Key Concepts and Formulas

• Absolute Value Function: The absolute value of a real number $x$, denoted by $|x|$, is defined as $x$ if $x \geq 0$ and $-x$ if $x < 0$.
• Local Minima and Maxima: A function $f(x)$ has a local minimum at $x=c$ if $f(c) \leq f(x)$ for all $x$ in some open interval containing $c$. Similarly, $f(x)$ has a local maximum at $x=c$ if $f(c) \geq f(x)$ for all $x$ in some open interval containing $c$.
• Piecewise Functions: Functions defined by different formulas on different intervals.

Step-by-Step Solution

Step 1: Analyze the inner absolute value $|x|$

We need to consider two cases for $|x|$: $x \geq 0$ and $x < 0$.

• Case 1: $x \geq 0$. Then $|x| = x$, and $f(x) = ||x+2| - 2x||$.
• Case 2: $x < 0$. Then $|x| = -x$, and $f(x) = ||x+2| + 2x||$.

Step 2: Analyze the absolute value $|x+2|$

Now we consider two cases for $|x+2|$ in each of the above cases.

• Case 1a: $x \geq 0$ and $x+2 \geq 0$ (which simplifies to $x \geq 0$). Then $|x+2| = x+2$, and $f(x) = |(x+2) - 2x| = |2-x|$.
• Case 1b: $x \geq 0$ and $x+2 < 0$. This case is impossible since $x$ cannot be both greater than or equal to 0 and less than -2.
• Case 2a: $x < 0$ and $x+2 \geq 0$ (which simplifies to $-2 \leq x < 0$). Then $|x+2| = x+2$, and $f(x) = |(x+2) + 2x| = |3x+2|$.
• Case 2b: $x < 0$ and $x+2 < 0$ (which simplifies to $x < -2$). Then $|x+2| = -(x+2)$, and $f(x) = |-(x+2) + 2x| = |x-2|$.

Step 3: Analyze the remaining absolute values

We now have three different expressions for $f(x)$ over different intervals.

• If $x \geq 0$, $f(x) = |2-x|$. Here, $2-x \geq 0$ if $x \leq 2$, and $2-x < 0$ if $x > 2$. Therefore, $f(x) = 2-x$ for $0 \leq x \leq 2$, and $f(x) = x-2$ for $x > 2$.
• If $-2 \leq x < 0$, $f(x) = |3x+2|$. Here, $3x+2 \geq 0$ if $x \geq -2/3$, and $3x+2 < 0$ if $x < -2/3$. Therefore, $f(x) = 3x+2$ for $-2/3 \leq x < 0$, and $f(x) = -3x-2$ for $-2 \leq x < -2/3$.
• If $x < -2$, $f(x) = |x-2|$. Since $x < -2$, $x-2 < -4$, so $|x-2| = -(x-2) = 2-x$.

Step 4: Summarize the piecewise function

We can write $f(x)$ as a piecewise function:

$$f(x) = \begin{cases} 2-x & \text{if } x < -2 \\ -3x-2 & \text{if } -2 \leq x < -2/3 \\ 3x+2 & \text{if } -2/3 \leq x < 0 \\ 2-x & \text{if } 0 \leq x \leq 2 \\ x-2 & \text{if } x > 2 \end{cases}$$

Step 5: Identify local minima and maxima by analyzing the graph

The graph of $f(x)$ consists of line segments. We examine the points where the function changes its definition.

• At $x=-2$, $f(x)$ changes from $2-x$ to $-3x-2$. The value at $x=-2$ is $2-(-2) = 4$ and $-3(-2)-2 = 4$, so the function is continuous at $x=-2$. The slope changes from $-1$ to $-3$, so there is no local extremum.
• At $x=-2/3$, $f(x)$ changes from $-3x-2$ to $3x+2$. The value at $x=-2/3$ is $-3(-2/3)-2 = 0$ and $3(-2/3)+2=0$, so the function is continuous at $x=-2/3$. The slope changes from $-3$ to $3$, so there is a local minimum at $x=-2/3$, with $f(-2/3) = 0$.
• At $x=0$, $f(x)$ changes from $3x+2$ to $2-x$. The value at $x=0$ is $3(0)+2 = 2$ and $2-0=2$, so the function is continuous at $x=0$. The slope changes from $3$ to $-1$, so there is a local maximum at $x=0$, with $f(0) = 2$.
• At $x=2$, $f(x)$ changes from $2-x$ to $x-2$. The value at $x=2$ is $2-2=0$ and $2-2=0$, so the function is continuous at $x=2$. The slope changes from $-1$ to $1$, so there is a local minimum at $x=2$, with $f(2) = 0$.

Therefore, there are two local minima at $x = -2/3$ and $x=2$, and one local maximum at $x=0$. So, $m=2$ and $n=1$.

Step 6: Calculate $m+n$

$m+n = 2 + 1 = 3$.

Common Mistakes & Tips

• Sign Errors: Be very careful with the signs when removing absolute values. It's easy to make a mistake, especially when there are multiple absolute values nested within each other.
• Checking Continuity: Always check if the piecewise function is continuous at the points where the definition changes. A discontinuity can also be a local extremum.
• Graphing: Sketching the graph, even a rough one, can be extremely helpful in visualizing the function and identifying local minima and maxima.

Summary

We systematically removed the absolute value signs to express the function as a piecewise function. Then, we analyzed the graph of the piecewise function to identify the local minima and maxima. There are two local minima and one local maximum, so $m=2$ and $n=1$, and $m+n=3$.

Final Answer

The final answer is \boxed{3}, which corresponds to option (A).