Suppose f:[-2,2] rightarrow R is defined by f | vedclass

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(C) Given the function $f(x) = \begin{cases} -1, & -2 \leq x \leq 0 \ x-1, & 0 < x \leq 2 \end{cases}$.We need to find the set of $x$ such that $x \l

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$\{-\frac{1}{2}\}$

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(C) Given the function $f(x) = \begin{cases} -1, & -2 \leq x \leq 0 \ x-1, & 0 We need to find the set of $x$ such that $x \leq 0$ and $f(|x|) = x$.Since $x \leq 0$,we have $|x| = -x$. Since $x \in [-2, 0]$,then $|x| \in [0, 2]$.For $0 Setting $f(|x|) = x$,we get $-x - 1 = x$,which implies $2x = -1$,so $x = -\frac{1}{2}$.Checking the condition $x \leq 0$,we see that $x = -\frac{1}{2}$ satisfies this.If $x = 0$,then $f(|0|) = f(0) = -1$. But $x = 0$,so $f(0) 
eq 0$.Thus,the set is $\{-\frac{1}{2}\}$.

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

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