The function \[f\left( x \right)=\left\{ \begin{align}
  & \left| x-3 \right|,\text{ }x\ge 1 \\
 & \dfrac{{{x}^{2}}}{4}-\dfrac{3x}{2}+\dfrac{13}{4},\text{ }x<1 \\
\end{align} \right.\] is
(a) continuous at \[x=1\]
(b) differentiable at \[x=1\]
(c) continuous at \[x=3\]
(d) differentiable at \[x=3\]

Hint: If the value of limit of the function at a point \[x=a\] is equal to the value of the function at \[x=a\] , the function is said to be continuous at \[x=a\]. A function is differentiable at \[x=a\] , if the left-hand derivative of the function is equal to the right hand derivative of the function at \[x=a\].

Complete step-by-step answer:
At \[x=3\], \[f\left( x \right)=\left| x-3 \right|\].
We know \[\left| x-a \right|\] is a function that is continuous everywhere but it is not differentiable at vertex i.e. at \[x=a\].
So , \[f\left( x \right)\]is continuous at \[x=3\] but not differentiable at \[x=3\].
Now , we will check if the function is continuous or differentiable at critical points of the function .
A function is differentiable at \[x=a\] , if the left-hand derivative of the function is equal to the right hand derivative of the function at \[x=a\].
We know , the left hand derivative of \[f\left( x \right)\] at \[x=a\] is given as \[{{L}^{'}}=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{f\left( a-h \right)-f\left( a \right)}{-h}\] and the right hand derivative of \[f\left( x \right)\]at \[x=a\] is given as \[{{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( a+h \right)-f\left( a \right)}{h}\].
We will consider the critical point \[x=1\]. To determine differentiability of \[f\left( x \right)\] at \[x=1\], we will evaluate its left-hand derivative.
\[{{L}^{'}}=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{f\left( 1-h \right)-f\left( 1 \right)}{-h}\]
\[=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \dfrac{\left( \dfrac{{{\left( 1-h \right)}^{2}}}{4}-\dfrac{3\left( 1-h \right)}{2}+\dfrac{13}{4} \right)-\left( \dfrac{{{1}^{2}}}{4}-\dfrac{3\left( 1 \right)}{2}+\dfrac{13}{4} \right)}{-h} \right]\]
\[=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \dfrac{\dfrac{{{h}^{2}}-2h+1}{4}-\dfrac{3-3h}{2}+\dfrac{13}{4}-\dfrac{1}{4}+\dfrac{3}{2}-\dfrac{13}{4}}{-h} \right]\]
\[=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \dfrac{{{h}^{2}}+4h}{-4h} \right]\]
\[=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \dfrac{h+4}{-4} \right]\]
\[=-1\]
Now , we will evaluate the right-hand derivative of \[f\left( x \right)\] at \[x=1\]
\[{{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( 1+h \right)-f\left( 1 \right)}{h}\]
\[{{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left| 1+h-3 \right|-\left| 1-3 \right|}{h}\]
Now, we know\[1+h<3\],
So \[\left| 1+h-3 \right|=\left| h-2 \right|=2-h\]
So, \[{{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{2-h-2}{h}\]
\[=-1\]
Clearly, the left hand derivative is equal to the right hand derivative.
Hence \[f\left( x \right)\] is differentiable at \[x=1\].
Now, we also know that if a function is differentiable at a point then it must be continuous at that point.
So, \[f\left( x \right)\] is continuous at \[x=1\].
Hence , the function is differentiable and continuous at \[x=1\] and the function is continuous but not differentiable at \[x=3\].
Answer is (a) , (b) , (c)

Note: A function which is differentiable at a point is necessarily continuous at that point. But a function which is continuous at a point, may or may not be differentiable at that point.