Let f(x) = begin{cases} |x|+3, & text{if } x | vedclass

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(B) For $x \leq -3$,$f(x) = |x| + 3 = -x + 3$. At $x = -3$: Left-hand limit: $\lim_{x 	o -3^-} f(x) = -(-3) + 3 = 6$. Right-hand limit: $\lim_{x 	o

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$f(x)$ is continuous at $x = -3$ but discontinuous at $x = 3$.

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(B) For $x \leq -3$,$f(x) = |x| + 3 = -x + 3$. At $x = -3$: Left-hand limit: $\lim_{x 	o -3^-} f(x) = -(-3) + 3 = 6$. Right-hand limit: $\lim_{x 	o -3^+} f(x) = -2(-3) = 6$. Value of function: $f(-3) = -(-3) + 3 = 6$. Since $\lim_{x 	o -3^-} f(x) = \lim_{x 	o -3^+} f(x) = f(-3)$,$f(x)$ is continuous at $x = -3$. At $x = 3$: Left-hand limit: $\lim_{x 	o 3^-} f(x) = -2(3) = -6$. Right-hand limit: $\lim_{x 	o 3^+} f(x) = 6(3) + 2 = 20$. Since $\lim_{x 	o 3^-} f(x) 
eq \lim_{x 	o 3^+} f(x)$,$f(x)$ is discontinuous at $x = 3$.

Let $f(x) = \begin{cases} |x|+3, & \text{if } x \leq -3 \\ -2x, & \text{if } -3 < x < 3 \\ 6x+2, & \text{if } x \geq 3 \end{cases}$. Determine the continuity of $f(x)$ at $x = -3$ and $x = 3$.

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