A function$f \left(x\right)$ is defined in the interval$[1, 4)$ as, follows:

$f\left(x\right)=\left\{\begin{array}{l}{\log }_e\left[x\right],\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\le x<3\\ \left|{\log }_{e}x\right|,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\le x<4\end{array}\right.$Then, the curve$y = f \left(x\right)$

 

We have,

$\begin{array}{r}f\left(x\right)=\left\{\begin{array}{ll}{\log }_e\left[x\right],& 1\le x<3\\ \left|{\log }_{e}x\right|,& 3\le x<4\end{array}\right.\\ \Rightarrow f\left(x\right)=\left\{\begin{array}{ll}0& 1\le x<2\\ {\log }_{e}2,& 2\le x<3\\ \left|{\log }_{e}x\right|,& 3\le x<4\end{array}\right.\end{array}$

Clearly, $f \left(x\right)$ is everywhere continuous and differentiable except possibly at $x = 2, 3.$

We observe that

$\underset{x\to 2^{-}}{lim} f\left(x\right)=\underset{x\to 2^{-}}{lim} 0=0$

and $\underset{x\to 2^{+}}{lim} f\left(x\right)=\underset{x\to 2^{+}}{lim} {\log }_{e}2={\log }_{e}2$

Clearly, $f \left(x\right)$  is not continuous at $x = 2.$

It can be easily seen that $f \left(x\right)$ is not continuous at $x = 3$.

 Hence, $f \left(x\right)$ is neither continuous nor differentiable at $x = 2, 3$. 

Thus, the curve $y = f \left(x\right)$ is broken at two points and it does not 

have a definite tangent at these points.