Question

Show that $f\left(x\right)=|x-5|$ is continuous but not differentiable at x=5

Answer 1

We have, $f\left(x\right)=|x-5|\therefore f\left(x\right)=\{\begin{array}{cc}-(x-5),& ifx<5\\ x-5& ifx\ge 5\end{array}$

For continuity at x=5,
LHL=$\underset{x\to 5^{-}}{lim}(-x+5)=\underset{h\to 0}{lim}[-(5-h)+5]=\underset{h\to 0}{lim}h=0$
RHL =$\underset{x\to 5^{+}}{lim}(x-5)=\underset{h\to 0}{lim}(5+h-5)=\underset{h\to 0}{lim}h=0\therefore f\left(5\right)=5-5=0$
$\Rightarrow$ LHL=RHL =f(5)
Hence, f(x) is continuous at x=5
Now,
$L{f}^{\prime }\left(5\right)=\underset{x\to 5^{-}}{lim}\frac{f\left(x\right)-f\left(5\right)}{x-5}=\underset{x\to 5^{-}}{lim}\frac{-x+5-0}{x-5}=-1R{f}^{\prime }\left(5\right)=\underset{x\to 5^{+}}{lim}\frac{f\left(x\right)-f\left(5\right)}{x-5}=\underset{x\to 5^{+}}{lim}\frac{x-5-0}{x-5}=1\therefore L{f}^{\prime }\left(5\right)\ne R{f}^{\prime }\left(5\right)$
So, f(x)=|x-5| is not differentiable at x=5