The function $\text{⨍ (x)=}\left\{\begin{array}{l}\left|\mathrm{x}-3\right|,\mathrm{x}\ge 1\\ \frac{{\mathrm{x}}^2}{4}\text{-}\frac{3\mathrm{x}}{2}\text{+}\frac{13}{4}\text{, x<1}\end{array}\right.$ is

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\left|\mathrm{x}-3\right| ; \mathrm{x}\ge 1\\ \frac{{\mathrm{x}}^2}{4}-\frac{3\mathrm{x}}{2}+\frac{13}{4} ; \mathrm{x}<1\end{array}\right. \\ =\left\{\begin{array}{l}\frac{{\mathrm{x}}^2}{4}-\frac{3\mathrm{x}}{2}+\frac{13}{4} ; \mathrm{x}<1\\ 3-\mathrm{x} ; 1\le \mathrm{x}<3 \\ \mathrm{x}-3 ; \mathrm{x}\ge 3\end{array}\right. \end{gathered}$$

$$\begin{gathered} \mathrm{f}\left(1^{-}\right)=\mathrm{f}\left(1^{+}\right)=\mathrm{f}\left(1\right)=2 \\ \mathrm{f}\left(3^{-}\right)=\mathrm{f}\left(3^{+}\right)=\mathrm{f}\left(3\right)=0 \\ \therefore \mathrm{f} \mathrm{is} \mathrm{continuous} \mathrm{function} \end{gathered}$$

If x<1, then ${\mathrm{f}}^1\left(\mathrm{x}\right)=\frac{\mathrm{x}}{2}-\frac{3}{2}$

$$\begin{gathered} \Rightarrow {\mathrm{f}}^1\left(1^{-}\right)=- \\ {\mathrm{f}}^1\left(1^{+}\right)=-1\Rightarrow \mathrm{f} \mathrm{is} \mathrm{differentiable} \mathrm{at} \mathrm{x}=1 \\ {\mathrm{f}}^1\left(3^{-}\right)=-1, {\mathrm{f}}^1\left(3^{+}\right)=1 \\ \Rightarrow \mathrm{f} \mathrm{is} \mathrm{not} \mathrm{differentiable} \mathrm{at} \mathrm{x}=3 \end{gathered}$$