The function \(f(x)=|x-2|+x\) is

\[
f(x)=|x-2|+x
\]
First we check the continuity
At \(\mathrm{x}=0\),
\(\begin{aligned} \text { RHL } f(0+h) &=\lim _{h \rightarrow 0}|0+h-2|+(0+h) \\ &=|-2|=2 \\ \text { LHL } f(0-h) &=\lim _{h \rightarrow 0}|0-h-2|+(0-h) \\ &=|-2|=2 \end{aligned}\)
At \(\mathrm{x}=2\),
\[
\begin{aligned}
\operatorname{RHL} \mathrm{f}(2+\mathrm{h}) &=\lim _{\mathrm{h} \rightarrow 0}|2+\mathrm{h}-2|+(2+\mathrm{h}) \\
&=0+2+0=2
\end{aligned}
\]
LHL \(f(2-h)=\lim _{h \rightarrow 0}|2-h-2|+(2+h)\)
\[
=0+2-0=2
\]
and \(\quad f(0)=2, f(2)=2\)
Hence, \(f(x)\), is continuous at \(x=0,2\)
Now, we check differentiability
\[
f(x)=\left\{\begin{array}{c}
(-x+2-x), x < 0 \\
(-x+2+x), 0 \leq x \leq 2 \\
(x-2+2), 2 \leq x
\end{array}\right.
\]
\[
f(x)=\left\{\begin{array}{c}
2-2 x, x < 0 \\
2,0 \leq x \leq 2 \\
2 x-2,2 \leq x
\end{array}\right.
\]
Now,\(f^{\prime}(x)=\left\{\begin{array}{c}-2, x < 0 \\ 0,0 \leq x \leq 2 \\ 2,2 \leq x\end{array}\right.\)
Hence, \(f(x)\) is not differentiable at \(x=0,2\).