Question:

<p><strong>Step 1: Identify candidate points for non-differentiability</strong></p><p>The function is defined piecewise with potential issues at x = 0 and x = 1 (the boundary points between pieces).</p><p><strong>Step 2: Analyze differentiability at x = 0</strong></p><p>Check continuity from the right: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \left(-\frac{x}{2}\right) = 0$</p><p>Check continuity from the left using the middle piece: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-\cos x) = -\cos(0) = -1$</p><p>Since $0 \neq -1$, the function is <strong>discontinuous at x = 0</strong>. Therefore, f(x) is <strong>not differentiable at x = 0</strong>.</p><p><strong>Step 3: Analyze differentiability at x = 1</strong></p><p>Check continuity from the left: $\lim_{x \to 1^-} f(x) = -\cos(1)$</p><p>Check continuity from the right: $\lim_{x \to 1^+} f(x) = \ln(1) = 0$</p><p>Since $-\cos(1) \approx -0.540 \neq 0$, the function is <strong>discontinuous at x = 1</strong>. Therefore, f(x) is <strong>not differentiable at x = 1</strong>.</p><p><strong>Step 4: Check differentiability within each piece</strong></p><p>• For $x > 0$ (and $x \neq 1$): $f(x) = -\frac{x}{2}$ is a linear function, differentiable everywhere in its domain.</p><p>• For $0 < x < 1$: $f(x) = -\cos x$ is a trigonometric function, differentiable everywhere.</p><p>• For $x > 1$: $f(x) = \ln x$ is a logarithmic function, differentiable for all $x > 1$.</p><p><strong>Step 5: Conclusion</strong></p><p>The function is not differentiable at exactly two points: x = 0 and x = 1, where discontinuities occur.</p><p>$\therefore$ Answer: <strong>2</strong></p>