The function $f(x) = 2|x| + |x + 2| - ||x + 2| - 2|x||$ has a local minimum or a local maximum at $x=$

$f(x)=2|x|+|x+2|-\big|\,|x+2|-2|x|\,\big|$ $\text{To find: points of local minimum / maximum}$ $\text{Options: } -2,\ -\dfrac{2}{3},\ 2,\ \dfrac{2}{3}$ $1.\ \text{simplify } f \text{ first}$ $2.\ \text{find the modulus break points}$ $3.\ \text{sign changes of } f' \Rightarrow \text{min / max}$ $a+b-|a-b|=2\min(a,b)$ $f(x)=2\min\big(|x+2|,\ 2|x|\big)$ $|x+2|=2|x| \Rightarrow (x+2)^2=4x^2$ $3x^2-4x-4=0$ $x=2 \quad \text{or} \quad x=-\dfrac{2}{3}$ $x\ge 2:\ f=2(x+2)$ $0\le x\le 2:\ f=4x; \quad -\dfrac{2}{3}\le x\le 0:\ f=-4x$ $x\le -\dfrac{2}{3}:\ f=2(x+2)\ (x\ge -2); \quad f=-2(x+2)\ (x\le -2)$ $f \text{ is piecewise linear with corners at } x=-2,\ -\tfrac23,\ 0,\ 2$ $x\ge 2:\ f=2x+4,\ f'=2>0$ $0\le x\le 2:\ f=4x,\ f'=4>0$ $-\dfrac{2}{3}\le x\le 0:\ f=-4x,\ f'=-4<0$ $-2\le x\le -\dfrac{2}{3}:\ f=2x+4,\ f'=2>0$ $x\le -2:\ f=-2x-4,\ f'=-2<0$ $f'<0:\ (-\infty,-2)\cup\left(-\dfrac23,0\right)$ $f'>0:\ \left(-2,-\dfrac23\right)\cup(0,\infty)$ $x=-2:\ f':-\to+ \Rightarrow \text{local min}$ $x=-\dfrac23:\ f':+\to- \Rightarrow \text{local max}$ $x=0:\ -\to+ \Rightarrow \text{min}; \quad x=2:\ +\to+ \Rightarrow \text{neither}$ $\text{(1) } -2:\ \text{local min}\ \checkmark \qquad \text{(2) } -\dfrac23:\ \text{local max}\ \checkmark$ $\text{(3) } 2:\ \times \qquad \text{(4) } \dfrac23:\ \times$ $\text{local min at } x=-2, \quad \text{local max at } x=-\dfrac23$ $\boxed{\text{Answer: (1) and (2)}}$