Question:

<div class="solution"><p><strong>Domain:</strong> Need <span class="math-inline">\(x\ge 0\)</span> for <span class="math-inline">\(\sqrt{x}\)</span> and <span class="math-inline">\(x\ge -1\)</span> for <span class="math-inline">\(\sqrt{x+1}\)</span>. Domain: <span class="math-inline">\(x\ge 0\)</span>.</p><p><strong>RHD at x=0:</strong><br><span class="math-block">\[\lim_{h\to 0^+}\frac{f(h)-f(0)}{h}=\lim_{h\to 0^+}(\sqrt{h}-\sqrt{h+1})=-1\]</span>So Rf'(0)=-1. (A) Rf'(0) exists ✓</p><p><strong>LHD at x=0:</strong> Domain only includes x≥0, so LHD doesn't exist in the usual sense. But we can check: Lf'(0) doesn't exist (not in domain). (B) says Lf'(0) exists but Rf'(0) doesn't — False.</p><p><strong>lim f(x) as x→0⁺:</strong> <span class="math-inline">\(f(0)=0,\ \lim_{x\to 0^+}f(x)=0\cdot(0-1)=0\)</span>. Limit exists. (C) ✓</p><p><strong>Differentiability at x=0:</strong> Since domain is [0,∞), f is differentiable at x=0 from right, meaning f is differentiable at x=0 in its domain. (D) ✓</p><p><strong>Answer: (A),(C),(D)</strong></p><div class="key-concept"><strong>Key Concept:</strong> For functions with restricted domain, differentiability means one-sided derivative</div></div>