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The question asks for the values of \( a \) and \( b \) such that the function \( f(x) \) is continuous on its domain. The function \( f(x) \) is defined piecewise as:

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

### Step 1: Simplify \( f(x) \) for \( x < 2 \)
The first piece is \( f(x) = \frac{x^2 - 4}{x - 2} \) for \( x < 2 \). Notice that:
\[
x^2 - 4 = (x - 2)(x + 2)
\]
Thus, \( f(x) \) for \( x < 2 \) simplifies to:
\[
f(x) = x + 2 \quad \text{for} \quad x < 2
\]
since the factor \( x - 2 \) cancels out.

### Step 2: Continuity at \( x = 2 \)
To ensure the function is continuous at \( x = 2 \), the value of the left-hand limit and right-hand limit must be equal at \( x = 2 \).

- Left-hand limit as \( x \to 2^- \) (from the first piece):
\[
\lim_{x \to 2^-} f(x) = 2 + 2 = 4
\]

- Right-hand limit as \( x \to 2^+ \) (from the second piece):
\[
\lim_{x \to 2^+} f(x) = a(2)^2 - b(2) + 3 = 4a - 2b + 3
\]

Setting these equal for continuity at \( x = 2 \):
\[
4 = 4a - 2b + 3
\]
\[
4a - 2b = 1 \quad \text{(Equation 1)}
\]

### Step 3: Continuity at \( x = 3 \)
Similarly, for the function to be continuous at \( x = 3 \), the values of the second and third pieces must match at \( x = 3 \).

- Left-hand limit as \( x \to 3^- \) (from the second piece):
\[
\lim_{x \to 3^-} f(x) = a(3)^2 - b(3) + 3 = 9a - 3b + 3
\]

- Right-hand limit as \( x \to 3^+ \) (from the third piece):
\[
\lim_{x \to 3^+} f(x) = 2(3) - a + b = 6 - a + b
\]

Setting these equal for continuity at \( x = 3 \):
\[
9a - 3b + 3 = 6 - a + b
\]
\[
10a - 4b = 3 \quad \text{(Equation 2)}
\]

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

We can multiply the first equation by 2 to align it with the second equation:
\[
8a - 4b = 2
\]

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

Substitute \( a = \frac{1}{2} \) into Equation 1:
\[
4\left(\frac{1}{2}\right) - 2b = 1
\]
\[
2 - 2b = 1
\]
\[
-2b = -1
\]
\[
b = \frac{1}{2}
\]

### Final Answer:
The values of \( a \) and \( b \) such that \( f(x) \) is continuous are:
\[
a = \frac{1}{2}, \quad b = \frac{1}{2}
\]

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

Here are 5 related questions to expand on this:

1. How do you determine continuity for piecewise functions?
2. Why is factoring useful in simplifying rational expressions?
3. What is the significance of left-hand and right-hand limits?
4. How do we verify the values of \( a \) and \( b \) ensure differentiability at \( x = 2 \) and \( x = 3 \)?
5. Can piecewise functions have discontinuities in their domain?

**Tip**: Always simplify piecewise expressions as much as possible before working with limits. This can make solving continuity conditions much easier!

Tentukan nilai a dan b agar f(x) kontinu pada domainnya. The piecewise function f(x) is defined as follows: 
- f(x) = (x^2 - 4)/(x - 2), for x < 2 
- f(x) = ax^2 - bx + 3, for 2 ≤ x < 3 
- f(x) = 2x - a + b, for x ≥ 3

This problem explores the concept of continuity in piecewise functions. Learn how to determine the values of a and b that make f(x) continuous over its domain, using limits and algebraic equations.

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