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The problem asks to find the values of \( A \) and \( B \) that make the given function continuous everywhere. The function \( f(x) \) is defined in piecewise form as:

\[
f(x) = 
\begin{cases} 
\frac{x^2 - 4}{x - 2} & \text{if } x < 2 \\
Ax^2 - Bx + 3 & \text{if } 2 \leq x \leq 3 \\
2x - A + B & \text{if } x \geq 3
\end{cases}
\]

### Step 1: Continuity at \( x = 2 \)
To ensure continuity at \( x = 2 \), the left-hand limit and the right-hand limit must be equal at that point. 

- For \( x < 2 \), we simplify the expression \( \frac{x^2 - 4}{x - 2} \):
  \[
  \frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2} = x + 2, \text{ for } x \neq 2.
  \]
  So as \( x \to 2^- \), \( f(x) \to 4 \).

- For \( x = 2 \), using the second part \( f(x) = Ax^2 - Bx + 3 \):
  \[
  f(2) = A(2)^2 - B(2) + 3 = 4A - 2B + 3.
  \]
  To ensure continuity, we set the left-hand and right-hand limits equal at \( x = 2 \):
  \[
  4A - 2B + 3 = 4.
  \]
  Simplifying this, we get:
  \[
  4A - 2B = 1. \tag{1}
  \]

### Step 2: Continuity at \( x = 3 \)
For \( x = 3 \), we again match the left-hand and right-hand limits:

- For \( x = 3 \), using the second part \( Ax^2 - Bx + 3 \):
  \[
  f(3) = A(3)^2 - B(3) + 3 = 9A - 3B + 3.
  \]
  
- For \( x = 3 \), using the third part \( f(x) = 2x - A + B \):
  \[
  f(3) = 2(3) - A + B = 6 - A + B.
  \]

  To ensure continuity, set the left-hand and right-hand limits equal at \( x = 3 \):
  \[
  9A - 3B + 3 = 6 - A + B.
  \]
  Simplifying this equation:
  \[
  10A - 4B = 3. \tag{2}
  \]

### Step 3: Solve the System of Equations
Now we have the system of equations:
1. \( 4A - 2B = 1 \)
2. \( 10A - 4B = 3 \)

Solve equation (1) for \( B \):
\[
2B = 4A - 1 \implies B = 2A - \frac{1}{2}.
\]

Substitute this into equation (2):
\[
10A - 4(2A - \frac{1}{2}) = 3.
\]
Simplify:
\[
10A - 8A + 2 = 3 \implies 2A + 2 = 3 \implies 2A = 1 \implies A = \frac{1}{2}.
\]

Now substitute \( A = \frac{1}{2} \) back into \( B = 2A - \frac{1}{2} \):
\[
B = 2\left(\frac{1}{2}\right) - \frac{1}{2} = 1 - \frac{1}{2} = \frac{1}{2}.
\]

### Final Answer:
The values of \( A \) and \( B \) that make the function continuous everywhere are:
\[
A = \frac{1}{2}, \quad B = \frac{1}{2}.
\]

Would you like further details or clarifications?

Here are some related questions for practice:
1. How do you determine if a piecewise function is continuous at a given point?
2. What happens if the function is not continuous at a point?
3. Can a function be continuous but not differentiable?
4. How would you adjust this problem if the intervals were different?
5. What is the significance of limits when analyzing piecewise functions?

**Tip:** Always check both the left-hand and right-hand limits to ensure continuity of piecewise functions at the given points.

Find the values of A & B that would make the given function continuous everywhere. The function f(x) is defined as: f(x) = { (x^2 - 4)/(x - 2), if x < 2; Ax^2 - Bx + 3, if 2 ≤ x ≤ 3; 2x - A + B, if x ≥ 3. }

This problem explores how to find the values of A and B to make a piecewise function continuous at all points. Concepts covered include limits, continuity, and solving a system of equations to ensure the function's smoothness across different intervals.

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