Question

Prove that the function f is given by f(x) = |x-1|, x $ϵ$R is not differentiable at x = 1.

Answer 1

Given, f(x) = |x - 1| = $\{\begin{array}{c}x-1,ifx-1\ge 0\\ -(x-1),ifx-1<0\end{array}$

We have to check the differentiability at x = 1

Here, f(1) = 1 - 1 =0

L$L{f}^{\prime }\left(1\right)=\underset{h\to 0^{-}}{lim}\frac{f(1-h)-f\left(1\right)}{-h}=\underset{h\to 0^{-}}{lim}\frac{1-(1-h)-0}{-h}=\underset{h\to 0^{-}}{lim}\frac{1-(1-h)-0}{-h}=\underset{h\to 0^{-}}{lim}\frac{+h}{-h}=-1$

[$\because whenx<1\Rightarrow f\left(x\right)=1-x$]

and $R{f}^{\prime }\left(1\right)=\underset{h\to 0^{+}}{lim}\frac{f(1+h)-f\left(1\right)}{-h}=\underset{h\to 0^{+}}{lim}\frac{(1+h)-f\left(1\right)}{h}=\underset{h\to 0^{+}}{lim}\frac{(1+h)-1-0}{h}=\underset{h\to 0^{+}}{lim}\frac{h}{h}=1$

[$\because whenx>1\Rightarrow f\left(x\right)=x-1$]

$\therefore$Lf ' (x) $\ne$Rf ' (x), Hence, f(x) is not differentiable at x = 1.