Let f : [-1,3] to R be defined as f(x) = begi | vedclass

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(A) We analyze the function $f(x)$ in different intervals:For $x \in [-1, 0)$,$|x| = -x$ and $[x] = -1$,so $f(x) = -x - 1$.For $x \in [0, 1)$,$|x| = x

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(A) We analyze the function $f(x)$ in different intervals:For $x \in [-1, 0)$,$|x| = -x$ and $[x] = -1$,so $f(x) = -x - 1$.For $x \in [0, 1)$,$|x| = x$ and $[x] = 0$,so $f(x) = x + 0 = x$.For $x \in [1, 2)$,$|x| = x$,so $f(x) = x + x = 2x$.For $x \in [2, 3]$,$|x| = x$,so $f(x) = x + x = 2x$.Thus,$f(x) = \begin{cases} -x-1, & -1 \leq x Checking continuity at $x=0$: $\lim_{x 	o 0^-} f(x) = -1$ and $f(0) = 0$. Since $-1 
eq 0$,$f$ is discontinuous at $x=0$.Checking continuity at $x=1$: $\lim_{x 	o 1^-} f(x) = 1$ and $f(1) = 2(1) = 2$. Since $1 
eq 2$,$f$ is discontinuous at $x=1$.Checking continuity at $x=2$: $\lim_{x 	o 2^-} f(x) = 2(2) = 4$ and $f(2) = 2(2) = 4$. Since the limit equals the function value,$f$ is continuous at $x=2$.Therefore,$f$ is discontinuous at only two points,$x=0$ and $x=1$.

Let $f : [-1,3] \to R$ be defined as $f(x) = \begin{cases} |x| + [x], & -1 \leq x < 1 \\ x + |x|, & 1 \leq x < 2 \\ x + |x|, & 2 \leq x \leq 3 \end{cases}$ where $[t]$ denotes the greatest integer less than or equal to $t$. Then,$f$ is discontinuous at:

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