Question:
If $f(x)$ be defined on $[-2,2]$ and is given by $f(x)=\left\{\begin{array}{rr}-1, & -2 \leq x \leq 0 \\ x-1, & 0
Solution:
Given:
$f(x)= \begin{cases}-1, & -2 \leqslant x \leqslant 0 \\ x-1, & 0
Thus,
$g(x)=f(|x|)+|f(x)|$
$= \begin{cases}x-1+1, & -2 \leqslant x \leqslant 0 \\ x-1+(-x+1), & 0
$= \begin{cases}x, & -2 \leqslant x \leqslant 0 \\ 0, & 0
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