If f(x) = begin{cases} 2+2x, & -1 leq x

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(C) Given $f(x) = \begin{cases} 2+2x, & -1 \leq x < 0 \ 1-\frac{x}{3}, & 0 \leq x \leq 3 \end{cases}$ and $g(x) = \begin{cases} -x, & -3 \leq x \leq

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$[0, 1]$

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(C) Given $f(x) = \begin{cases} 2+2x, & -1 \leq x We need to find the range of $f(g(x))$.Case $1$: $-3 \leq x \leq 0$,then $g(x) = -x$. Since $-3 \leq x \leq 0$,we have $0 \leq g(x) \leq 3$.For $0 \leq g(x) \leq 3$,$f(g(x)) = 1 - \frac{g(x)}{3} = 1 - \frac{-x}{3} = 1 + \frac{x}{3}$.As $x$ varies from $-3$ to $0$,$f(g(x))$ varies from $1 + \frac{-3}{3} = 0$ to $1 + \frac{0}{3} = 1$.So,the range for this part is $[0, 1]$.Case $2$: $0 For $0 As $x$ varies from $0$ to $1$,$f(g(x))$ varies from $1 - \frac{0}{3} = 1$ to $1 - \frac{1}{3} = \frac{2}{3}$.So,the range for this part is $[\frac{2}{3}, 1)$.Combining both cases,the range of $f(g(x))$ is $[0, 1]$.

If $f(x) = \begin{cases} 2+2x, & -1 \leq x < 0 \\ 1-\frac{x}{3}, & 0 \leq x \leq 3 \end{cases}$ and $g(x) = \begin{cases} -x, & -3 \leq x \leq 0 \\ x, & 0 < x \leq 1 \end{cases}$,then the range of $(f \circ g)(x)$ is:

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