Question - Finding the Domain and Range of a Given Function

The given function is \( f(x) = \frac{1}{\sqrt{9-x^2}} \).

To find the domain, we need to identify all values of \( x \) for which the function is defined. The denominator cannot be zero, and the expression under the square root must be non-negative because the root must be a real number.

We have the inequality \( 9 - x^2 \geq 0 \) which leads to \( -3 \leq x \leq 3 \).

Hence, the domain of \( f(x) \) is \( [-3, 3] \).

To find the range, we know that the function is always positive because it is the reciprocal of the square root of a positive number, and the square root function only returns non-negative values.

Since the square root function \( \sqrt{9-x^2} \) will have values ranging from \( 0 \) (not included because it would make the denominator zero) to \( 3 \), the reciprocal of this function will range from \( \frac{1}{3} \) to infinity.

Hence, the range of \( f(x) \) is \( (0, \infty) \).