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The piecewise function you have provided is:

\[
f(x) = \begin{cases} 
ax + b + a, & x < -2, \\
(a - b)x - 2a, & -2 \leq x \leq 1, \\
(a - 1)x + a + 4, & x > 1.
\end{cases}
\]

To analyze this function, we would typically need to ensure it's continuous across the given intervals at \( x = -2 \) and \( x = 1 \). Continuity requires that the function values from the left and right at these points match up. 

Let's break this down:

### 1. Continuity at \( x = -2 \)
To ensure continuity at \( x = -2 \), we need the left-hand limit to equal the right-hand limit:

- Left-hand limit (\( x < -2 \)): 
\[
f(x) = ax + b + a.
\]
Substituting \( x = -2 \):
\[
f(-2^-) = a(-2) + b + a = -2a + b + a = -a + b.
\]

- Right-hand limit (\( x \geq -2 \)): 
\[
f(x) = (a - b)x - 2a.
\]
Substituting \( x = -2 \):
\[
f(-2^+) = (a - b)(-2) - 2a = -2(a - b) - 2a = -2a + 2b - 2a = -4a + 2b.
\]

For continuity at \( x = -2 \):
\[
-a + b = -4a + 2b.
\]
This simplifies to:
\[
3a = b.
\]

### 2. Continuity at \( x = 1 \)
Now, let's ensure continuity at \( x = 1 \).

- Left-hand limit (\( x \leq 1 \)): 
\[
f(x) = (a - b)x - 2a.
\]
Substituting \( x = 1 \):
\[
f(1^-) = (a - b)(1) - 2a = a - b - 2a = -a - b.
\]

- Right-hand limit (\( x > 1 \)): 
\[
f(x) = (a - 1)x + a + 4.
\]
Substituting \( x = 1 \):
\[
f(1^+) = (a - 1)(1) + a + 4 = a - 1 + a + 4 = 2a + 3.
\]

For continuity at \( x = 1 \):
\[
-a - b = 2a + 3.
\]
This simplifies to:
\[
-3a - b = 3.
\]

### System of Equations
We now have two equations from the continuity conditions:

1. \( 3a = b \),
2. \( -3a - b = 3 \).

Substitute \( b = 3a \) into the second equation:
\[
-3a - (3a) = 3,
\]
\[
-6a = 3,
\]
\[
a = -\frac{1}{2}.
\]

Substituting \( a = -\frac{1}{2} \) into \( b = 3a \):
\[
b = 3 \left( -\frac{1}{2} \right) = -\frac{3}{2}.
\]

### Final Values
Thus, the values of \( a \) and \( b \) that ensure continuity are:
\[
a = -\frac{1}{2}, \quad b = -\frac{3}{2}.
\]

Would you like further details or have any questions?

Here are 5 related questions to explore further:

1. How do we check differentiability of the piecewise function at \( x = -2 \) and \( x = 1 \)?
2. How would the function behave if \( a \) and \( b \) were different?
3. Can you verify the continuity of the function by graphing it?
4. How would discontinuities affect the nature of the function?
5. Can we determine if the function is differentiable everywhere, or just continuous?

**Tip:** Continuity at a point doesn't always imply differentiability. Checking the derivative at the transition points is important for smoothness.

Given a piecewise function:

f(x) = {
  ax + b + a,    x < -2
  (a - b)x - 2a,  -2 ≤ x ≤ 1
  (a - 1)x + a + 4,   x > 1

Determine the values of 'a' and 'b' to ensure continuity at x = -2 and x = 1.

Learn how to find the values of 'a' and 'b' that ensure the continuity of a piecewise function. This problem involves analyzing left and right-hand limits for x = -2 and x = 1 and solving a system of equations to guarantee continuous behavior across the intervals. Suitable for undergraduate students studying calculus.

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