Question:

<div class="solution"><p>Piecewise analysis at breakpoints -3,-1,1,3:</p><p>x\le-3: f=0; -3\lex\le-1: f=2x+6 (0 to 4); -1\lex\le1: f=4; 1\lex\le3: f=-2x+6 (4 to 0); x\ge3: f=0</p><p>Range=[0,4]. So (A) false, (B) true ✓</p><p>f=4 on [-1,1] \to infinitely many ✓; f=0 on (-\infty,-3]\cup[3,\infty) \to infinitely many ✓</p><p><strong>Answer: (B),(C),(D)</strong></p><div class="trap-box"><strong>Trap:</strong> Range is [0,4], not (-\infty,4]. The function is bounded below by 0.<div class="key-concept"><strong>Key Concept:</strong> Absolute value piecewise -- symmetric breakpoints often give plateau regions