Consider the following piecewise function f (x), where A and B are constants.
f (x) =



Ax + B if x < −2
x2 + 2Ax − B if − 2 ≤ x < 1
4 if x > 1
Determine all values of the constants A and B so that lim
x→−2 f (x) and lim
x→1 f (x) both exist.

To determine the values of constants \( A \) and \( B \) so that the limits of the piecewise function \( f(x) \) at \( x = -2 \) and \( x = 1 \) both exist, we need to analyze the continuity of the function at these points.

### Given Piecewise Function:
\[
f(x) =
\begin{cases}
Ax + B & \text{if } x < -2 \\
x^2 + 2Ax - B & \text{if } -2 \leq x < 1 \\
4 & \text{if } x > 1
\end{cases}
\]

### 1. Continuity at \( x = -2 \):

For the limit \( \lim_{x \to -2} f(x) \) to exist, the function must approach the same value from both sides of \( x = -2 \).

- **Left-hand limit at \( x = -2 \)** (using the first piece \( Ax + B \)):

\[
\lim_{x \to -2^-} f(x) = A(-2) + B = -2A + B
\]

- **Right-hand limit at \( x = -2 \)** (using the second piece \( x^2 + 2Ax - B \)):

\[
\lim_{x \to -2^+} f(x) = (-2)^2 + 2A(-2) - B = 4 - 4A - B
\]

For continuity at \( x = -2 \), these two limits must be equal:

\[
-2A + B = 4 - 4A - B
\]

Simplifying this equation:

\[
-2A + B = 4 - 4A - B
\]
\[
-2A + B + 4A + B = 4
\]
\[
2A + 2B = 4
\]
\[
A + B = 2 \quad \text{(Equation 1)}
\]

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

For the limit \( \lim_{x \to 1} f(x) \) to exist, the function must again approach the same value from both sides of \( x = 1 \).

- **Left-hand limit at \( x = 1 \)** (using the second piece \( x^2 + 2Ax - B \)):

\[
\lim_{x \to 1^-} f(x) = 1^2 + 2A(1) - B = 1 + 2A - B
\]

- **Right-hand limit at \( x = 1 \)** (using the constant value 4):

\[
\lim_{x \to 1^+} f(x) = 4
\]

For continuity at \( x = 1 \), these two limits must be equal:

\[
1 + 2A - B = 4
\]

Simplifying this equation:

\[
2A - B = 3 \quad \text{(Equation 2)}
\]

### 3. Solving for \( A \) and \( B \):

We now have a system of two equations:

1. \( A + B = 2 \)
2. \( 2A - B = 3 \)

Adding these two equations:

\[
(A + B) + (2A - B) = 2 + 3
\]
\[
3A = 5
\]
\[
A = \frac{5}{3}
\]

Substitute \( A = \frac{5}{3} \) into \( A + B = 2 \):

\[
\frac{5}{3} + B = 2
\]
\[
B = 2 - \frac{5}{3} = \frac{6}{3} - \frac{5}{3} = \frac{1}{3}
\]

### Final Answer:
The values of the constants \( A \) and \( B \) that ensure the limits at both \( x = -2 \) and \( x = 1 \) exist are:
\[
A = \frac{5}{3}, \quad B = \frac{1}{3}
\]

Would you like further clarification on any step or additional details?

Here are some related questions:
1. What happens if \( A \) and \( B \) are different from the values found?
2. Can you explain the importance of continuity in piecewise functions?
3. How do we interpret one-sided limits in real-world applications?
4. What happens to the function \( f(x) \) if \( x = -2 \) or \( x = 1 \)?
5. What is the role of the constant value in the third piece of the function?

**Tip:** Continuity at a point requires that the left-hand and right-hand limits at that point are equal and that the function is defined at that point.

Consider the following piecewise function f(x), where A and B are constants. f(x) = { Ax + B if x < −2; x^2 + 2Ax − B if −2 ≤ x < 1; 4 if x > 1. Determine all values of the constants A and B so that lim x→−2 f(x) and lim x→1 f(x) both exist.

Learn how to find the constants A and B that ensure continuity in a piecewise function. This solution covers continuity at points, limits from both sides, and solving a system of linear equations. Suitable for advanced high school and college-level students.

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