Local Extrema of Absolute Value Function

To find the points at which the function attains a local minimum or a local maximum, we first simplify the given expression.

Recall the algebraic identity for the minimum of two real numbers: $\min(a, b) = \frac{a + b - |a - b|}{2} \implies 2\min(a, b) = a + b - |a - b|$

Let $a = |x+2|$ and $b = 2|x|$. The given function is: $f(x) = 2|x| + |x+2| - ||x+2| - 2|x||$ Using the identity, this can be written compactly as: $f(x) = 2 \min(|x+2|, 2|x|)$

To determine the behavior of $f(x)$, we find the points of intersection of $|x+2|$ and $2|x|$: $|x+2| = 2|x| \implies (x+2)^2 = 4x^2 \implies 3x^2 - 4x - 4 = 0$ $(3x + 2)(x - 2) = 0 \implies x = -\frac{2}{3} \quad \text{or} \quad x = 2$

Along with the roots of the individual absolute value components, $x = -2$ and $x = 0$, we analyze $f(x)$ over the real line partitioned into five intervals:

1. For $x \le -2$:

   • $|x+2| = -(x+2)$
   • $2|x| = -2x$
   • Since $x \le -2$, we have $-(x+2) \le -2x \iff -x - 2 \le -2x \iff x \le 2$ (which is true).
   • Thus, $\min(|x+2|, 2|x|) = |x+2| = -x-2$. $f(x) = 2(-x-2) = -2x - 4$ Here, the slope of $f(x)$ is $-2$ (strictly decreasing).

2. For $-2 < x \le -\frac{2}{3}$:

   • $|x+2| = x+2$
   • $2|x| = -2x$
   • Since $x \le -\frac{2}{3}$, we have $x+2 \le -2x \iff 3x \le -2$.
   • Thus, $\min(|x+2|, 2|x|) = |x+2| = x+2$. $f(x) = 2(x+2) = 2x + 4$ Here, the slope of $f(x)$ is $+2$ (strictly increasing).

3. For $-\frac{2}{3} < x \le 0$:

   • $|x+2| = x+2$
   • $2|x| = -2x$
   • Since $x > -\frac{2}{3}$, we have $x+2 > -2x$.
   • Thus, $\min(|x+2|, 2|x|) = 2|x| = -2x$. $f(x) = 2(-2x) = -4x$ Here, the slope of $f(x)$ is $-4$ (strictly decreasing).

4. For $0 < x \le 2$:

   • $|x+2| = x+2$
   • $2|x| = 2x$
   • Since $x \le 2$, we have $2x \le x+2$.
   • Thus, $\min(|x+2|, 2|x|) = 2|x| = 2x$. $f(x) = 2(2x) = 4x$ Here, the slope of $f(x)$ is $+4$ (strictly increasing).

5. For $x > 2$:

   • $|x+2| = x+2$
   • $2|x| = 2x$
   • Since $x > 2$, we have $x+2 < 2x$.
   • Thus, $\min(|x+2|, 2|x|) = |x+2| = x+2$. $f(x) = 2(x+2) = 2x + 4$ Here, the slope of $f(x)$ is $+2$ (strictly increasing).

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Analysis of Local Extrema:

• At $x = -2$:

  • For $x < -2$, $f'(x) = -2 < 0$ (function decreases).
  • For $-2 < x < -\frac{2}{3}$, $f'(x) = +2 > 0$ (function increases).
  • Therefore, $f(x)$ attains a local minimum at $x = -2$.

• At $x = -\frac{2}{3}$:

  • For $-2 < x < -\frac{2}{3}$, $f'(x) = +2 > 0$ (function increases).
  • For $-\frac{2}{3} < x < 0$, $f'(x) = -4 < 0$ (function decreases).
  • Therefore, $f(x)$ attains a local maximum at $x = -\frac{2}{3}$.

• At $x = 0$:

  • $f(x)$ attains a local minimum (not among the options).

• At $x = 2$:

  • For $0 < x < 2$, $f'(x) = +4 > 0$.
  • For $x > 2$, $f'(x) = +2 > 0$.
  • Since $f(x)$ is strictly increasing in the neighborhood of $x = 2$, it is neither a local maximum nor a local minimum.

• At $x = \frac{2}{3}$:

  • In the neighborhood around $\frac{2}{3}$, $f'(x) = 4 > 0$, so $f(x)$ is strictly increasing and has no extremum.

Hence, the function attains a local extremum at $x = -2$ and $x = -\frac{2}{3}$.

Correct Options: (A) and (B)