question_answer

Show that \[f(x)=\,\,|x-2|+|x-3|\] is not differentiable at x = 2.

Answer:

Given, \[f(x)=\,\,|x-2|+|x-3|\]   Now,     \[f(2)=\,\,|2-2|+|2-3|\,\,=0+1=1\] \[\therefore \]      \[LHD=\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\frac{f(x)-f(2)}{x-2}\]             \[=\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\frac{|x-2|+|x-3|-1}{x-2}\]             \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{|2-h-2|+|2-h-3|-1}{2-h-2}\] [put \[x=2-h;\] when \[x\to 2,\] then \[h\to 0\]] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{h+1+h-1}{-h}=\underset{h\to \,\,0}{\mathop{\lim }}\,\frac{2h}{-h}=-\,2\] and       \[RHD=\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\frac{f(x)-f(2)}{x-2}\] \[=\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\frac{f|x\,-2|+|x\,-3|-\,1}{x-2}\]             \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{f|2+h-2|+|2+h-3|-\,1}{2+h-2}\] [put \[x=2\,+h;\] when \[x\to {{2}^{+}},\] then \[h\to 0\]] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{|h|+|h\,-1|-\,1}{h}\]             \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{h+1-h\,-1}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{0}{h}=0\] \[\because \]       \[LHD\ne RHD\] Hence, f(x) is not differentiable at x = 2.   Hence proved.