# Examine the differentialibilty of the function f defined by

Question:

Examine the differentialibilty of the function f defined by

$f(x)=\left\{\begin{array}{l}2 x+3 \text { if }-3 \leq x \leq-2 \\ x+1 \quad \text { if }-2 \leq x<0 \\ x+2 \quad \text { if } 0 \leq x \leq 1\end{array}\right.$

Solution:

$f(x)=\left\{\begin{array}{cc}2 x+3 & \text { if }-3 \leq x \leq-2 \\ x+1 & \text { if }-2 \leq x<0 \\ x+2 & \text { if } 0 \leq x \leq 1\end{array}\right.$

$\Rightarrow f^{\prime}(x)=\left\{\begin{array}{cc}2 & \text { if }-3 \leq x \leq-2 \\ 1 & \text { if }-2 \leq x<0 \\ 1 & \text { if } 0 \leq x \leq 1\end{array}\right.$

Now,

$\mathrm{LHL}=\lim _{x \rightarrow-2^{-}} f^{\prime}(x)=\lim _{x \rightarrow-2^{-}} 2=2$

$\mathrm{RHL}=\lim _{x \rightarrow-2^{+}} f^{\prime}(x)=\lim _{x \rightarrow-2^{+}} 1=1$

Since, at $x=-2, \mathrm{LHL} \neq \mathrm{RHL}$

Hence, $f(x)$ is not differentiable at $x=-2$

Again,

$\mathrm{LHL}=\lim _{x \rightarrow 0^{-}} f^{\prime}(x)=\lim _{x \rightarrow 0^{-}} 1=1$

$\mathrm{RHL}=\lim _{x \rightarrow 0^{+}} f^{\prime}(x)=\lim _{x \rightarrow 0^{+}} 1=1$

Since, at $x=0, \mathrm{LHL}=\mathrm{RHL}$

Hence, $f(x)$ is differentiable at $x=0$