1. **Problem:** Find the Domain and Range of the function $$h(x) = \frac{\sqrt{4 - x^2}}{x - 3}$$ using interval notation. 2. **Step 1: Understand the domain restrictions.** - The expression under the square root must be non-negative: $$4 - x^2 \geq 0$$. - The denominator cannot be zero: $$x - 3 \neq 0$$. 3. **Step 2: Solve the inequality for the square root.** $$4 - x^2 \geq 0 \implies x^2 \leq 4 \implies -2 \leq x \leq 2$$. 4. **Step 3: Exclude values that make the denominator zero.** Since $$x - 3 = 0$$ at $$x = 3$$, exclude $$x=3$$. But $$3$$ is not in the interval $$[-2, 2]$$, so no exclusion needed here. 5. **Step 4: Domain conclusion.** Domain is all $$x$$ such that $$-2 \leq x \leq 2$$. 6. **Step 5: Find the range.** - The numerator $$\sqrt{4 - x^2}$$ ranges from $$0$$ to $$2$$ (since max of $$4 - x^2$$ is 4). - The denominator $$x - 3$$ ranges from $$-5$$ to $$-1$$ on $$[-2, 2]$$. - Since numerator is non-negative and denominator is negative in domain, $$h(x)$$ is non-positive. - The maximum value of $$h(x)$$ approaches $$0$$ from below. - The minimum value is $$\frac{2}{-1} = -2$$. 7. **Step 6: Range in interval notation:** $$[-2, 0]$$. **Final answer:** - Domain: $$[-2, 2]$$ - Range: $$[-2, 0]$$