The function  $\mathrm{f}\left(\mathrm{x}\right)=2\left|\mathrm{x}\right|+|\mathrm{x}+2|-\left|\right|\mathrm{x}+2|-2|\mathrm{x}\parallel$ has a local minimum (or) a local maximum at x=

$y=2\left|x\right|+|x+2|-\parallel x+2|-2|x\parallel$
$\text{ Case}\left(\text{1}\right)\text{: }\mathrm{x}<-2$
$\begin{array}{r} y=-2x-x-2-|-x-2+2x|\\ y=-2x-4\dots \dots \dots \dots \left(1\right)\end{array}$
$\text{ Case}\left(\text{2}\right)\text{: }-2<x<0$
$y=-x+2-|3x+2|$
$\text{ So, for }x<\frac{-2}{3}$
$y=2x+4$
$\text{ And for }x\in \left[\frac{-2}{3},0\right]$
$y=-4x$
$\text{ Case 3: }x>0$
$y=3x+2-|x-2|$
$\text{ For }0<x<2\text{, }$
$y=4x$
$\text{ And for }\mathrm{x}>2$
$y=2x+4\dots \dots \dots \dots \left(5\right)$
$$\begin{gathered} \text{ After drawing this graph, we can conclude }\mathrm{x}=-2,0\text{ are} \\ \text{ point of minimum and } \end{gathered}$$$\mathrm{x}=\frac{-2}{3}\text{ is a point of maxima }$