If f:[0,3] rightarrow [0,3] is defined by f(x | vedclass

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

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Discontinuous at $x=1$ and $x=2$

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(C) Let $g(x) = f(f(x))$.Given $f(x) = \begin{cases} 1+x, & 0 \leq x \leq 2 \ 3-x, & 2 For $0 \leq x \leq 2$,$f(x) = 1+x$. Since $0 \leq x \leq 2$,$1 \leq 1+x \leq 3$. If $1 \leq 1+x \leq 2$ (i.e.,$0 \leq x \leq 1$),then $f(f(x)) = f(1+x) = 1+(1+x) = 2+x$.If $2 For $2 Thus,$f(f(x)) = f(3-x) = 1+(3-x) = 4-x$.So,$g(x) = \begin{cases} 2+x, & 0 \leq x \leq 1 \ 2-x, & 1 Checking continuity at $x=1$:$LHL = \lim_{x 	o 1^-} (2+x) = 3$,$RHL = \lim_{x 	o 1^+} (2-x) = 1$. Since $LHL 
eq RHL$,it is discontinuous at $x=1$.Checking continuity at $x=2$:$LHL = \lim_{x 	o 2^-} (2-x) = 0$,$RHL = \lim_{x 	o 2^+} (4-x) = 2$. Since $LHL 
eq RHL$,it is discontinuous at $x=2$.Therefore,$f(f(x))$ is discontinuous at $x=1$ and $x=2$.

If $f:[0,3] \rightarrow [0,3]$ is defined by $f(x) = \begin{cases} 1+x, & 0 \leq x \leq 2 \\ 3-x, & 2 < x \leq 3 \end{cases}$,then $f(f(x))$ is:

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