What is the interval in which the function f(x) =`sqrt(9-x^(2))` is increasing ? `(f(x)gt0)`

Options

A. `0 lt x lt 3`
B. `-3 lt x lt 0`
C. `0 lt x lt 9`
D. `-3 lt x lt 3`

Explanation

To find the interval in which the function \( f(x) = \sqrt{9 - x^2} \) is increasing, we will follow these steps:

### Step 1: Find the derivative of the function
The first step is to differentiate the function \( f(x) \).

\[ f(x) = \sqrt{9 - x^2} \]

Using the chain rule, the derivative \( f'(x) \) is given by:

\[ f'(x) = \frac{1}{2\sqrt{9 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{9 - x^2}} \]

### Step 2: Determine where the derivative is positive
To find where the function is increasing, we need to determine where the derivative \( f'(x) \) is greater than zero:

\[ f'(x) > 0 \implies \frac{-x}{\sqrt{9 - x^2}} > 0 \]

Since the square root in the denominator is always positive for \( x^2 < 9 \) (i.e., \( -3 < x < 3 \)), we can focus on the numerator:

\[ -x > 0 \implies x < 0 \]

### Step 3: Find the domain of the function
Next, we need to find the domain of the function \( f(x) \). The expression under the square root must be non-negative:

\[ 9 - x^2 \geq 0 \implies x^2 \leq 9 \implies -3 \leq x \leq 3 \]

### Step 4: Combine the results
Now we combine the results from Step 2 and Step 3. The function is increasing where \( x < 0 \) and within the domain \( -3 \leq x \leq 3 \). Therefore, the interval where the function is increasing is:

\[ [-3, 0) \]

### Final Answer
The function \( f(x) = \sqrt{9 - x^2} \) is increasing in the interval \( [-3, 0) \).

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