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 $\underset{h\to 0}{\mathop{\lim }}\,f\left( 0-h \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( -h \right)$ .
$=\underset{h\to 0}{\mathop{\lim }}\,\left| \left( -h+1 \right)\left[ -h \right] \right|$

& =\left| \left( 0+1 \right)\left[ -1 \right] \right| \\\ & =1 \\\ \end{aligned}$$ Now, the right-hand limit of $$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{aligned} & =\left| \left( 2 \right)\left[ 0 \right] \right| \\\ & =0 \\\ \end{aligned}$$ Now, the right-hand limit of $$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{aligned} & =\left| \left( 3 \right)\left( 1 \right) \right| \\\ & =3 \\\ \end{aligned}$$ The value of the function at $$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.