<p>For <span class="math-inline">\(f(x)=|x+3|-|x+1|-|x-1|+|x-3|\)</span>, which are correct?</p>
Step-by-Step Solution
Key Concept: General
<div class="solution"><p>Piecewise analysis at breakpoints -3,-1,1,3:</p><p>x≤-3: f=0; -3≤x≤-1: f=2x+6 (0 to 4); -1≤x≤1: f=4; 1≤x≤3: f=-2x+6 (4 to 0); x≥3: f=0</p><p>Range=[0,4]. So (A) false, (B) true ✓</p><p>f=4 on [-1,1] → infinitely many ✓; f=0 on (-∞,-3]∪[3,∞) → infinitely many ✓</p><p><strong>Answer: (B),(C),(D)</strong></p><div class="trap-box"><strong>Trap:</strong> Range is [0,4], not (-∞,4]. The function is bounded below by 0.</div><div class="key-concept"><strong>Key Concept:</strong> Absolute value piecewise — symmetric breakpoints often give plateau regions</div></div>
Correct Answer: B,C,D