Discuss the continuity and differentiability

Solution:

Given : $f(x)=|x|+|x-1|$

$|x|=-x$ for $x<0$

$|x|=x$ for $x>0$

$|x-1|=-(x-1)=-x+1$ for $x-1<0$ or $x<1$

$|x-1|=x-1$ for $x-1>0$ or $x>1$

Now,

$f(x)=-x-x+1=-2 x+1 \quad x \in(-1,0)$

Or

$f(x)=x-x+1=1 \quad x \in(0,1)$

Or

$f(x)=x+x-1=2 x-1 \quad x \in(1,2)$

Now,

$\mathrm{LHL}=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}-2 x+1=0+1=1$

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

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

Again,

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

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

Hence, at $x=1, \mathrm{LHL}=\mathrm{RHL}$

Now,

$f(x)=-x-x+1=-2 x+1 \quad x \in(-1,0)$

$\Rightarrow f^{\prime}(x)=-2 \quad x \in(-1,0)$

or

$f(x)=x-x+1=1 \quad x \in(0,1)$

$\Rightarrow f^{\prime}(x)=0 \quad x \in(0,1)$

or

$f(x)=x+x-1=2 x-1 \quad x \in(1,2)$

$\Rightarrow f^{\prime}(x)=2 \quad x \in(1,2)$

Now,

$\mathrm{LHL}=\lim _{x \rightarrow 0^{-}} f^{\prime}(x)=\lim _{x \rightarrow 0^{-}}-2=-2$ $\mathrm{RHL}=\lim _{x \rightarrow 0^{+}} f^{\prime}(x)=\lim _{x \rightarrow 0^{+}} 0=0$

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

Hence, $f(x)$ is not differentiable at $x=0$

Again,

$\mathrm{LHL}=\lim _{x \rightarrow 1^{-}} f^{\prime}(x)=\lim _{x \rightarrow 1^{-}} 0=0$ $\mathrm{RHL}=\lim _{x \rightarrow 1^{+}} f^{\prime}(x)=\lim _{x \rightarrow 1^{+}} 2=2$

Since, at $x=1, \mathrm{LHL} \neq \mathrm{RHL}$

Hence, $f(x)$ is not differentiable at $x=1$