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# 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 $x=3$

Last updated date: 20th Mar 2023
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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 $x=a$.

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.