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(A) Given $f(x) = 3(1 - \frac{|x|}{2})$ for $|x| \leq 2$ and $0$ otherwise.$f(x+2) = 3(1 - \frac{|x+2|}{2})$ for $|x+2| \leq 2$ (i.e.,$-4 \leq x \leq

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$4$

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(A) Given $f(x) = 3(1 - \frac{|x|}{2})$ for $|x| \leq 2$ and $0$ otherwise.$f(x+2) = 3(1 - \frac{|x+2|}{2})$ for $|x+2| \leq 2$ (i.e.,$-4 \leq x \leq 0$) and $0$ otherwise.$f(x-2) = 3(1 - \frac{|x-2|}{2})$ for $|x-2| \leq 2$ (i.e.,$0 \leq x \leq 4$) and $0$ otherwise.Thus,$g(x) = f(x+2) - f(x-2)$ is defined as:$g(x) = \begin{cases} 3(1 - \frac{|x+2|}{2}) & -4 \leq x Simplifying $g(x)$:$g(x) = \begin{cases} \frac{3x}{2} + 6 & -4 \leq x \leq -2 \ -\frac{3x}{2} & -2 Checking continuity: The function $g(x)$ is continuous everywhere because the limits at $x = -4, -2, 2, 4$ match the function values (all are $0$ at boundaries).Thus,$n = 0$.Checking differentiability: The function $g(x)$ is non-differentiable at points where the slope changes abruptly: $x = -4, -2, 2, 4$.Thus,$m = 4$.Therefore,$n + m = 0 + 4 = 4$.

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