f(x) = begin{cases} 3-x, & -1 leqslant x

Question

(C) To find the range of $(f \circ g)(x)$,we first determine the range of $g(x)$.For $g(x) = \begin{cases} -x, & -2 \leqslant x \leqslant 3 \ x, & 0

Vedclass · Accepted Answer

$[-4, \frac{13}{3}]$

Vedclass · Answer

(C) To find the range of $(f \circ g)(x)$,we first determine the range of $g(x)$.For $g(x) = \begin{cases} -x, & -2 \leqslant x \leqslant 3 \ x, & 0 \leqslant x \leqslant 1 \end{cases}$,the range of $g(x)$ is $[-3, 2]$.Now,we evaluate $f(u)$ where $u \in [-3, 2]$.Given $f(u) = \begin{cases} 3-u, & -1 \leqslant u Since the domain of $f$ is $[-3, 2]$,we consider the values of $f(u)$ for $u \in [-3, 2]$.For $u \in [-3, 2]$,$f(u) = 1 + \frac{5u}{3}$.The minimum value is $f(-3) = 1 + \frac{5(-3)}{3} = 1 - 5 = -4$.The maximum value is $f(2) = 1 + \frac{5(2)}{3} = 1 + \frac{10}{3} = \frac{13}{3}$.Thus,the range of $(f \circ g)(x)$ is $[-4, \frac{13}{3}]$.

$f(x) = \begin{cases} 3-x, & -1 \leqslant x < 0 \\ 1+\frac{5x}{3}, & -3 \leqslant x \leqslant 2 \end{cases}$ and $g(x) = \begin{cases} -x, & -2 \leqslant x \leqslant 3 \\ x, & 0 \leqslant x \leqslant 1 \end{cases}$. Find the range of $(f \circ g)(x)$.

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