Limits, Continuity & Differentiability
Non-differentiability of composite functions
Grade 12
<p>Let \(f(x)=x\), \(g(x)=|1-f(x)|\), \(h(x)=2-g(x)\), \(L(x)=h(|x|)+|h(x)|\). Find the number of points where \(L(x)\) is non-differentiable.</p>
Step-by-Step Solution
Key Concept: Track the absolute value operations through compositions f→g→h→L to identify all critical points where the piecewise definition changes sign or structure, then test differentiability at each candidate point using left and right derivatives.
<p><strong>Step 1: Build the composite function</strong></p><p>f(x) = x</p><p>g(x) = |1 - x|, which is: 1-x for x≤1, and x-1 for x>1</p><p><strong>Step 2: Find h(x) = 2 - g(x)</strong></p><p>h(x) = 2 - |1-x| = {x+1 for x≤1; 3-x for x>1}</p><p><strong>Step 3: Analyze L(x) = h(|x|) + |h(x)|</strong></p><p>We need to consider regions based on where x and |x| cause sign changes in h.</p><p><strong>For x ≥ 0:</strong> h(|x|) = h(x) = {x+1 for 0≤x≤1; 3-x for x>1}</p><p>For h(x): h(x)≥0 when -1≤x≤3, so |h(x)| = h(x) for x∈[0,3]</p><p>Thus L(x) = 2h(x) for x∈[0,1]: L(x) = 2(x+1) = 2x+2</p><p>And L(x) = 2h(x) for x∈(1,∞): L(x) = 2(3-x) = 6-2x</p><p><strong>For x < 0:</strong> h(|x|) = h(-x) = {-x+1 for -x≤1 (i.e., x≥-1); 3-(-x)=3+x for x<-1}</p><p>For x∈[-1,0): h(x) = x+1 ∈[0,1], so |h(x)| = x+1</p><p>Thus L(x) = (-x+1) + (x+1) = 2 (constant)</p><p>For x<-1: h(x) = x+1 < 0, so |h(x)| = -(x+1) = -x-1</p><p>Thus L(x) = (3+x) + (-x-1) = 2 (constant)</p><p><strong>Step 4: Check differentiability at critical points</strong></p><p>L(x) = {2 for x<0; 2x+2 for 0≤x≤1; 6-2x for x>1}</p><p>At x=0: Left derivative = 0, Right derivative = 2. Non-differentiable ✓</p><p>At x=1: Left derivative = 2, Right derivative = -2. Non-differentiable ✓</p><p>∴ <strong>Answer: 2 points</strong></p>
Correct Answer: 2