Question

Show that $f\left(x\right)=|x-2|+|x-3|$ is not differentiable at $x=2$.

Answer 1

Given,

$f\left(x\right)=|x-2|+|x-3|$

Now,

$f\left(2\right)=|2-2|+|2-3|=0+1=1$

Now,

$LHD=li{m}_{x\to 2^{-}}\frac{f\left(x\right)-f\left(2\right)}{x-2}$

$=li{m}_{x\to 2^{-}}\frac{|x-2|+|x-3|-1}{x-2}$

$=li{m}_{h\to 0}\frac{|2-h-2|+|2-h-3|-1}{2-h-2}$

$=li{m}_{h\to 0}\frac{|-h|+|-h-1|-1}{-h}$

$=li{m}_{h\to 0}\frac{h+1+h-1}{-h}=li{m}_{h\to 0}\frac{2h}{-h}=-2$

And, $RHD=li{m}_{x\to 2^{+}}\frac{f\left(x\right)-f\left(2\right)}{x-2}$

$=li{m}_{x\to 2^{+}}\frac{|x-2|+|x-3|-1}{x-2}$

$=li{m}_{h\to 0}\frac{|2+h-2|+|2+h-3|-1}{2+h-2}$

$=li{m}_{h\to 0}\frac{|h|+|h-1|-1}{h}$

$=li{m}_{h\to 0}\frac{h+1-h-1}{h}=li{m}_{h\to 0}\frac{0}{h}=0$

$LHD\ne RHD$

Hence, f(x) is not differentiable at $x=2$.