If \(\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}\mathrm{x}, & \text { for } \mathrm{x} \leq 0 \ 0, & \text { for } \mathrm{x}>0\end{array}\right.\), then the function \(\mathrm{f}(\mathrm{x})\) at \(\mathrm{x}=0\) is

\(\begin{aligned}
& \mathrm{f}(\mathrm{x})=\mathrm{x}, \quad \text { if } \mathrm{x} \leq 0 \\
& =0, \quad \text { if } x>0 \\
& \therefore \quad \lim _{x \rightarrow 0^{-}} f(x)=\lim _{n \rightarrow 0} x=0 \text { and } \lim _{x \rightarrow 0^{+}} f(x)=0 \\
& \mathrm{f}(0)=0 \\
& \mathrm{f}^{\prime}(\mathrm{x})=1, \quad \text { if } \mathrm{x} \leq 0 \\
& =0, \quad \text { if } x>0 \\
&
\end{aligned}\)
Thus \(f(x)\) is continuous at \(x=0\)
\(\begin{aligned}
\mathrm{f}^{\prime}(\mathrm{x}) & =1, & & \text { if } \mathrm{x} \leq 0 \\
& =0, & & \text { if } \mathrm{x}>0
\end{aligned}\)
Thus \(\mathrm{f}(\mathrm{x})\) is not differentiable at \(\mathrm{x}=0\).