Question

Show that the function $f\left(x\right)=|x-1|+|x+1|$, for all $x\in R$, is not differentiable at the points $x=-1$ and $x=1$.

Answer 1

From the definition of absloute value we have:

$f\left(x\right)=\{\begin{array}{c}-x-1+1-x,x<-1\\ x+1+1-x,-1\le x<1\\ x-1+x+1,x\ge 1\end{array}=\{\begin{array}{c}-2x,x<-1\\ 2,-1\le x<1\\ 2x,x\ge 1\end{array}\Rightarrow f^{{}^{\prime }}\left(x\right)=\{\begin{array}{c}-2,x<-1\\ 0,-1\le x<1\\ 2,x\ge 1\end{array}\Rightarrow f^{{}^{\prime }}\left({-1}^{-}\right)=-2,f^{{}^{\prime }}\left({-1}^{+}\right)=0\Rightarrow f^{{}^{\prime }}\left(1^{-}\right)=0,f^{{}^{\prime }}\left(1^{+}\right)=2$

The above derivative calculations can be done from the first principle also.

Since the left hand and right hand derivatives are not equal at the points $1$ and $-1,$ the function is not differentiable at those points.