1. **Problem Statement:** Find the domain and range of the function $f(x) = \sqrt{4 - x^2}$.\n\n2. **Understanding the function:** The function involves a square root, so the expression inside the root must be non-negative for the function to be real-valued.\n\n3. **Domain determination:** Set the radicand (expression inside the square root) greater than or equal to zero: $$4 - x^2 \geq 0$$\n\n4. **Solve the inequality:**\n$$4 \geq x^2$$\nThis means $$-2 \leq x \leq 2$$ because $x^2 \leq 4$ implies $x$ is between $-2$ and $2$.\n\n5. **Domain:** The domain is all real numbers $x$ such that $$-2 \leq x \leq 2$$.\n\n6. **Range determination:** Since $f(x) = \sqrt{4 - x^2}$, the smallest value inside the root is 0 (when $x = \pm 2$), and the largest is 4 (when $x=0$).\n\n7. **Calculate range values:**\n- Minimum value of $f(x)$ is $\sqrt{0} = 0$.\n- Maximum value of $f(x)$ is $\sqrt{4} = 2$.\n\n8. **Range:** Therefore, the range is $$0 \leq f(x) \leq 2$$.\n\n**Final answer:**\n- Domain: $$[-2, 2]$$\n- Range: $$[0, 2]$$