Question

Let $g:[-2,2]\to R$ be defined as $g\left(x\right)=|x+1|(|x|+|1-x|).$ Then

• A. $g\left(x\right)$ is continuous for all $x\in [-2,2].$
• B. $g\left(x\right)$ is not differentiable at three points in $[-2,2].$
• C. $g\left(x\right)$ attains the least value equal to $0$ for $x\in [-2,2].$
• D. $g\left(x\right)$ attains the greatest value equal to $9$ for $x\in [-2,2].$

Answer 1

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

$g\left(x\right)$ attains the greatest value equal to $9$ for $x\in [-2,2].$

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


From the graph, it is clear that $g\left(x\right)$ is continuous for all $x\in [-2,2]$.
$g\left(x\right)$ is not differentiable at $x=-1,0,1$ i.e., three points in $[-2,2]$.
$g\left(x\right)$ attains the least value equal to $0$ at $x=-1.$
$g\left(x\right)$ attains the greatest value equal to $9$ at $x=2.$