$\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{ll}|\mathrm{x}-4|& \mathrm{for} \mathrm{x}\ge 1\\ \frac{{\mathrm{x}}^3}{2}-{\mathrm{x}}^2+3\mathrm{x}+\frac{1}{2}& \mathrm{for} \mathrm{x}<1\end{array}, \mathrm{then}\right.$

Since $\mathrm{g}\left(\mathrm{x}\right)=\left|\mathrm{x}\right|$ is a continuous function and 

 ${\lim }_{\mathrm{x}\to 1+} f\left(\mathrm{x}\right)=3={\lim }_{\mathrm{x}\to 1-} \mathrm{f}\left(\mathrm{x}\right)$ , so f is continuous function. 

In particular f is continuous at $\mathrm{x}=1$ and $\mathrm{x}=4$) 

f is clearly not differentiable at $\mathrm{x}=4)$ Since $\mathrm{g}\left(\mathrm{x}\right)=\left|\mathrm{x}\right|$  is not differentiable at  x = 0. Now 

$\begin{array}{rl} & {\mathrm{f}}^{\mathrm{\text{'}}}(1+)\underset{\mathrm{h}\to 0+}{\lim } \frac{\mathrm{f}(1+\mathrm{h})-\mathrm{f}\left(1\right)}{\mathrm{h}}=\underset{\mathrm{h}\to 0+}{\lim } \frac{|-3+\mathrm{h}|-3}{\mathrm{h}}=\underset{\mathrm{h}\to 0+}{\lim } \frac{3-h-3}{\mathrm{h}}-1\\ & {\mathrm{f}}^{\mathrm{\text{'}}}(1-)\underset{\mathrm{h}\to 0-}{\lim } \frac{(1/2)(1+\mathrm{h}{)}^3-(1+\mathrm{h}{)}^2+3(1+\mathrm{h})+(1/2)-3}{\mathrm{h}}\\ & \mathrm{ }=\underset{\mathrm{h}\to 0-}{\lim } \frac{(1/2)\left({\mathrm{h}}^3+3{\mathrm{h}}^2+3\mathrm{h}\right)-\left({\mathrm{h}}^2+2\mathrm{h}\right)+3\mathrm{h}}{\mathrm{h}}=\frac{5}{2}\end{array}$