Question

Let $f\left(x\right)=x^2-2|x|$ and $g\left(x\right)=\left\{\begin{array}{cc}\text{min}\left\{f\right(t\left)\right\},& -2\le t\le x,-2\le x<0\\ \text{max}\left\{f\right(t\left)\right\},& 0\le t\le x,0\le x\le 2,\\ f\left(x\right),& x>2\end{array}\right\}$
then $g\left(x\right)$ is not differentiable at

• A. $x=0$
• B. $x=-1$
• C. $x=2$
• D. $x=1$

Answer 1

Answer: (A), (C)

The correct options are
A $x=0$
C $x=2$

$f\left(x\right)=|x{|}^2-2|x|$
$f\left(x\right)$ $\to$


$g\left(x\right)=\left\{\begin{array}{cc}f\left(x\right),& -2\le x<-1\\ f(-1),& -1\le x<0\\ f\left(0\right),& 0\le x\le 2\\ f\left(x\right),& x>2\end{array}\right\}$
$g\left(x\right)$ $\to$


$g\left(x\right)$ is not differentiable at $x=0,x=2$