Let g: [-2, 2] rightarrow R and f: [-2, 2] ri | vedclass

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(B) First, we determine $|g(x)|$: $|g(x)| = \begin{cases} |-1| = 1, & -2 \le x < 0 \ |x^2 - 1|, & 0 \le x \le 2 \end{cases} = \begin{cases} 1, & -2 \

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(B) First, we determine $|g(x)|$: $|g(x)| = \begin{cases} |-1| = 1, & -2 \le x Next, we determine $g(|x|)$: Since $|x| \ge 0$ for all $x \in [-2, 2]$, we use the second part of $g(x)$: $g(|x|) = |x|^2 - 1 = x^2 - 1$ for all $x \in [-2, 2]$. Now, calculate $f(x) = |g(x)| + g(|x|) + 2$: For $-2 \le x For $0 \le x For $1 \le x \le 2$: $f(x) = (x^2 - 1) + (x^2 - 1) + 2 = 2x^2$. Thus, $f(x) = \begin{cases} x^2 + 2, & -2 \le x Checking differentiability at $x = 0$: $LHD = \lim_{x 	o 0^-} \frac{d}{dx}(x^2 + 2) = 2(0) = 0$. $RHD = \lim_{x 	o 0^+} \frac{d}{dx}(2) = 0$. Since $LHD = RHD$, $f$ is differentiable at $x = 0$. Checking differentiability at $x = 1$: $LHD = \lim_{x 	o 1^-} \frac{d}{dx}(2) = 0$. $RHD = \lim_{x 	o 1^+} \frac{d}{dx}(2x^2) = 4(1) = 4$. Since $LHD 
eq RHD$, $f$ is not differentiable at $x = 1$.

Let $g: [-2, 2] \rightarrow R$ and $f: [-2, 2] \rightarrow R$ be two functions defined as $g(x) = \begin{cases} -1, & \text{if } -2 \le x < 0 \\ x^2 - 1, & \text{if } 0 \le x \le 2 \end{cases}$ and $f(x) = |g(x)| + g(|x|) + 2$. In the interval $(-2, 2)$, $f$ is not differentiable at $x = $

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