<p>Find the domain and range of the function
\(f(x) = \sqrt{2 - x} + \sqrt{1 + x}\).</p>
Step-by-Step Solution
Key Concept: For f(x) = √(2-x) + √(1+x) to be defined, both expressions under the square roots must be non-negative simultaneously. The domain is their intersection: 2-x ≥ 0 AND 1+x ≥ 0, giving [-1, 2]. For range, treat f as a function on this domain and find its minimum and maximum by calculus or algebraic manipulation.
<p><strong>Step 1: Find Domain</strong></p><p>For f(x) = √(2-x) + √(1+x) to be defined:</p><p>• 2 - x ≥ 0 ⟹ x ≤ 2</p><p>• 1 + x ≥ 0 ⟹ x ≥ -1</p><p>∴ Domain = [-1, 2]</p><p><strong>Step 2: Find Range using Calculus</strong></p><p>f'(x) = -1/(2√(2-x)) + 1/(2√(1+x))</p><p>Setting f'(x) = 0:</p><p>1/(2√(1+x)) = 1/(2√(2-x))</p><p>√(2-x) = √(1+x)</p><p>2 - x = 1 + x ⟹ x = 1/2</p><p><strong>Step 3: Evaluate at Critical Point and Endpoints</strong></p><p>• f(-1) = √(2-(-1)) + √(1-1) = √3 + 0 = √3</p><p>• f(1/2) = √(2-1/2) + √(1+1/2) = √(3/2) + √(3/2) = 2√(3/2) = √6</p><p>• f(2) = √(2-2) + √(1+2) = 0 + √3 = √3</p><p>Minimum value = √3 (at x = -1 and x = 2)</p><p>Maximum value = √6 (at x = 1/2)</p><p>∴ Range = [√3, √6]</p>
Correct Answer: Domain = [-1, 2], Range = [√3, √6]