If f(x) = begin{cases} |x - 3|, & x geqslant | vedclass

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(C) To check continuity at $x = 1$: $f(1) = |1 - 3| = |-2| = 2$. Left-hand limit: $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^-} (\frac{x^2}{4} - \frac{3x}

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Continuous at $x = 1$ and $x = 3$

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(C) To check continuity at $x = 1$: $f(1) = |1 - 3| = |-2| = 2$. Left-hand limit: $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^-} (\frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4}) = \frac{1}{4} - \frac{3}{2} + \frac{13}{4} = \frac{1 - 6 + 13}{4} = \frac{8}{4} = 2$. Right-hand limit: $\lim_{x 	o 1^+} f(x) = \lim_{x 	o 1^+} |x - 3| = |1 - 3| = 2$. Since $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1) = 2$,the function is continuous at $x = 1$. To check continuity at $x = 3$: The function is defined as $f(x) = |x - 3|$ in the neighborhood of $x = 3$. Since $|x - 3|$ is a modulus function,it is continuous everywhere. Thus,$f(x)$ is continuous at $x = 1$ and $x = 3$.

$A$. $\|x\| + \|x - 2\|$	$I$. Right hand limit does not exist at $x = 2$.
$B$. $\text{cosech } x$	$II$. Continuous only for non-zero real values of $x$.
$C$. $x - [x]$	$III$. Limit is zero for all real $x$.
$D$. $\sqrt{2 - x}$	$IV$. Continuous for all real value of $x$.
	$V$. Discontinuous at all integral values of $x$.

If $f(x) = \begin{cases} |x - 3|, & x \geqslant 1 \\ \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4}, & x < 1 \end{cases}$,then $f(x)$ is:

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