum value of \( y \)**: The minimum value occurs when \( x^2 \) is maximized, which is at \( x = \pm 3 \):
\[ y = \sqrt{9 - 3^2} = \sqrt{0} = 0 \]

4. **Conclusion about the range**: Therefore, the range of \( f(x) \) is:
\[ \text{Range} = [0, 3] \]

### Final Answer

- **Domain**: \([-3, 3]\)
- **Range**: \([0, 3]\)