Question

The domain of the function \[ f(x) = \frac{1}{\sqrt{9 - x^2}} \] is:

• A. \( -3 \leq x \leq 3 \)
• B. \( -3<x<3 \)
• C. \( -9 \leq x \leq 9 \)
• D. \( -9<x<9 \)

Hint:

For a function \( f(x) = \frac{1}{\sqrt{g(x)}} \), ensure \( g(x)>0 \) since square root values in the denominator must be strictly positive.

Answer 1

The Correct Option is B

Step 1: Identifying restrictions on the function. 
- The function contains a square root in the denominator, meaning the expression inside the square root must be strictly positive: \[ 9 - x^2>0 \] - Solving for \( x \): \[ 9>x^2 \] \[ -3<x<3 \] Step 2: Understanding why \( x = \pm3 \) is excluded. 
- If \( x = 3 \) or \( x = -3 \), then \[ 9 - x^2 = 0 \] - This makes the denominator zero, which is undefined in real numbers. - Hence, the function is not defined at \( x = \pm3 \). 
Step 3: Identifying the correct answer. 
- The domain must be strictly between -3 and 3, so the correct answer is (B) \( -3<x<3 \).