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We are asked to find the values of \( a \) and \( b \) such that the piecewise function \( f(x) \) is continuous everywhere.

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

### Step 1: Simplifying the first piece of the function
For \( x < 2 \), the expression is \( \frac{x^2 - 4}{x - 2} \). Notice that:
\[
x^2 - 4 = (x - 2)(x + 2)
\]
So, the function simplifies to:
\[
\frac{(x - 2)(x + 2)}{x - 2} = x + 2 \quad \text{for} \quad x \neq 2
\]
Thus, for \( x < 2 \), the function is \( f(x) = x + 2 \).

### Step 2: Ensuring continuity at \( x = 2 \)
For the function to be continuous at \( x = 2 \), the limit of \( f(x) \) as \( x \) approaches 2 from both sides must be equal to \( f(2) \).

- As \( x \to 2^- \) (from the left), \( f(x) = x + 2 \), so \( \lim_{x \to 2^-} f(x) = 2 + 2 = 4 \).
- At \( x = 2 \), from the second piece \( f(x) = ax^2 - bx + 3 \). Substituting \( x = 2 \) into this gives:
  \[
  f(2) = a(2)^2 - b(2) + 3 = 4a - 2b + 3
  \]

To ensure continuity at \( x = 2 \), we set \( 4a - 2b + 3 = 4 \). Simplifying:
\[
4a - 2b = 1 \quad \text{(Equation 1)}
\]

### Step 3: Ensuring continuity at \( x = 3 \)
For the function to be continuous at \( x = 3 \), the limit of \( f(x) \) as \( x \) approaches 3 from both sides must also be equal.

- As \( x \to 3^- \) (from the left), \( f(x) = ax^2 - bx + 3 \). Substituting \( x = 3 \) gives:
  \[
  f(3) = a(3)^2 - b(3) + 3 = 9a - 3b + 3
  \]
- As \( x \to 3^+ \) (from the right), \( f(x) = 2x - a + b \). Substituting \( x = 3 \) gives:
  \[
  f(3) = 2(3) - a + b = 6 - a + b
  \]

To ensure continuity at \( x = 3 \), set \( 9a - 3b + 3 = 6 - a + b \). Simplifying this:
\[
9a - 3b + 3 = 6 - a + b
\]
\[
10a - 4b = 3 \quad \text{(Equation 2)}
\]

### Step 4: Solving the system of equations
We now solve the system of two equations:
1. \( 4a - 2b = 1 \)
2. \( 10a - 4b = 3 \)

We can solve this system using substitution or elimination. I'll use elimination here.

Multiply the first equation by 2:
\[
8a - 4b = 2
\]
Now subtract the second equation from this:
\[
(8a - 4b) - (10a - 4b) = 2 - 3
\]
\[
-2a = -1 \quad \Rightarrow \quad a = \frac{1}{2}
\]

Substitute \( a = \frac{1}{2} \) into the first equation:
\[
4\left(\frac{1}{2}\right) - 2b = 1
\]
\[
2 - 2b = 1 \quad \Rightarrow \quad -2b = -1 \quad \Rightarrow \quad b = \frac{1}{2}
\]

### 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 more details or have any questions about the steps? Here are five related questions to explore:

1. How does one check differentiability for a piecewise function?
2. Can you explain how limits work when verifying continuity?
3. What are common techniques for solving systems of linear equations?
4. What are the conditions required for a function to be continuous?
5. How would the process change if \( f(x) \) were defined differently for one of the intervals?

**Tip**: Always check for removable discontinuities when dealing with rational functions!

Find the values of a and b that make the function f continuous everywhere. The function is defined as a piecewise function with three cases: (x^2 - 4)/(x - 2) if x < 2, ax^2 - bx + 3 if 2 <= x < 3, and 2x - a + b if x >= 3.

Learn how to determine the values of a and b to make the function continuous everywhere. This problem involves solving for continuity across different intervals in a piecewise function, factoring quadratics, and applying limit laws.

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