Question

Let $f\left(x\right)=min\{|x-1|,|x+1|,1\}$. Find the number of points where it is not differentiable.

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

Answer: (C)

$5$

$f\left(x\right)=min\{|x-1|,|x+1|,1\}$.
$\therefore f\left(x\right)=\{\begin{array}{c}1;-\infty <x\le -2\\ -x-1;-2<x\le -1\\ x+1;-1<x\le 0\\ 1-x;0<x\le 1\\ x-1;1<x\le 2\\ 1;2<x<\infty \end{array}$
$f^{\prime }(-2^{-})=0$ while $f^{\prime }(-2^{+})=-1$
$f^{\prime }(-1^{-})=-1$ while $f^{\prime }(-1^{+})=1$
$f^{\prime }\left(0^{-}\right)=1$ while $f^{\prime }\left(0^{+}\right)=-1$
$f^{\prime }\left(1^{-}\right)=-1$ while $f^{\prime }\left(1^{+}\right)=1$
$f^{\prime }\left(2^{-}\right)=1$ while $f^{\prime }\left(2^{+}\right)=0$
Hence we have 5 non differentiable points.