Question:

<div class="solution"><p><strong>Key Idea:</strong> Let <span class="math-inline">\(a=\sqrt{x+2},\ b=\sqrt{2-x}\)</span>. Then <span class="math-inline">\(a^2+b^2=4\)</span>.</p><p><strong>Step 1:</strong> Domain: <span class="math-inline">\(x\in[-2,2]\)</span></p><p><strong>Step 2:</strong> Since <span class="math-inline">\(a^2+b^2=4\)</span>, the sum <span class="math-inline">\(a+b\)</span> satisfies <span class="math-block">\[\sqrt{2} \le a+b \le 2\sqrt{2}\]</span></p><p><strong>Step 3:</strong> <span class="math-block">\[\frac{4}{a+b} \in [\sqrt{2},\ 2]\]</span></p><p><strong>Step 4:</strong> <span class="math-inline">\(f(x) = \log_2\)</span> of that, giving range <span class="math-inline">\([\frac{1}{2}, 1]\)</span></p><p><strong>Answer: <span class="math-inline">\(\left[\frac{1}{2},1\right]\)</span></strong></p><div class="trap-box"><strong>Trap:</strong> Do not differentiate immediately. The fixed square-sum <span class="math-inline">\(a^2+b^2=4\)</span> makes a range argument much cleaner.</div><div class="key-concept"><strong>Key Concept:</strong> Range via substitution + AM-QM on constrained variables</div></div>