Question

Discuss the continuity and differentiability of the $f\left(x\right)=\left|x\right|+\left|x-1\right|\mathrm{in}\mathrm{the}\mathrm{interval}\left(-1,2\right)$

Answer 1

$$\begin{gathered} \mathrm{Given}:f\left(x\right)=\left|x\right|+\left|x-1\right| \\ \left|x\right|=-x\mathrm{for}x<0 \\ \left|x\right|=x\mathrm{for}x>0 \\ \left|x-1\right|=-\left(x-1\right)=-x+1\mathrm{for}x-1<0\mathrm{or}x<1 \\ \left|x-1\right|=x-1\mathrm{for}x-1>0\mathrm{or}x>1 \\ \mathrm{Now}, \\ f\left(x\right)=-x-x+1=-2x+1x\in \left(-1,0\right) \\ \mathrm{or} \\ f\left(x\right)=x-x+1=1x\in \left(0,1\right) \\ \mathrm{or} \\ f\left(x\right)=x+x-1=2x-1x\in \left(1,2\right) \end{gathered}$$
$$\begin{gathered} \mathrm{Now}, \\ \mathrm{LHL}=\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{-}}{\lim }-2x+1=0+1=1 \\ \mathrm{RHL}=\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }1=1 \\ \mathrm{Hence},\mathrm{at}x=0,\mathrm{LHL}=\mathrm{RHL} \\ \mathrm{Again}, \\ \mathrm{LHL}=\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{-}}{\lim }1=1 \\ \mathrm{RHL}=\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }2x-1=2-1=1 \\ \mathrm{Hence},\mathrm{at}x=1,\mathrm{LHL}=\mathrm{RHL} \end{gathered}$$
Now,
$$\begin{gathered} f\left(x\right)=-x-x+1=-2x+1x\in \left(-1,0\right) \\ \Rightarrow f\text{'}\left(x\right)=-2x\in \left(-1,0\right) \\ \mathrm{or} \\ f\left(x\right)=x-x+1=1x\in \left(0,1\right) \\ \Rightarrow f\text{'}\left(x\right)=0x\in \left(0,1\right) \\ \mathrm{or} \\ f\left(x\right)=x+x-1=2x-1x\in \left(1,2\right) \\ \Rightarrow f\text{'}\left(x\right)=2x\in \left(1,2\right) \end{gathered}$$
$$\begin{gathered} \mathrm{Now}, \\ \mathrm{LHL}=\underset{x\to 0^{-}}{\lim }f\text{'}\left(x\right)=\underset{x\to 0^{-}}{\lim }-2=-2 \\ \mathrm{RHL}=\underset{x\to 0^{+}}{\lim }f\text{'}\left(x\right)=\underset{x\to 0^{+}}{\lim }0=0 \\ \mathrm{Since},\mathrm{at}x=0,\mathrm{LHL}\ne \mathrm{RHL} \\ \mathrm{Hence},f\left(x\right)\mathrm{is}\mathrm{not}\mathrm{differentiable}\mathrm{a}tx=0 \\ \mathrm{Again}, \\ \mathrm{LHL}=\underset{x\to 1^{-}}{\lim }f\text{'}\left(x\right)=\underset{x\to 1^{-}}{\lim }0=0 \\ \mathrm{RHL}=\underset{x\to 1^{+}}{\lim }f\text{'}\left(x\right)=\underset{x\to 1^{+}}{\lim }2=2 \\ \mathrm{Since},\mathrm{at}x=1,\mathrm{LHL}\ne \mathrm{RHL} \\ \mathrm{Hence},f\left(x\right)\mathrm{is}\mathrm{not}\mathrm{differentiable}\mathrm{a}tx=1 \end{gathered}$$