Question:

<p><strong>Step 1: Build f, g, h sequentially</strong></p><p>f(x) = x</p><p>g(x) = |1 - x|, critical point at x = 1</p><p>h(x) = 2 - |1 - x| = {3 - x for x ≤ 1; x + 1 for x > 1}</p><p>Critical point: x = 1 (where slope changes)</p><p><strong>Step 2: Evaluate h at critical point and zeros</strong></p><p>h(1) = 2, h(0) = 3, h(-1) = 4</p><p>h(x) = 0 gives: 3 - x = 0 → x = 3 (x ≤ 1? No) or x + 1 = 0 → x = -1</p><p>So h(x) = 0 at x = -1</p><p><strong>Step 3: Construct L(x) = h(|x|) + |h(x)|</strong></p><p>Critical points to check: x = 0 (from |x|), x = 1 (from h's non-differentiability), x = -1 (from h(x) = 0)</p><p><strong>Step 4: Analyze each critical point</strong></p><p><strong>At x = -1:</strong> h(-1) = 0, h(|-1|) = h(1) = 2</p><p>Left derivative: h'(1⁻) = -1, |h|'(-1⁻) = -h' = 1 → L'(-1⁻) = -1 + 1 = 0</p><p>Right derivative: h'(1⁺) = 1, |h|'(-1⁺) = h' = -1 → L'(-1⁺) = 1 - 1 = 0</p><p>Differentiable at x = -1</p><p><strong>At x = 0:</strong> h(0) = 3, h(|0|) = 3</p><p>Left: L'(0⁻) = -h'(0) + h'(0) = 0</p><p>Right: L'(0⁺) = h'(0) + h'(0) = 2</p><p>Non-differentiable ✓</p><p><strong>At x = 1:</strong> h(1) = 2, h(|1|) = 2</p><p>Left: L'(1⁻) = -1 + (-1) = -2</p><p>Right: L'(1⁺) = 1 + 1 = 2</p><p>Non-differentiable ✓</p><p>∴ <strong>Answer: 2 points (x = 0 and x = 1)</strong></p>