Question

Consider the function \[f\left( x \right)=\left| \left( x+1 \right)\left[ x \right] \right|\] for \[-1\le x\le 2\] where \[\left[ x \right]\] is the integral part of\[x\] . Then \[f\] is
(a) right continuous at \[x=-1\]
(b) not continuous at \[x=0\]
(c) continuous at \[x=1\]
(d) not left continuous at \[x=2\]

Answer 1

Hint: The given problem is related to the continuity of a function. If the value of the 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 \[x=a\] .

Complete step-by-step answer:
 The given function is \[f\left( x \right)=\left| \left( x+1 \right)\left[ x \right] \right|\] .
Now, from the domain of the function, we can see \[f\left( x \right)\] does not exist to the left of \[x=-1\] and to the right of \[x=2\] . So, the right derivative and right continuity do not exist at \[x=2\] and the left derivative and left continuity do not exist at \[x=-1\] .
We will check if the function is continuous at critical points of the function.
First, we will check the continuity of the function at \[x=-1\] .
The right-hand limit of \[f\left( x \right)\] at \[x=-1\] is given as \[\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,f\left( -1+h \right)=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\left| \left( -1+h+1 \right)\left[ -1+h \right] \right|\] .
\[=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\left| \left( h \right)\left( -1 \right) \right|\]
\[=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,h\]
\[=0\]
Also, the value of the function at \[x=-1\] is \[f\left( -1 \right)=\left| \left( -1+1 \right)\left[ -1 \right] \right|\] .
\[=0\]
The right-hand limit is equal to the value of \[f\left( x \right)\]at\[x=-1\] .
So, the function is right continuous at \[x=-1\] .
Now, we will consider the point \[x=0\] .
The left-hand limit of \[f\left( x \right)\] at \[x=0\] is given as \[\underset{h\to 0}{\mathop{\lim }}\,f\left( 0-h \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( -h \right)\] .
\[=\underset{h\to 0}{\mathop{\lim }}\,\left| \left( -h+1 \right)\left[ -h \right] \right|\]
\[\begin{align}
  & =\left| \left( 0+1 \right)\left[ -1 \right] \right| \\
 & =1 \\
\end{align}\]
Now, the right-hand limit of \[f\left( x \right)\] at \[x=0\] is given as \[\underset{h\to 0}{\mathop{\lim }}\,f\left( 0+h \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( h \right)\] .
\[=\underset{h\to 0}{\mathop{\lim }}\,\left| \left( 1+h \right)\left[ h \right] \right|\]
\[=\left| \left( 1+0 \right)\left[ 0 \right] \right|\]
\[=0\]
Now, the value of the function at \[x=0\] is given as \[f\left( 0 \right)=\left| \left( 0+1 \right)\left[ 0 \right] \right|\] .
\[=0\]
The left-hand limit is not equal to the right-hand limit of the function at \[x=0\] . So, the function is not continuous at \[x=0\] .
Now, let’s consider the point \[x=1\] .
The left-hand limit of \[f\left( x \right)\] at \[x=1\] is given by \[\underset{h\to 0}{\mathop{\lim }}\,f\left( 1-h \right)=\underset{h\to 0}{\mathop{\lim }}\,\left| \left( 1-h+1 \right)\left[ 1-h \right] \right|\] .
\[\begin{align}
  & =\left| \left( 2 \right)\left[ 0 \right] \right| \\
 & =0 \\
\end{align}\]
Now, the right-hand limit of \[f\left( x \right)\] at \[x=1\] is given by \[\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,f\left( 1+h \right)=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\left| \left( 1+h+1 \right)\left[ 1+h \right] \right|\] .
\[=\left| \left( 2 \right)\left[ 1 \right] \right|\]
\[=2\]
The left-hand limit is not equal to the right-hand limit. Hence, \[f\left( x \right)\] is not continuous at \[x=1\] .
Now, let’s consider the point \[x=2\] .
The left-hand limit of \[f\left( x \right)\] at \[x=2\] is given as \[\underset{h\to 0}{\mathop{\lim }}\,f\left( 2-h \right)=\underset{h\to 0}{\mathop{\lim }}\,\left| \left( 2-h+1 \right)\left[ 2-h \right] \right|\] .
\[\begin{align}
  & =\left| \left( 3 \right)\left( 1 \right) \right| \\
 & =3 \\
\end{align}\]
The value of the function at \[x=2\] is given as \[f\left( 2 \right)=\left| \left( 2+1 \right)\left[ 2 \right] \right|\] .
\[=\left| 3\left( 2 \right) \right|\]
\[=6\]
The left-hand limit is not equal to the value of the function at \[x=2\] .
So, we can say that the function is not left continuous at \[x=2\] .
Hence, the function is right continuous at \[x=-1\], not continuous at \[x=0\], not continuous at \[x=1\] and not left continuous at \[x=2\] .
Answer is (a), (b) and (d)

Note: While checking the continuity of the function, make sure to evaluate the left and right-hand limits carefully. The possibility of error is high in these areas.