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The problem in the image asks to determine the values of \( a \) and \( b \) such that the given function \( f(x) \) is continuous for all real numbers. The function is defined piecewise as follows:

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

### Approach to solve:

For the function to be continuous at \( x = 2 \) and \( x = 3 \), the left-hand limit and the right-hand limit at these points must equal the function values.

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

For \( f(x) \) to be continuous at \( x = 2 \):

1. The left-hand limit as \( x \to 2^- \) (from the first piece):
   \[
   \lim_{x \to 2^-} f(x) = 2^2 - 4 = 4 - 4 = 0
   \]

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

Since the function is continuous at \( x = 2 \), these limits must be equal:
\[
0 = 2a - b + 3
\]
\[
2a - b = -3 \quad \text{(Equation 1)}
\]

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

For \( f(x) \) to be continuous at \( x = 3 \):

1. The left-hand limit as \( x \to 3^- \) (from the second piece):
   \[
   \lim_{x \to 3^-} f(x) = a(3) - b + 3 = 3a - b + 3
   \]

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

Since the function is continuous at \( x = 3 \), these limits must be equal:
\[
3a - b + 3 = 6 - a
\]
Simplifying this equation:
\[
3a - b + 3 = 6 - a
\]
\[
4a - b = 3 \quad \text{(Equation 2)}
\]

#### Step 3: Solving the system of equations

Now, we have the following system of equations:
1. \( 2a - b = -3 \)
2. \( 4a - b = 3 \)

We can subtract Equation 1 from Equation 2 to eliminate \( b \):
\[
(4a - b) - (2a - b) = 3 - (-3)
\]
\[
2a = 6
\]
\[
a = 3
\]

Substitute \( a = 3 \) into Equation 1:
\[
2(3) - b = -3
\]
\[
6 - b = -3
\]
\[
b = 9
\]

### Final Answer:
The values of \( a \) and \( b \) are:
\[
a = 3, \quad b = 9
\]

Let me know if you have any questions or would like further clarification.

---

Here are 5 related questions to expand your understanding:

1. What does it mean for a function to be continuous at a point?
2. How can we verify the continuity of a function at different points on a piecewise function?
3. What is the difference between a left-hand limit and a right-hand limit?
4. Can a function be continuous but not differentiable at some points? Why?
5. How does one solve a system of linear equations using substitution or elimination?

**Tip:** Always check the left-hand and right-hand limits separately to ensure the continuity of a piecewise function at boundary points.

Determine the values of a and b for which the following function is continuous over all real numbers:

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

Solve for a and b in a piecewise function to ensure continuity. This problem covers key concepts in calculus such as continuity, limits, and solving systems of equations. Step-by-step solution provided.

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