Math Problem Statement
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.
Solution
To determine the values of constants and so that the limits of the piecewise function at and both exist, we need to analyze the continuity of the function at these points.
Given Piecewise Function:
\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.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity
Limits
Piecewise Functions
Systems of Equations
Formulas
Left-hand limit: lim_{x→-2^-} f(x) = -2A + B
Right-hand limit: lim_{x→-2^+} f(x) = 4 - 4A - B
Left-hand limit: lim_{x→1^-} f(x) = 1 + 2A - B
Right-hand limit: lim_{x→1^+} f(x) = 4
Theorems
Continuity Theorem
Limits Definition
System of Linear Equations
Suitable Grade Level
College/Advanced High School
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