Question

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 1

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 \[x=a\]..