<p>Let \(f(x)=\sqrt{x-2}+\sqrt{4-x}\). Choose domain \(X\) and codomain \(Y\) so that \(f:X\to Y\) is bijective.</p>
Step-by-Step Solution
<div class="solution"><p><strong>Key Idea:</strong> The natural domain is [2,4] but f rises then falls. Restrict to one monotone side.</p><p><strong>Step 1:</strong> $f'(x) = \frac{1}{2\sqrt{x-2}} - \frac{1}{2\sqrt{4-x}}$ -- positive on [2,3), negative on (3,4]. So f is strictly decreasing on [3,4].</p><p><strong>Step 2:</strong> On [3,4], range is $[f(4),f(3)] = [\sqrt{2}, 2]$</p><p><strong>Answer: $X=[3,4],\ Y=[\sqrt{2},2]$</strong></p><div class="trap-box"><strong>Trap:</strong> Full domain [2,4] gives correct range but f(2)=f(4), so not injective.<div class="key-concept"><strong>Key Concept:</strong> Restrict domain to monotone branch, then match codomain to that branch's range
Correct Answer: X=[3,4], Y=[√2,2]