https://file.aiquickdraw.com/mathbot/file/1734139656711slz8m36x.jpg

To analyze the problem:

The given function is:

\[
f(x) = \frac{3x(x-1)}{x^2 - 3x + 2}, \quad \text{for } x \neq 1, 2,
\]
with the additional conditions:
\[
f(1) = -3 \quad \text{and} \quad f(2) = 4.
\]

The problem is to determine where \( f(x) \) is continuous.

### Step 1: Analyze the Denominator
The denominator \( x^2 - 3x + 2 \) factors as:
\[
x^2 - 3x + 2 = (x - 1)(x - 2).
\]

This means the function \( f(x) \) is undefined at \( x = 1 \) and \( x = 2 \), unless those points are specifically defined.

### Step 2: Check Continuity at \( x = 1 \) and \( x = 2 \)
- The function is explicitly defined as \( f(1) = -3 \) and \( f(2) = 4 \). However, for \( f(x) \) to be continuous at these points:
  - The limit of \( f(x) \) as \( x \to 1 \) must equal \( f(1) = -3 \).
  - The limit of \( f(x) \) as \( x \to 2 \) must equal \( f(2) = 4 \).

#### For \( x = 1 \):
The limit of \( f(x) \) as \( x \to 1 \) is:
\[
\lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{3x(x-1)}{(x-1)(x-2)}.
\]
Canceling \( (x-1) \) in the numerator and denominator:
\[
\lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{3x}{x-2}.
\]
Substituting \( x = 1 \):
\[
\frac{3(1)}{1-2} = -3.
\]
This equals \( f(1) = -3 \), so \( f(x) \) is continuous at \( x = 1 \).

#### For \( x = 2 \):
The limit of \( f(x) \) as \( x \to 2 \) is:
\[
\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{3x(x-1)}{(x-1)(x-2)}.
\]
Canceling \( (x-2) \) results in:
\[
\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{3x}{x-1}.
\]
Substituting \( x = 2 \):
\[
\frac{3(2)}{2-1} = 6.
\]
This does not equal \( f(2) = 4 \), so \( f(x) \) is not continuous at \( x = 2 \).

### Step 3: Check Continuity at Other Points
For \( x \neq 1, 2 \), \( f(x) \) is a rational function, which is continuous wherever its denominator is nonzero. Since \( x^2 - 3x + 2 \neq 0 \) for \( x \neq 1, 2 \), \( f(x) \) is continuous at all other points.

### Conclusion
The function \( f(x) \) is continuous **except at \( x = 2 \)**.

### Correct Answer:
\[
\boxed{\text{except at } x = 2.}
\]

Let me know if you want further clarification or additional steps explained!  

---

#### Follow-up Questions:
1. How do we define continuity for a piecewise function?
2. Why is the factorization of the denominator important in rational functions?
3. How do we compute limits of functions with removable discontinuities?
4. What are common reasons for discontinuities in rational functions?
5. What is the difference between continuity and differentiability?

#### Tip:
When checking for continuity at a point, always compare the limit of the function at that point with its actual value.

Given f(x) = 3x(x-1)/(x^2 - 3x + 2) for x ≠ 1, 2, with f(1) = -3 and f(2) = 4, determine where f(x) is continuous.

Explore the continuity of the piecewise rational function f(x) = 3x(x-1)/(x^2 - 3x + 2) with defined values at x = 1 and x = 2. This problem uses concepts of limits, continuity, and rational functions. Suitable for advanced high school or college-level students.

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