Suppose f: [-2, 2] rightarrow R is defined by | 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 values $x$ such t

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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 values $x$ such that $x \leq 0$ and $f(|x|) = x$.Since $x \leq 0$, we have $|x| = -x$.Substituting this into the condition $f(|x|) = x$, we get $f(-x) = x$.Since $x \leq 0$, it follows that $-x \geq 0$.If $-x = 0$, then $x = 0$. Checking the condition: $f(|0|) = f(0) = -1$. But $x = 0$, so $-1 
eq 0$.If $-x > 0$, then $-x$ falls in the interval $(0, 2]$.Using the definition $f(t) = t - 1$ for $t \in (0, 2]$, we have $f(-x) = (-x) - 1$.Setting this equal to $x$, we get $-x - 1 = x$.Solving for $x$: $2x = -1$, which gives $x = -\frac{1}{2}$.Since $-\frac{1}{2} \leq 0$, this value is valid.Thus, the set is $\{-\frac{1}{2}\}$.

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

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