Question

Discuss the continuity and differentiability of the function
f(x) = |x| + |x - 1| in the interval (-1, 2).

Answer 1

$f\left(x\right)=\left|x\right|+|x-1|$ in $(-1,2)$

$\Rightarrow f\left(x\right)=\{-2x+1-1<x\le 0\}$

$\{10<x<1\}$

$\{2x-11\le x<2\}$

at $x=0$

$\Rightarrow f\left(x\right)=0+10-11=1$

$\Rightarrow {lim}_{x\to 0^{+}}-2x+1=1$

${lim}_{x\to 0^{-}}1=1$

$\therefore f\left(x\right)$ is continuous at $x=0$

$\Rightarrow f\left(1\right)=1+|1-1|=1+0=1$

$\Rightarrow {lim}_{x\to 1^{-}}f\left(x\right)={lim}_{x\to 1^{-}}2x-1=2\times 1-1=1$

$\Rightarrow {lim}_{x\to 1^{+}}f\left(x\right)=1$

$f\left(x\right)$ is also cont. at $x=1$

Hence, solved.