Question
Examine the differentialiblilty of the function f defined by $\text{f(x)}=\begin{cases}2\text{x}+3 & \text{if}-3\leq\text{x}\leq-2\\\text{x}+1 & \text{if} -2\leq\text{x}\leq0\\\text{x}+2&\text{if}\ 0\leq\text{x}\leq1\end{cases}$

Answer

$\text{f(x)}=\begin{cases}2\text{x}+3 & \text{if}-3\leq\text{x}\leq-2\\\text{x}+1 & \text{if} -2\leq\text{x}\leq0\\\text{x}+2&\text{if}\ 0\leq\text{x}\leq1\end{cases}$
$\text{f}'(\text{x})=\begin{cases}2 & \text{if}-3\leq\text{x}\leq-2,\\1 & \text{if} -2\leq\text{x}\leq0\\1&\text{if}\ 0\leq\text{x}\leq1\end{cases}$
Now,
$\text{LHL}=\lim_\limits{\text{x}\rightarrow2^{-}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow2^{-}}2=2$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow2^{+}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow2^{+}}1=1$
Since, at $\text{x}=-2,\text{LHL}\neq\text{RHL}$
Hence, f(x) is not differentiable at x = -2
Again,
$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^{-}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow0^{-}}1=1$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow0^{+}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow0^{+}}1=1$
Since, at $\text{x}=0,$
$\text{LHL}=\text{RHL}$
Hence, f(x) is differentiable at x = 0

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