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(D) First,we rewrite the function $f(x)$ by analyzing the absolute value $|x - 3|$:$|x - 3| = x - 3$ if $x \ge 3$ and $-(x - 3) = 3 - x$ if $x < 3$.Th

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All the above

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(D) First,we rewrite the function $f(x)$ by analyzing the absolute value $|x - 3|$:$|x - 3| = x - 3$ if $x \ge 3$ and $-(x - 3) = 3 - x$ if $x Thus,the function is defined as:$f(x) = \begin{cases} \frac{1}{4}x^2 - \frac{3}{2}x + \frac{13}{4}, & x Check continuity at $x = 1$:$f(1) = 3 - 1 = 2$.$LHL = \lim_{x 	o 1^-} (\frac{1}{4}x^2 - \frac{3}{2}x + \frac{13}{4}) = \frac{1}{4} - \frac{3}{2} + \frac{13}{4} = \frac{14}{4} - \frac{6}{4} = 2$.$RHL = \lim_{x 	o 1^+} (3 - x) = 2$.Since $LHL = RHL = f(1)$,the function is continuous at $x = 1$.Check continuity at $x = 3$:$f(3) = 3 - 3 = 0$.$LHL = \lim_{x 	o 3^-} (3 - x) = 0$.$RHL = \lim_{x 	o 3^+} (x - 3) = 0$.Since $LHL = RHL = f(3)$,the function is continuous at $x = 3$.Check differentiability at $x = 1$:$LHD = \lim_{h 	o 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h 	o 0^-} \frac{\frac{1}{4}(1+h)^2 - \frac{3}{2}(1+h) + \frac{13}{4} - 2}{h} = \lim_{h 	o 0^-} \frac{\frac{1}{4}(1+2h+h^2) - \frac{3}{2} - \frac{3}{2}h + \frac{13}{4} - 2}{h} = \lim_{h 	o 0^-} \frac{\frac{1}{4}h^2 + \frac{1}{2}h - \frac{3}{2}h}{h} = \lim_{h 	o 0^-} (\frac{1}{4}h - 1) = -1$.$RHD = \lim_{h 	o 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h 	o 0^+} \frac{3 - (1+h) - 2}{h} = \lim_{h 	o 0^+} \frac{-h}{h} = -1$.Since $LHD = RHD$,the function is differentiable at $x = 1$.Therefore,all statements are correct.

The function defined by $f(x) = \begin{cases} |x - 3|, & x \ge 1 \\ \frac{1}{4}x^2 - \frac{3}{2}x + \frac{13}{4}, & x < 1 \end{cases}$ is

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