To solve this problem systematically, we need to understand the relationships and operations on the sets $M_1$, $M_2$, and their images under the function $f(x) = x^2$. Here’s the detailed step-by-step approach:

Definitions and Steps:

1. Given:

   • $M_1 = \{-1, -1, 1, 1, 2, 2\}$
   • $M_2 = \{1, 1, -2, -2, -2\}$
   • $f(x) = x^2$

2. Simplify $M_1$ and $M_2$:

   • Sets inherently have no repeated elements:
     • $M_1 = \{-1, 1, 2\}$
     • $M_2 = \{1, -2\}$

3. Function $f(x) = x^2$ applies:

   • $f(M_1) = f(\{-1, 1, 2\}) = \{1, 1, 4\} = \{1, 4\}$
   • $f(M_2) = f(\{1, -2\}) = \{1, 4\}$

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Required Calculations:

1. Union $M_1 \cup M_2$:

   • $M_1 \cup M_2 = \{-1, 1, 2, -2\}$
   • $f(M_1 \cup M_2) = f(\{-1, 1, 2, -2\}) = \{1, 4\}$

2. Union of Images $f(M_1) \cup f(M_2)$:

   • $f(M_1) \cup f(M_2) = \{1, 4\}$

3. Intersection $M_1 \cap M_2$:

   • $M_1 \cap M_2 = \{1\}$
   • $f(M_1 \cap M_2) = f(\{1\}) = \{1\}$

4. Symmetric Difference $M_1 \oplus M_2$:

   • $M_1 \oplus M_2 = (M_1 \cup M_2) \setminus (M_1 \cap M_2) = \{-1, 2, -2\}$
   • $f(M_1 \oplus M_2) = f(\{-1, 2, -2\}) = \{1, 4\}$

5. Difference $M_1 \setminus M_2$:

   • $M_1 \setminus M_2 = \{-1, 2\}$
   • $f(M_1 \setminus M_2) = f(\{-1, 2\}) = \{1, 4\}$

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Final Results:

• $f(M_1 \cup M_2) = \{1, 4\}$
• $f(M_1) \cup f(M_2) = \{1, 4\}$
• $f(M_1 \cap M_2) = \{1\}$
• $f(M_1 \oplus M_2) = \{1, 4\}$
• $f(M_1 \setminus M_2) = \{1, 4\}$

Would you like a further breakdown of any specific part?

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Related Questions:

1. What is the difference between $f(M_1 \cap M_2)$ and $f(M_1) \cap f(M_2)$?
2. Can you generalize results for any $f(x)$ where $f$ is non-linear?
3. How does the symmetric difference property change if $M_1$ and $M_2$ overlap completely?
4. How does $f(x) = x^3$ affect the calculations for $f(M_1 \oplus M_2)$?
5. Could we compute the Cartesian product $f(M_1 \times M_2)$ and what would that represent?

Tip:

Always simplify the sets by removing duplicates and focus on the nature of the function $f(x)$ to ensure correct results.