Define f(x) = begin{cases} 1 + x, & 0 leq x l | vedclass

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(D) We have, $f(x) = \begin{cases} 1 + x, & 0 \leq x \leq 2 \ 3 - x, & 2 < x \leq 3 \end{cases}$.$f(f(x))$ is defined as:For $0 \leq f(x) \leq 2$, $f

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$8$

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(D) We have, $f(x) = \begin{cases} 1 + x, & 0 \leq x \leq 2 \ 3 - x, & 2 $f(f(x))$ is defined as:For $0 \leq f(x) \leq 2$, $f(f(x)) = 1 + f(x)$.For $2 Evaluating for $x \in [0, 3]$:If $0 \leq x \leq 1$, then $1 \leq f(x) \leq 2$, so $f(f(x)) = 1 + (1 + x) = 2 + x$.If $1 If $2 Thus, $f(f(x)) = \begin{cases} 2 + x, & 0 \leq x \leq 1 \ 2 - x, & 1 Checking continuity at $x = 1$: $\lim_{x 	o 1^-} f(f(x)) = 3$, $\lim_{x 	o 1^+} f(f(x)) = 1$. Since $3 
eq 1$, it is discontinuous at $x = 1$.Checking continuity at $x = 2$: $\lim_{x 	o 2^-} f(f(x)) = 0$, $\lim_{x 	o 2^+} f(f(x)) = 2$. Since $0 
eq 2$, it is discontinuous at $x = 2$.Therefore, $a = 1$ and $b = 2$.Calculating $2a + 3b = 2(1) + 3(2) = 2 + 6 = 8$.

Define $f(x) = \begin{cases} 1 + x, & 0 \leq x \leq 2 \\ 3 - x, & 2 < x \leq 3 \end{cases}$. If $f \circ f(x)$ is discontinuous at $a$ and $b$ in $[0, 3]$ and $a < b$, then $2 a + 3 b = $

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