Question
Examine the continuity and differentiability of function $f(x)=|x|+|x-1|$ in interval $(-1,2)$.
✓
Answer
at $x = 0 f(0)=1$
$f(0+h)=1$ and $f(0-h)=1-2(0-h)=1+ 2 h$
$\text{RHL} =\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} 1=1$
$LHL =\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0}(1+2 h)=1$
$\because f(0)= \text{RHL} = LHL =1$ hence function is continuous at $x=0$.
$R f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0}\left(\frac{1-1}{h}\right)=0$
$L f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0} \frac{(1+2 h-1)}{-h}=-2$
$\because R f^{\prime}(0) \neq L f^{\prime}(0)$, hence function is not differentiable at $x=0$
again at $x=1$
$f(1)=2-1=1$
$f(1+h)=2(1+h)-1=1+2 h \text { and } f(1-h)=1$
$\text { RHL }=\lim _{h \rightarrow 0} f(1+2 h)=\lim _{h \rightarrow 0}(1+2 h)=1$
$\text { LHL }=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}(1)=1$
$\because f(1)= \text{RHL} = \text{LHL},$ hence function is continuous at $x=1$.
$R f^{\prime}(1) =\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$
$ =\lim _{h \rightarrow 0} \frac{(1+2 h-1)}{h}=2$
$L f^{\prime}(1) =\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}=\lim _{h \rightarrow 0} \frac{1-1}{-h}=0$
$R f^{\prime}(1) \neq L f^{\prime}(1)$, hence function is not differentiable at $x=1$.
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