# 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 \[x=0,1,2,3\]

Answer

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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 \[x=2\].

Answer is (a), (b), (d)

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 \[x=a\]..

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 \[x=2\].

Answer is (a), (b), (d)

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 \[x=a\]..

Last updated date: 26th Sep 2023

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