Let f(x) = begin{cases} -1, & -2 le x

Question

(D) Given $f(x) = \begin{cases} -1, & -2 \le x < 0 \ x^2 - 1, & 0 \le x \le 2 \end{cases}$.First,find $|f(x)|$:$|f(x)| = \begin{cases} |-1| = 1, & -2

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not differentiable at one point

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(D) Given $f(x) = \begin{cases} -1, & -2 \le x First,find $|f(x)|$:$|f(x)| = \begin{cases} |-1| = 1, & -2 \le x Next,find $f(|x|)$:Since $|x| \ge 0$ for all $x$,$f(|x|) = |x|^2 - 1 = x^2 - 1$ for $x \in [-2, 2]$.Now,$g(x) = |f(x)| + f(|x|)$:For $x \in [-2, 0)$,$g(x) = 1 + (x^2 - 1) = x^2$.For $x \in [0, 2]$,$g(x) = |x^2 - 1| + (x^2 - 1)$.Expanding $g(x)$:$g(x) = \begin{cases} x^2, & -2 \le x Check continuity and differentiability at $x=0$:$g(0^-) = 0^2 = 0$,$g(0^+) = 0$. Continuous at $x=0$.$g'(0^-) = 2x|_{x=0} = 0$,$g'(0^+) = 0$. Differentiable at $x=0$.Check continuity and differentiability at $x=1$:$g(1^-) = 0$,$g(1^+) = 2(1^2 - 1) = 0$. Continuous at $x=1$.$g'(1^-) = 0$,$g'(1^+) = 4x|_{x=1} = 4$.Since $g'(1^-) 
eq g'(1^+)$,$g$ is not differentiable at $x=1$.Thus,$g$ is not differentiable at one point.

Let $f(x) = \begin{cases} -1, & -2 \le x < 0 \\ x^2 - 1, & 0 \le x \le 2 \end{cases}$ and $g(x) = |f(x)| + f(|x|)$. Then,in the interval $(-2, 2)$,$g$ is

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