Question

If $f\left(x\right)=\{\begin{array}{cc}|x+1|& x\le 0;\\ x& x>0;\end{array}$ and
$g\left(x\right)=\{\begin{array}{cc}|x|+2& x<2;\\ -|x-2|& x\ge 2;\end{array}$ then $f\left(x\right)-g\left(x\right)$ is continuous at

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

Answer 1

The correct option is B $x=-1,x=1$

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

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

$f\left(x\right)-g\left(x\right)=\{\begin{array}{cc}-3& x\le -1;\\ 2x-1& -1<x\le 0;\\ -2& 0<x<2;\\ 2x-2& x\ge 2;\end{array}$

Therefore, $f\left(x\right)-g\left(x\right)$ is continuous everywwhere except at $x=0,x=2$