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We are asked to find the values of \( a \) and \( b \) that make the given piecewise function \( f(x) \) continuous everywhere. The function is defined as follows:

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

### Step 1: Simplifying the first part of \( f(x) \)
For \( x < 2 \), we have:
\[
f(x) = \frac{x^2 - 4}{x - 2}
\]
Notice that \( x^2 - 4 \) is a difference of squares, so it can be factored:
\[
\frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2}
\]
For \( x \neq 2 \), we can cancel \( x - 2 \) from the numerator and denominator:
\[
f(x) = x + 2 \quad \text{for } x < 2
\]
Thus, \( f(x) = x + 2 \) for \( x < 2 \).

### Step 2: Continuity at \( x = 2 \)
For \( f(x) \) to be continuous at \( x = 2 \), the limit from the left must equal the value of the function at \( x = 2 \) in the middle piece.

From the first piece, as \( x \to 2^- \):
\[
\lim_{x \to 2^-} f(x) = 2 + 2 = 4
\]

For the second piece \( ax^2 - bx + 3 \) at \( x = 2 \), we need:
\[
f(2) = a(2^2) - b(2) + 3 = 4a - 2b + 3
\]
To ensure continuity at \( x = 2 \), set this equal to 4:
\[
4a - 2b + 3 = 4
\]
This simplifies to:
\[
4a - 2b = 1 \quad \text{(Equation 1)}
\]

### Step 3: Continuity at \( x = 3 \)
Now, for \( f(x) \) to be continuous at \( x = 3 \), the value of the second piece must match the value of the third piece at \( x = 3 \).

For the second piece \( ax^2 - bx + 3 \) at \( x = 3 \), we have:
\[
f(3) = a(3^2) - b(3) + 3 = 9a - 3b + 3
\]

For the third piece \( 4x - a + b \) at \( x = 3 \), we get:
\[
f(3) = 4(3) - a + b = 12 - a + b
\]

To ensure continuity at \( x = 3 \), set these equal to each other:
\[
9a - 3b + 3 = 12 - a + b
\]
Simplifying this equation:
\[
9a - 3b + 3 = 12 - a + b
\]
\[
10a - 4b = 9 \quad \text{(Equation 2)}
\]

### Step 4: Solving the system of equations
We now have the system:
1. \( 4a - 2b = 1 \)
2. \( 10a - 4b = 9 \)

Multiply the first equation by 2:
\[
8a - 4b = 2
\]
Now subtract this from the second equation:
\[
(10a - 4b) - (8a - 4b) = 9 - 2
\]
\[
2a = 7
\]
\[
a = \frac{7}{2}
\]

Substitute \( a = \frac{7}{2} \) into the first equation:
\[
4\left(\frac{7}{2}\right) - 2b = 1
\]
\[
14 - 2b = 1
\]
\[
2b = 13
\]
\[
b = \frac{13}{2}
\]

### Final Answer:
The values of \( a \) and \( b \) that make \( f(x) \) continuous everywhere are:
\[
a = \frac{7}{2}, \quad b = \frac{13}{2}
\]

Would you like more details, or do you have any questions about the steps?

Here are 5 related questions to explore further:
1. What does it mean for a function to be continuous at a point?
2. How would the solution change if the piecewise function had a different boundary at \( x = 2 \)?
3. How do you solve piecewise functions with more than two points of continuity?
4. What is the significance of the difference of squares factorization in this problem?
5. How would discontinuities appear graphically for a piecewise function?

**Tip:** Always ensure that the limits from both sides of a point match when checking for continuity in piecewise functions.

Find the values of a and b that make f continuous everywhere for the given piecewise function:

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

This problem requires finding the values of a and b that make a piecewise function continuous everywhere. It involves concepts of algebra, limits, and the difference of squares. Learn how to solve for continuity across the boundaries of the piecewise function.

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