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We are given a piecewise function:

\[
f(x) =
\begin{cases} 
ax^2 + x - b, & x \leq 2 \\
ax + b, & 2 < x < 5 \\
2ax - 7, & x \geq 5
\end{cases}
\]

We need to find the values of \(a\) and \(b\) such that \(f(x)\) is continuous on the entire real number line. For continuity, the left-hand limit and right-hand limit at \(x = 2\) and \(x = 5\) must be equal.

### Step 1: Continuity at \(x = 2\)
We require that:
\[
\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)
\]
The left-hand limit as \(x \to 2\) is:
\[
\lim_{x \to 2^-} f(x) = a(2)^2 + 2 - b = 4a + 2 - b
\]
The right-hand limit as \(x \to 2^+\) is:
\[
\lim_{x \to 2^+} f(x) = 2a + b
\]
For continuity at \(x = 2\), we set these equal:
\[
4a + 2 - b = 2a + b
\]
Simplifying this equation:
\[
4a + 2 - b = 2a + b \implies 2a + 2 = 2b \implies a + 1 = b
\]
Thus, we have the relation:
\[
b = a + 1
\]

### Step 2: Continuity at \(x = 5\)
We require that:
\[
\lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x)
\]
The left-hand limit as \(x \to 5\) is:
\[
\lim_{x \to 5^-} f(x) = 5a + b
\]
The right-hand limit as \(x \to 5^+\) is:
\[
\lim_{x \to 5^+} f(x) = 2a(5) - 7 = 10a - 7
\]
For continuity at \(x = 5\), we set these equal:
\[
5a + b = 10a - 7
\]
Substitute \(b = a + 1\) from the previous step:
\[
5a + (a + 1) = 10a - 7
\]
Simplifying this equation:
\[
6a + 1 = 10a - 7
\]
\[
6a - 10a = -7 - 1
\]
\[
-4a = -8
\]
\[
a = 2
\]

### Step 3: Solve for \(b\)
Now that we know \(a = 2\), substitute into the equation \(b = a + 1\):
\[
b = 2 + 1 = 3
\]

### Final Answer:
The values of \(a\) and \(b\) that make the function continuous are:
\[
a = 2, \quad b = 3
\]

Would you like further clarification or details on any part of the solution?

Here are 5 related questions for practice:
1. What does it mean for a function to be continuous at a point?
2. How would the solution change if the function had a different form in one of the intervals?
3. Can a function be continuous but not differentiable at a point? Give an example.
4. How would you verify the differentiability of this function at \(x = 2\) or \(x = 5\)?
5. If one of the pieces of the function was cubic instead of quadratic, how would that affect the process?

**Tip:** When solving continuity problems with piecewise functions, always check the limits from both sides of the breakpoints.

Let a and b represent real numbers. Find the values of a and b such that f is continuous on the entire real number line, where f(x) is defined as: f(x) = { ax^2 + x - b, x ≤ 2; ax + b, 2 < x < 5; 2ax - 7, x ≥ 5 }.

This solution explores finding values of a and b such that the piecewise function f(x) is continuous across the entire real number line. It involves checking the limits at the breakpoints x = 2 and x = 5 and solving for the unknowns using continuity conditions.

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