Let f(x) = begin{cases} max {|x|, x^2}, & |x| | vedclass

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(C) Given $f(x) = \begin{cases} \max \{|x|, x^2\}, & |x| \le 2 \ 8 - 2|x|, & 2 < |x| \le 4 \end{cases}$.For $|x| \le 2$,$f(x) = \begin{cases} x^2, &

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equals $\{-2, -1, 1, 2\}$

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(C) Given $f(x) = \begin{cases} \max \{|x|, x^2\}, & |x| \le 2 \ 8 - 2|x|, & 2 For $|x| \le 2$,$f(x) = \begin{cases} x^2, & |x| \le 1 \ |x|, & 1 Combining these,$f(x) = \begin{cases} 8+2x, & -4 Checking differentiability at critical points:At $x = -2$: $f(-2) = 4$. Left derivative is $-2$,right derivative is $-1$. Not differentiable.At $x = -1$: $f(-1) = 1$. Left derivative is $-1$,right derivative is $-2(-1) = -2$. Not differentiable.At $x = 0$: $f(0) = 0$. Left derivative is $2(0) = 0$,right derivative is $2(0) = 0$. Differentiable.At $x = 1$: $f(1) = 1$. Left derivative is $2(1) = 2$,right derivative is $1$. Not differentiable.At $x = 2$: $f(2) = 2$. Left derivative is $1$,right derivative is $-2$. Not differentiable.Thus,$S = \{-2, -1, 1, 2\}$.

Let $f(x) = \begin{cases} \max \{|x|, x^2\}, & |x| \le 2 \\ 8 - 2|x|, & 2 < |x| \le 4 \end{cases}$. Let $S$ be the set of points in the interval $(-4, 4)$ at which $f$ is not differentiable. Then $S$

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