Question

The range of the real valued function \( f(x) = \sqrt{9 - x^2} \) is:

• A. \( [-3, 3] \)
• B. \( [-3, 0] \)
• C. \( [0, 3] \)
• D. \( [-2, 2] \)

Hint:

For square root functions \( \sqrt{a - x^2} \), ensure the inside stays non-negative. The range will always be from 0 up to the square root of the maximum value inside.

Answer 1

The Correct Option is C

Step 1: Understanding the domain of \( f(x) = \sqrt{9 - x^2 \)}
The function \( f(x) \) is defined only when the expression inside the square root is non-negative: \[ 9 - x^2 \geq 0 \Rightarrow -3 \leq x \leq 3. \]
Step 2: Determining the range
Now, \( f(x) = \sqrt{9 - x^2} \). The maximum value of \( f(x) \) occurs when \( x = 0 \), giving \( f(0) = \sqrt{9} = 3 \). The minimum value of \( f(x) \) is at the boundaries \( x = \pm 3 \), where \( f(x) = \sqrt{0} = 0 \). \[ \Rightarrow \text{Range of } f(x) = [0, 3] \]