The fuction `f(x)=2|x|+|x+2|-||x+2|-2|x||` has a local minimum or a local maximum respectively at x =

Options

A. -2
B. `(-2)/(3)`
C. 2
D. `2//3`

Answer

To determine the local minima and maxima of the function \( f(x) = 2|x| + |x+2| - ||x+2| - 2|x|| \), we will analyze the function piecewise based on the critical points where the absolute value expressions change.

### Step 1: Identify critical points
The critical points occur when the expressions inside the absolute values change sign. The expressions are \( |x| \), \( |x+2| \), and \( ||x+2| - 2|x| \). The critical points are:
1. \( x = -2 \)
2. \( x = 0 \)

### Step 2: Analyze intervals
We will analyze the function in the intervals defined by these critical points:
1. \( x < -2 \)
2. \( -2 \leq x < 0 \)
3. \( x \geq 0 \)

### Step 3: Case 1: \( x < -2 \)
In this interval, we have:
- \( |x| = -x \)
- \( |x+2| = -(x+2) = -x - 2 \)
- \( ||x+2| - 2|x| = |-x - 2 - 2(-x)| = |-x - 2 + 2x| = |x - 2| = -x + 2 \) (since \( x < -2 \))

Substituting these into the function:
\[ f(x) = 2(-x) + (-x - 2) - (-x + 2) = -2x - x - 2 + x - 2 = -2x - 4 \]

### Step 4: Case 2: \( -2 \leq x < 0 \)
In this interval:
- \( |x| = -x \)
- \( |x+2| = x + 2 \)
- \( ||x+2| - 2|x| = |(x + 2) - 2(-x)| = |x + 2 + 2x| = |3x + 2| = 3x + 2 \) (since \( x \geq -2 \))

Substituting these into the function:
\[ f(x) = 2(-x) + (x + 2) - (3x + 2) = -2x + x + 2 - 3x - 2 = -4x \]

### Step 5: Case 3: \( x \geq 0 \)
In this interval:
- \( |x| = x \)
- \( |x+2| = x + 2 \)
- \( ||x+2| - 2|x| = |(x + 2) - 2x| = |2 - x| \)

Substituting these into the function:
\[ f(x) = 2x + (x + 2) - |2 - x| \]
For \( x < 2 \):
\[ f(x) = 2x + x + 2 - (2 - x) = 4x \]
For \( x \geq 2 \):
\[ f(x) = 2x + x + 2 - (x - 2) = 4x + 4 \]

### Step 6: Evaluate at critical points
Now we evaluate \( f(x) \) at the critical points:
1. At \( x = -2 \):
\[ f(-2) = -2(-2) - 4 = 4 - 4 = 0 \]
2. At \( x = 0 \):
\[ f(0) = -4(0) = 0 \]
3. At \( x = -\frac{2}{3} \):
\[ f\left(-\frac{2}{3}\right) = -4\left(-\frac{2}{3}\right) = \frac{8}{3} \]
4. At \( x = 2 \):
\[ f(2) = 4(2) + 4 = 8 + 4 = 12 \]

### Step 7: Determine local minima and maxima
From the evaluations:
- \( f(-2) = 0 \) and \( f(0) = 0 \) are local minima.
- \( f\left(-\frac{2}{3}\right) = \frac{8}{3} \) and \( f(2) = 12 \) are local maxima.

### Conclusion
The function has local minima at \( x = -2 \) and \( x = 0 \), and local maxima at \( x = -\frac{2}{3} \) and \( x = 2 \).