Question

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

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Solution

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

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