1. **State the problem:** Find the domain and range of the function $$f(x) = \frac{3}{\sqrt{9 - x^2}}$$. 2. **Understand the domain constraints:** The expression under the square root, called the radicand, must be positive because the denominator cannot be zero and the square root must be defined for real numbers. 3. **Set the radicand greater than zero:** $$9 - x^2 > 0$$ 4. **Solve the inequality:** $$9 > x^2$$ $$-3 < x < 3$$ 5. **Domain:** The function is defined for all real numbers $x$ such that $$-3 < x < 3$$. 6. **Find the range:** Since the denominator is $$\sqrt{9 - x^2}$$, it is positive and approaches zero as $x$ approaches ±3, making the function approach infinity. 7. **Minimum value of denominator:** The maximum value of $$\sqrt{9 - x^2}$$ is when $x=0$, which is $$\sqrt{9} = 3$$. 8. **Maximum value of function:** When denominator is largest (3), $$f(0) = \frac{3}{3} = 1$$. 9. **Range:** Since the denominator decreases from 3 to 0 as $x$ moves from 0 to ±3, the function increases from 1 to infinity. 10. **Final range:** $$f(x) \in (1, \infty)$$. **Answer:** - Domain: $$(-3, 3)$$ - Range: $$(1, \infty)$$