<p>Find the minimum value of \(2^{\sin x} + 2^{\cos x}\).</p>
Step-by-Step Solution
Key Concept: Use AM-GM inequality on 2^(sin x) and 2^(cos x), then minimize the constraint sin x + cos x = √2·sin(x + π/4) which ranges in [-√2, √2]. The minimum of f(t) = 2^t + 2^(√2-t) occurs at the boundary t = sin x where sin x + cos x is minimized.
<p><strong>Step 1:</strong> Let y = 2^(sin x) + 2^(cos x). Since sin x, cos x ∈ [-1, 1], we need to find the minimum over this range with the constraint sin²x + cos²x = 1.</p><p><strong>Step 2:</strong> Let sin x + cos x = t. Then sin x · cos x = (t² - 1)/2, where t ∈ [-√2, √2].</p><p><strong>Step 3:</strong> We have 2^(sin x) + 2^(cos x) ≥ 2√(2^(sin x)·2^(cos x)) = 2·2^((sin x + cos x)/2) = 2·2^(t/2) by AM-GM.</p><p><strong>Step 4:</strong> However, equality requires 2^(sin x) = 2^(cos x), so sin x = cos x = 1/√2 (impossible for real sin, cos). So we check boundary behavior.</p><p><strong>Step 5:</strong> The minimum occurs when sin x + cos x is minimized, at sin x + cos x = -√2 (when sin x = -1/√2, cos x = -1/√2). But we need to verify the actual minimum by checking critical points.</p><p><strong>Step 6:</strong> Taking derivative: dy/dx = 2^(sin x)·ln 2·cos x + 2^(cos x)·ln 2·(-sin x) = 0. This gives cos x·2^(sin x) = sin x·2^(cos x).</p><p><strong>Step 7:</strong> When sin x = -1/√2 and cos x = 1/√2: y = 2^(-1/√2) + 2^(1/√2) = 2^(-1/√2) + 2·2^(-1/√2) = 2^(1 - 1/√2).</p><p>∴ Answer: 2^(1 - 1/√2)</p>
Correct Answer: \(2^{1 - \frac{1}{\sqrt{2}}}\)