Question:

<p><strong>Step 1: Understanding g(x) = f(|x|)</strong></p><p>Since g(x) = f(|x|), we have:</p><p>• For x ≥ 0: g(x) = f(x)</p><p>• For x < 0: g(x) = f(-x)</p><p>This makes g(x) an even function.</p><p><strong>Step 2: Identify points to check for differentiability</strong></p><p>Non-differentiability can occur at:</p><p>• x = 0 (junction point of the absolute value)</p><p>• Any point where f(x) is non-differentiable in its original domain</p><p><strong>Step 3: Check differentiability at x = 0</strong></p><p>For g(x) to be differentiable at x = 0, we need:</p><p>• Left derivative: g'(0⁻) = lim[h→0⁻] [g(h) - g(0)]/h = lim[h→0⁻] [f(-h) - f(0)]/h</p><p>• Right derivative: g'(0⁺) = lim[h→0⁺] [g(h) - g(0)]/h = lim[h→0⁺] [f(h) - f(0)]/h</p><p>For differentiability at x = 0: g'(0⁻) must equal g'(0⁺)</p><p>If f'(0⁺) exists, then g'(0⁺) = f'(0⁺), and g'(0⁻) = -f'(0⁺)</p><p>These are equal only if f'(0⁺) = 0.</p><p><strong>Step 4: Analyze the given function</strong></p><p>Based on the incomplete problem statement mentioning f(x) = cos(x³)/3 for x ≥ 0 (and an unspecified condition for x < 0):</p><p>For x ≥ 0: f'(x) = -sin(x³) · x²</p><p>At x = 0: f'(0⁺) = 0</p><p>This means g(x) IS differentiable at x = 0 if the function definition is symmetric or appropriately defined for x < 0.</p><p><strong>Step 5: Check interior points</strong></p><p>For x > 0, f(x) = cos(x³)/3 is composed of differentiable functions, so f is differentiable for all x > 0.</p><p>Therefore, g(x) is differentiable for all x ≠ 0.</p><p>At x = 0, since f'(0⁺) = 0, g'(0) exists and equals 0.</p><p><strong>Step 6: Final count</strong></p><p>Without additional discontinuities in the definition of f for x < 0, and given that f is differentiable at x = 0 with f'(0) = 0, the function g(x) = f(|x|) has <strong>0 points of non-differentiability</strong>.</p><p><strong>∴ Answer: 0</strong></p>