`f(x)=sqrt(9-x^(2))`. find range of f(x).

Answer

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

### Step 1: Identify the domain of the function
The expression under the square root, \( 9 - x^2 \), must be non-negative for \( f(x) \) to be defined. Therefore, we need to solve the inequality:
\[ 9 - x^2 \geq 0 \]
This can be rearranged to:
\[ x^2 \leq 9 \]
Taking the square root of both sides gives:
\[ -3 \leq x \leq 3 \]
Thus, the domain of \( f(x) \) is \( x \in [-3, 3] \).

### Step 2: Determine the maximum and minimum values of \( f(x) \)
Next, we will evaluate \( f(x) \) at the endpoints of the domain:
- At \( x = -3 \):
\[ f(-3) = \sqrt{9 - (-3)^2} = \sqrt{9 - 9} = \sqrt{0} = 0 \]
- At \( x = 3 \):
\[ f(3) = \sqrt{9 - 3^2} = \sqrt{9 - 9} = \sqrt{0} = 0 \]
- At \( x = 0 \):
\[ f(0) = \sqrt{9 - 0^2} = \sqrt{9} = 3 \]

### Step 3: Analyze the function's behavior
Since \( f(x) = \sqrt{9 - x^2} \) is a continuous function and it achieves its maximum value at \( x = 0 \) and minimum values at \( x = -3 \) and \( x = 3 \), we can conclude that:
- The minimum value of \( f(x) \) is \( 0 \).
- The maximum value of \( f(x) \) is \( 3 \).

### Step 4: State the range of the function
Combining the minimum and maximum values, we find that the range of \( f(x) \) is:
\[ \text{Range of } f(x) = [0, 3] \]

### Final Answer:
The range of the function \( f(x) = \sqrt{9 - x^2} \) is \( [0, 3] \).

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