Question

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

• A. -2
• B. $\frac{-2}{3}$
• C. 2
• D. $\frac{2}{3}$

Answer 1

Answer: (A), (B)

$\frac{-2}{3}$

For $f\left(x\right)=2\left|x\right|+|x+2|-||x+2|-2\left|x\right||$
the critical points can be obtained by solving $\left|x\right|=0,$
$|x+2|=0$ and $||x+2|-2\left|x\right||=0$
we get x=0,-2,2, $-\frac{2}{3}$
Then we can write
$f\left(x\right)\{\begin{array}{cc}& \text{-2x-4,}x\le -2\\ & \text{2x+4,}-2<x\le -\frac{2}{3}\\ & \text{-4x,}-\frac{2}{3}<x\le 0\\ & \text{4x},0<x\le 2\\ & \text{2x+4},x>2\end{array}$
The graph of y = f(x) is as follows

From graph f(x) has local minimum at -2, 0 and local maximum at $-\frac{2}{3}$