Question

The function $f\left(x\right)=min\{|x|,|x-2|,2-|x-1|\}$ is non-differentiable at $x$ equals

• A. $-0.5$
• B. $0$
• C. $2$
• D. $2.5$

Answer 1

Answer: (A), (B), (C), (D)

$2.5$

$f\left(x\right)=min\{|x|,|x-2|,2-|x-1|\}$

$|x|=\{\begin{array}{cc}x,& x\ge 0\\ -x,& x<0\end{array}$

$|x-2|=\{\begin{array}{cc}x-2,& x\ge 2\\ 2-x,& x<2\end{array}$

$2-|x-1|=\{\begin{array}{cc}3-x,& x\ge 1\\ 1+x,& x<1\end{array}$

$\Rightarrow f\left(x\right)=\{\begin{array}{cc}1+x,& -1\le x<-\frac{1}{2}\\ -x,& -\frac{1}{2}\le x<0\\ x,& 0\le x<1\\ 2-x,& 1\le x<2\\ x-2,& 2\le x<\frac{5}{2}\\ 3-x& \frac{5}{2}\le x\end{array}$


We know that, derivative at sharp point does not exist.
From graph, we see that there are $5$ sharp point i.e., $-\frac{1}{2},0,1,2,\frac{5}{2}.$