Suppose f:[-2, 2] to R is defined by f(x) = b | vedclass

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(C) We are given the function $f(x) = \begin{cases} -1 & 	ext{for } -2 \le x \le 0 \ x - 1 & 	ext{for } 0 < x \le 2 \end{cases}$.We need to find $x

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$\{-1/2\}$

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(C) We are given the function $f(x) = \begin{cases} -1 & 	ext{for } -2 \le x \le 0 \ x - 1 & 	ext{for } 0 We need to find $x \in (-2, 2)$ such that $x \le 0$ and $f(|x|) = x$.Since $x \le 0$,we have $|x| = -x$. Since $x \in (-2, 2)$ and $x \le 0$,it follows that $|x| \in [0, 2)$.If $x = 0$,then $f(|0|) = f(0) = -1$. But $x = 0$,so $f(0) 
eq 0$.If $x \in (-2, 0)$,then $|x| \in (0, 2)$.For $|x| \in (0, 2)$,the function is defined as $f(|x|) = |x| - 1$.We set $f(|x|) = x$,which gives $|x| - 1 = x$.Since $x This simplifies to $2x = -1$,or $x = -1/2$.Since $-1/2 \in (-2, 0)$,this is a valid solution.Thus,the set is $\{-1/2\}$.

Suppose $f:[-2, 2] \to R$ is defined by $f(x) = \begin{cases} -1 & \text{for } -2 \le x \le 0 \\ x - 1 & \text{for } 0 < x \le 2 \end{cases}$,then find the set $\{ x \in (-2, 2) : x \le 0 \text{ and } f(|x|) = x \}$.

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