<p>Find the domain of the function
\(f(x) = \sqrt{3 - 2^x - 2^{1-x}} + \sqrt{\sin^{-1}x}\).</p>
Step-by-Step Solution
Key Concept: The domain requires both radicands non-negative: (1) 3 - 2^x - 2^(1-x) ≥ 0 must be solved by substituting t = 2^x to get a quadratic inequality, and (2) sin⁻¹(x) requires -1 ≤ x ≤ 1. The intersection of these constraints gives the final domain.
<p><strong>Step 1: Analyze the second radical constraint</strong></p><p>For √(sin⁻¹x) to be defined: sin⁻¹(x) ≥ 0 and x ∈ [-1, 1]</p><p>sin⁻¹(x) ≥ 0 when x ≥ 0</p><p>Combined: x ∈ [0, 1] ... (Constraint A)</p><p><strong>Step 2: Analyze the first radical constraint</strong></p><p>For √(3 - 2^x - 2^(1-x)) to be defined:</p><p>3 - 2^x - 2·2^(-x) ≥ 0</p><p><strong>Step 3: Substitute t = 2^x (where t > 0)</strong></p><p>3 - t - 2/t ≥ 0</p><p>Multiply by t > 0: 3t - t² - 2 ≥ 0</p><p>t² - 3t + 2 ≤ 0</p><p>(t - 1)(t - 2) ≤ 0</p><p>Therefore: t ∈ [1, 2], which means 2^x ∈ [1, 2]</p><p>So: x ∈ [0, 1] ... (Constraint B)</p><p><strong>Step 4: Find intersection</strong></p><p>Constraint A ∩ Constraint B = [0, 1] ∩ [0, 1] = [0, 1]</p><p>∴ Domain = <strong>[0, 1]</strong></p>
Correct Answer: [0, 1]