The function $f\left(x\right)$ defined by $f\left(x\right)=\left\{\begin{array}{l}\begin{array}{ll}{\log }_{(4x-3)}(x^2-2x+5),& \frac{3}{4}<x<1\text{and}x>1\end{array}\\ 4 , if x=1\end{array}\right.\begin{array}{} \end{array}$

The given function $f\left(x\right)=\left\{\begin{array}{l}\begin{array}{ll}{\log }_{(4x-3)}(x^2-2x+5),& \frac{3}{4}<x<1\text{and}x>1\end{array}\\ 4 , if x=1\end{array}\right.\begin{array}{} \end{array}$

$$\begin{gathered} f\left(1^{+}\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right) \\ =\underset{x\to 1^{+}}{\lim }{\log }_{(4x-3)}(x^2-2x+5) \\ = \underset{x\to 1^{+}}{\lim }\frac{\log (x^2-2x+5)}{\log (4x-3)} \\ =\underset{h\to 0}{\lim }\frac{\log ({(1+h)}^2-2(1+h)+5)}{\log \left(4\right(1+h)-3)} \\ =\underset{h\to 0^{+}}{\lim }\frac{\log (4+h^2)}{{\displaystyle \frac{\log (1+4h)}{4h}}\times 4h} \\ =\frac{\log 4}{\left(1\right)4\left(0\right)} =+\infty \\ Similarly f\text{'}\left(1^{-}\right)=-\infty \\ So,f\left(x\right) is discontinuous at x=1 \sin ce neither f\left(1^{+}\right) nor f\left(1^{-}\right) exists. \end{gathered}$$