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The problem you're asking for is:

Given the function \( f: \mathbb{R} \to \mathbb{R} \) such that:

\[
(x - 2)^2 \cdot f(x) = x^2 - 4x + 2 \quad \text{for every} \quad x \neq 2
\]

You are asked to prove that \( f \) is not continuous at \( x = 2 \).

### Solution:

To prove that \( f \) is not continuous at \( x = 2 \), we need to check the conditions for continuity:

1. **Existence of \( f(2) \):** We need to check if \( f(2) \) is defined.
2. **Existence of the limit:** We need to check if the limit \( \lim_{x \to 2} f(x) \) exists.
3. **Continuity condition:** We need to check if \( \lim_{x \to 2} f(x) = f(2) \).

#### Step 1: Express \( f(x) \)

The given equation is:

\[
(x - 2)^2 \cdot f(x) = x^2 - 4x + 2
\]

Let's simplify the right-hand side:

\[
x^2 - 4x + 2 = (x - 2)^2 - 2
\]

Thus, we have:

\[
(x - 2)^2 \cdot f(x) = (x - 2)^2 - 2
\]

Now, divide both sides by \( (x - 2)^2 \) (for \( x \neq 2 \)):

\[
f(x) = 1 - \frac{2}{(x - 2)^2}
\]

#### Step 2: Investigate the limit of \( f(x) \) as \( x \to 2 \)

To determine whether \( f(x) \) is continuous at \( x = 2 \), we compute the limit of \( f(x) \) as \( x \to 2 \):

\[
\lim_{x \to 2} f(x) = \lim_{x \to 2} \left( 1 - \frac{2}{(x - 2)^2} \right)
\]

As \( x \to 2 \), \( (x - 2)^2 \to 0 \). Therefore, \( \frac{2}{(x - 2)^2} \to \infty \), and thus:

\[
\lim_{x \to 2} f(x) = 1 - \infty = -\infty
\]

#### Step 3: Check if \( f(2) \) is defined

From the given problem, the function is only defined by the equation for \( x \neq 2 \). We don't have information about \( f(2) \), so we assume that \( f(2) \) is either undefined or arbitrary.

#### Conclusion:

Since the limit \( \lim_{x \to 2} f(x) = -\infty \) does not exist (it diverges), \( f(x) \) is not continuous at \( x = 2 \).

---

Do you have any further questions or would you like more details? Here are some related questions:

1. How would we redefine \( f(x) \) to make it continuous at \( x = 2 \)?
2. What would happen to the continuity if \( f(2) \) was specifically defined as a finite value?
3. How does the behavior of \( (x - 2)^2 \) near \( x = 2 \) influence the continuity of \( f(x) \)?
4. Could we find an interval around \( x = 2 \) where \( f(x) \) remains continuous?
5. What are the practical applications of determining continuity in real-world problems?

**Tip:** When analyzing limits at points of discontinuity, look at how the function behaves on either side of the point to identify any jumps or divergences.

Δίνεται η συνάρτηση f: R → R τέτοια ώστε (x−2)^2 * f(x) = x^2 − 4x + 2 για κάθε x ≠ 2. Να αποδείξετε ότι η f δεν είναι συνεχής στο x = 2.

Solve a problem involving the function f defined by (x−2)^2 * f(x) = x^2 − 4x + 2 for x ≠ 2. This advanced problem demonstrates how to prove discontinuity at x=2 using limit analysis and algebraic manipulation.

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