The function f(x) = 2|x| + |x + 2| - ||x + 2| - 2|x|| has a local...

f\left(x\right) = 2\left|x\right| + \left|x + 2\right| - \left|\left|x + 2\right| - 2\left|x\right|\right|$</p><p><u>CASE 1:</u></p><p>For$x \le - 2,$</p><p>\left|x\right| = - x$ and \left|x + 2\right| = - \left(x + 2\right)$( ∵ x, x + 2 < 0 )$

$⇒ f ( x ) = - 3x - 2 - \left|x - 2\right|$

$= - 3x - 2 + ( x - 2 ) ( ∵ x - 2 < 0 )$

$= - 2x - 4 . . . ( 1 )$

CASE 2:

For $- 2 < x \le 0,$

\left|x\right| = - x\left(∵ x 0 )$</p><p>$⇒ f ( x ) = - 2x + ( x + 2 ) - \left|( x + 2 ) + 2x\right| = - x + 2 - \left|3x + 2\right|$</p><p>$= \left\{$$\begin{array}{ll}- x + 2 + ( 3x + 2 ) , - 2 < x \le \dfrac{- 2}{3} \\ - x + 2 - ( 3x + 2 ) , \dfrac{- 2}{3} < x \le 0\end{array}$$\right\}$</p><p>$= \left\{$$\begin{array}{ll}2x + 4, - 2 < x \le \dfrac{- 2}{3} \\ - 4x, \dfrac{- 2}{3} < x \le 0\end{array}$$\right\} . . . ( 2 )$</p><p><u>CASE 3:</u></p><p>For$x > 0,$</p><p>\left|x\right| = x \text{ and } \left|x + 2\right| = ( x + 2 )$

$⇒ f ( x ) = 2x + ( x + 2 ) - \left|( x + 2 ) - 2x\right| = 3x + 2 - \left|- x + 2\right|$

= \left\{$\begin{array}{ll}3x + 2 - ( - x + 2 ) , 0 < x \le 2 \\ 3x + 2 + ( - x + 2 ) , 2 < x\end{array}$\right\}

= \left\{$\begin{array}{ll}4x, 0 < x \le 2 \\ 2x + 4, 2 < x\end{array}$\right\} . . . ( 3 )

From $( 1 ) , ( 2 ) \text{ and } ( 3 ) ,$ we get
\text{f}\text{(}\text{x}\text{)=}\left\{\begin{cases}- 2x - 4, & x \le - 2 \\ 2x + 4, & - 2 2\end{cases}\right\}

From graph, $y = f ( x )$ has local minima at $x = - 2, 0$ and local maxima at $x = \dfrac{- 2}{3} .$