Question

The number of extremum point(s) for $f\left(x\right)=3|x|+|x+1|-||x+1|-3|x||$ is

Answer 1

Given : $f\left(x\right)=3|x|+|x+1|-||x+1|-3|x||$
We know, $f\left(x\right)+g\left(x\right)-|f\left(x\right)-g\left(x\right)|=min\left\{2f\right(x),2g(x\left)\right\}$
Here, $f\left(x\right)=|x+1|,g\left(x\right)=3|x|$
where:
$2x+2=0$ gives $x=-1,$
$6x=0$ gives $x=0,$
$2x+2=6x$ gives $x=\frac{1}{2}$ and
$2x+2=-6x$ gives $x=-\frac{1}{4}$
Now, plotting the graph of $y=min\{|2x+2|,6|x|\}:$
Clearly, it has $3$ extremum points.