Question-34
Consider the function:
\[ \begin{equation*} f( x) =\begin{cases} \frac{|( x-1)( x-3) |}{x-3} , & x\neq 3\\ 2, & x=3 \end{cases} \end{equation*} \]
Break the piece-wise function by looking at important intervals.
We can break down the function as follows:
\[ \begin{equation*} f( x) =\begin{cases} \frac{( x-1)( x-3)}{x-3} , & x >3\\ 2, & x=3\\ \frac{-( x-1)( x-3)}{x-3} , & 1< x< 3\\ \frac{( x-1)( x-3)}{x-3} , & x\leqslant 1 \end{cases} \end{equation*} \]
This simplifies to:
\[ \begin{equation*} f( x) =\begin{cases} x-1, & x >3\\ 2, & x=3\\ 1-x, & 1< x< 3\\ x-1, & x\leqslant 1 \end{cases} \end{equation*} \]
At \(\displaystyle x=3\) we have:
\[ \begin{equation*} \begin{aligned} \lim\limits _{x\rightarrow 3^{+}} f( x) & =2\\ & \\ \lim\limits _{x\rightarrow 3^{-}} f( x) & =-2 \end{aligned} \end{equation*} \]
The limit does not exist at \(\displaystyle x=3\). Hence, \(\displaystyle f\) is not continuous at \(\displaystyle x=3\) and not differentiable at \(\displaystyle x=3\).
At \(\displaystyle x=1\) we have:
\[ \begin{gather*} \lim\limits _{x\rightarrow 1^{+}} f( x) =0\\ \\ \lim\limits _{x\rightarrow 1^{-}} \ f( x) =0 \end{gather*} \]
The limit exists at \(\displaystyle x=1\) and is equal to \(\displaystyle 0\). Since \(\displaystyle f( 1) =0\), \(\displaystyle f\) is continuous at \(\displaystyle x=1\). The LHD at \(\displaystyle x=1\) is \(\displaystyle 1\) and the RHD at \(\displaystyle x=1\) is \(\displaystyle -1\). So \(\displaystyle f\) is not differentiable at \(\displaystyle x=3\).