Question:

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