Solution: The expression under the square root must satisfy $ 4 - x^2 \geq 0 $, so $ x \in [-2, 2] $. The maximum value of $ \sqrt{4 - x^2} $ is $ 2 $ (at $ x = 0 $), and the minimum is $ 0 $ (at $ x = \pm 2 $). The range is $ \boxed{[0, 2]} $.<think>

["The Complete Guide to Analyzing the Square Root Expression $ \sqrt{4 - x^2} $", "Understanding the behavior of square root functions is essential in algebra, calculus, and real-world applications like physics and engineering. One key expression to study is $ \sqrt{4 - x^2} $, which introduces critical concepts like domain restrictions and optimization. Let’s explore why the domain of this expression must lie within $ [-2, 2] $, its maximum and minimum values, and its full range.", "### Understanding the Domain: Why $ 4 - x^2 \geq 0 $?", "The expression inside the square root, $ 4 - x^2 $, must be non-negative for the square root to yield a real number:\n$$ 4 - x^2 \geq 0 $$\nRewriting this inequality:\n$$ x^2 \leq 4 $$\nTaking square roots gives:\n$$ -2 \leq x \leq 2 $$\nThus, the domain of $ \sqrt{4 - x^2} $ is $ \boxed{[-2, 2]} $. This restriction ensures the function remains defined in the real number system. Without it, the square root of a negative number would produce imaginary results.", "---", "### Finding the Maximum and Minimum Values", "The function $ f(x) = \sqrt{4 - x^2} $ reaches its maximum and minimum values over the interval $ x \in [-2, 2] $.", "- Maximum occurs when $ 4 - x^2 $ is largest. Since $ x^2 $ increases from 0 to 4 across $ [-2, 2] $, $ 4 - x^2 $ is maximized when $ x = 0 $:\n $$ f(0) = \sqrt{4 - 0^2} = \sqrt{4} = 2 $$\n So, the maximum value is 2.", "- Minimum occurs when $ 4 - x^2 $ is smallest. At $ x = \pm 2 $, $ x^2 = 4 $, making $ 4 - x^2 = 0 $, so:\n $$ f(\pm 2) = \sqrt{0} = 0 $$\n Therefore, the minimum value is 0.", "---", "### The Full Range of $ \sqrt{4 - x^2} $", "Combining the maximum and minimum values, the output values of $ \sqrt{4 - x^2} $ span continuously from 0 to 2 as $ x $ moves from $ -2 $ to $ 0 $ to $ +2 $. Because the function is continuous and symmetric about $ x = 0 $, the range covers all real numbers from minimum to maximum:\n$$ \boxed{[0, 2]} $$", "---", "### Practical Insight: Geometric Interpretation", "The equation $ y = \sqrt{4 - x^2} $ describes the upper half of a circle with radius 2 centered at the origin. This confirms:\n- The expression is defined for $ |x| \leq 2 $,\n- The maximum vertical height (radius) is 2 (attained at $ x = 0 $),\n- The lowest points ($ y = 0 $) lie at $ x = \pm 2 $.", "---", "### Summary", "- Domain: $ \boxed{[-2, 2]} $ due to $ 4 - x^2 \geq 0 $.\n- Maximum value: $ 2 $ at $ x = 0 $.\n- Minimum value: $ 0 $ at $ x = \pm 2 $.\n- Full range: $ \boxed{[0, 2]} $.", "Mastering this function helps solidify core algebraic and calculus concepts while preparing you for more advanced equations involving radicals and symmetry."]