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To solve the problem, we are asked to find the values of \( a \) and \( b \) that make the piecewise function continuous everywhere. The function is defined as:

\[
f(x) = 
\begin{cases}
\frac{x^2 - 4}{x - 2} & \text{if } x < 2 \\
ax^2 - bx + 1 & \text{if } 2 \leq x < 3 \\
4x - a + b & \text{if } x \geq 3
\end{cases}
\]

To ensure continuity, the left-hand and right-hand limits at \( x = 2 \) and \( x = 3 \) must be equal to the function's value at these points.

### Step 1: Continuity at \( x = 2 \)

For \( x < 2 \), the first expression is \( \frac{x^2 - 4}{x - 2} \). We can simplify this expression:

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

Thus, as \( x \to 2^- \), the limit is \( 2 + 2 = 4 \).

For \( x \geq 2 \), the function is \( ax^2 - bx + 1 \). To ensure continuity at \( x = 2 \), we must have:

\[
a(2^2) - b(2) + 1 = 4
\]

This simplifies to:

\[
4a - 2b + 1 = 4
\]

Which gives the equation:

\[
4a - 2b = 3 \quad \text{(Equation 1)}
\]

### Step 2: Continuity at \( x = 3 \)

For \( x \to 3^- \), the expression is \( ax^2 - bx + 1 \), so:

\[
a(3^2) - b(3) + 1 = 9a - 3b + 1
\]

For \( x \geq 3 \), the expression is \( 4x - a + b \), and at \( x = 3 \), this becomes:

\[
4(3) - a + b = 12 - a + b
\]

To ensure continuity at \( x = 3 \), we set the two expressions equal to each other:

\[
9a - 3b + 1 = 12 - a + b
\]

Simplifying:

\[
9a - 3b + 1 = 12 - a + b
\]
\[
9a + a = 12 + 3b - b - 1
\]
\[
10a = 11 + 2b
\]

This gives the second equation:

\[
10a - 2b = 11 \quad \text{(Equation 2)}
\]

### Step 3: Solve the system of equations

We now solve the system of linear equations:
1. \( 4a - 2b = 3 \)
2. \( 10a - 2b = 11 \)

We can subtract Equation 1 from Equation 2 to eliminate \( b \):

\[
(10a - 2b) - (4a - 2b) = 11 - 3
\]
\[
6a = 8
\]
\[
a = \frac{8}{6} = \frac{4}{3}
\]

Substitute \( a = \frac{4}{3} \) into Equation 1:

\[
4\left(\frac{4}{3}\right) - 2b = 3
\]
\[
\frac{16}{3} - 2b = 3
\]
Multiply through by 3 to clear the denominator:

\[
16 - 6b = 9
\]
\[
-6b = -7
\]
\[
b = \frac{7}{6}
\]

### Final Answer:
The values of \( a \) and \( b \) that make the function continuous everywhere are:

\[
a = \frac{4}{3}, \quad b = \frac{7}{6}
\]

Would you like more details on any part of the solution?

Here are some related questions:
1. How do we check differentiability for a piecewise function?
2. What is the difference between continuity and differentiability?
3. What is the significance of simplifying rational expressions for limits?
4. How do limits at jump discontinuities behave?
5. What are the common techniques for solving systems of equations?

**Tip:** When simplifying limits for piecewise functions, always factor out terms before directly substituting values at potential discontinuities.

Find the values of a and b that make the function continuous everywhere for the piecewise function f(x) defined as: 

f(x) = { (x^2 - 4) / (x - 2) if x < 2, ax^2 - bx + 1 if 2 ≤ x < 3, 4x - a + b if x ≥ 3 }

Solve for the values of a and b to ensure continuity in a piecewise function with quadratic and linear segments. This involves limit calculations and solving a system of equations. Appropriate for Grades 11-12 students learning about limits, continuity, and algebraic systems.

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