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,

& -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{aligned} \right.$$ Now, we also know, $$\left| 1-x \right|=\left\\{ \begin{aligned} & 1-x,\text{ }x\le 1 \\\ & x-1,\text{ }x\ge 1 \\\ \end{aligned} \right.$$ So, we can rewrite $$f\left( x \right)=\left\\{ \begin{aligned} & -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{aligned} \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{aligned} & =\underset{h\to 0}{\mathop{\lim }}\,\left( h \right) \\\ & =0 \\\ \end{aligned}$$ Right hand limit of $$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{aligned} & =\underset{h\to 0}{\mathop{\lim }}\,\left( h \right) \\\ & =0 \\\ \end{aligned}$$ Right hand limit of $$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$$..