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# Let $f\left( x \right)=\left[ x \right]+\left| 1-x \right|$ for $-1\le x\le 3$, where $\left[ x \right]$denotes the integer part of $x$. Then.(a) In the open interval $\left( -1,3 \right)$, $f$ has three points of discontinuity(b) $f$ is right continuous at $x=-1$ and has right derivative at $x=-1$(c) $f$ is left continuous at $x=3$ and has left derivative at $x=3$(d) $f$ has right derivative at $x=-1$ and is not differentiable at $x=0,1,2,3$

Last updated date: 17th Jul 2024
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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 $x=a$.

Let us consider $\left[ x \right]$ first.
We know,
\left[ x \right]=\left\{ \begin{align} & -1,\text{ }-1\le x<0 \\ & 0,\text{ }0\le x<1 \\ & 1,\text{ }1\le x<2 \\ & 2,\text{ }2\le x<3 \\ & 3,\text{ }x=3 \\ \end{align} \right.
Now, we also know,
\left| 1-x \right|=\left\{ \begin{align} & 1-x,\text{ }x\le 1 \\ & x-1,\text{ }x\ge 1 \\ \end{align} \right.
So, we can rewrite
f\left( x \right)=\left\{ \begin{align} & -1+1-x=-x,\text{ }-1\le x<0 \\ & 0+1-x=1-x,\text{ for 0}\le x<1 \\ & 1+x-1=x,\text{ for 1}\le x<2 \\ & 2+x-1=1+x,\text{ for }2\le x<3 \\ & 3+x-1=2+x,\text{ for }x=3 \\ \end{align} \right.
First of all, let’s consider the point $x=3$.
All the points greater than $x=3$ are not in the domain of function.
So, the right derivative and right continuity does not exist at $x=3$.
Now, the left-hand limit of $f\left( x \right)$ at $x=3$ is given as
$\underset{h\to 0}{\mathop{\lim }}\,f\left( 3-h \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 1+\left( 3-h \right) \right)$
$=4-0=4$
And the value of the function at $x=3$ is given as$f\left( 3 \right)=2+3=5$.
Since the left-hand limit is not equal to value of function at $x=3$, hence , it is not left continuous at $x=3$.
Hence , it is also not left differentiable at $x=3$.
Now , we will consider the point $x=-1$.
The function does not exist to the left of this point.
We know, the right-hand derivative of $f\left( x \right)$ at $x=a$ is given as ${{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( a+h \right)-f\left( a \right)}{h}$.
So, the right-hand derivative of $f\left( x \right)$ at $x=-1$ is given as
$R'=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( -1+h \right)-f\left( -1 \right)}{h}$
$=\dfrac{-(-1+h)-(-(-1))}{h}$
$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{1-h-1}{h}$
$=-1$
So , the right-hand derivative $f\left( x \right)$ exists at $x=-1$and hence ,it will also be right continuous.
Now, we will consider the point $x=0$.
Left hand limit of function $f\left( x \right)$at $x=0$ is given as
$\underset{h\to 0}{\mathop{\lim }}\,f\left( 0-h \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( -h \right)$
$=\underset{h\to 0}{\mathop{\lim }}\,-\left( -h \right)$
\begin{align} & =\underset{h\to 0}{\mathop{\lim }}\,\left( h \right) \\ & =0 \\ \end{align}
Right hand limit of $f\left( x \right)$at $x=0$ is given as
$\underset{h\to 0}{\mathop{\lim }}\,f\left( 0+h \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( h \right)$
$=\underset{h\to 0}{\mathop{\lim }}\,\left( 1-h \right)$
$=1$
Clearly, the left hand limit is not equal to the right hand limit. Hence, the function $f\left( x \right)$ is discontinuous at $x=0$ and therefore is not differentiable at $x=0$.
Now, let’s consider the point $x=1$.
Left hand limit of function $f\left( x \right)$at $x=1$ is given as
$\underset{h\to 0}{\mathop{\lim }}\,f\left( 1-h \right)$
$=\underset{h\to 0}{\mathop{\lim }}\,\left( 1-1+h \right)$
\begin{align} & =\underset{h\to 0}{\mathop{\lim }}\,\left( h \right) \\ & =0 \\ \end{align}
Right hand limit of $f\left( x \right)$at $x=1$is given as
$\underset{h\to 0}{\mathop{\lim }}\,f\left( 1+h \right)$
$=1$
Clearly, the left hand limit of $f\left( x \right)$ is not equal to the right hand limit.
Hence, the function $f\left( x \right)$ is discontinuous at $x=1$ and therefore is not differentiable at $x=1$.
Now, let’s consider the point $x=2$.
Left hand limit of function $f\left( x \right)$ at $x=2$ is given as
$\underset{h\to 0}{\mathop{\lim }}\,f\left( 2-h \right)$
$=\underset{h\to 0}{\mathop{\lim }}\,\left( 2-h \right)$
$=2$
Right hand limit of $f\left( x \right)$at $x=2$ is given as
$\underset{h\to 0}{\mathop{\lim }}\,f\left( 2+h \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 1+\left( 2+h \right) \right)$
$=1+2=3$
Clearly, the left hand limit of $f\left( x \right)$ is not equal to the right hand limit.
Hence, the function $f\left( x \right)$ is not continuous at $x=2$ and therefore is not differentiable at $x=2$.
Note: If a function is continuous at $x=a$, then it may or may not be differentiable at $x=a$. But if a function is discontinuous at $x=a$, then it is definitely not differentiable at $x=a$..