Math Problem Statement
Determine the constants a and b such that the piecewise function f(x) is continuous for all x.
Solution
The piecewise function provided is:
x & \text{for } x < -1 \text{ and } x > 1 \\ x^2 + ax + b & \text{for } -1 \leq x \leq 1 \end{cases}$$ This function is defined differently for different intervals of $$x$$. Let's explore its properties, especially continuity and differentiability, which often require matching conditions at the boundaries ($$x = -1$$ and $$x = 1$$). ### Continuity Check: To ensure continuity at $$x = -1$$ and $$x = 1$$, the function values from both intervals must match. 1. **At $$x = -1$$**: - From $$x < -1$$: $$f(-1) = -1$$ - From $$-1 \leq x \leq 1$$: $$f(-1) = (-1)^2 + a(-1) + b = 1 - a + b$$ For continuity at $$x = -1$$, $$-1 = 1 - a + b \implies a - b = 2$$ 2. **At $$x = 1$$**: - From $$x > 1$$: $$f(1) = 1$$ - From $$-1 \leq x \leq 1$$: $$f(1) = 1^2 + a(1) + b = 1 + a + b$$ For continuity at $$x = 1$$, $$1 = 1 + a + b \implies a + b = 0$$ ### Solving for $$a$$ and $$b$$: From the two equations: 1. $$a - b = 2$$ 2. $$a + b = 0$$ Adding and subtracting these equations: - $$2a = 2 \implies a = 1$$ - Substituting $$a = 1$$ into $$a + b = 0$$: $$1 + b = 0 \implies b = -1$$ Thus, $$a = 1$$ and $$b = -1$$. ### Final Function: $$f(x) = \begin{cases} x & \text{for } x < -1 \text{ and } x > 1 \\ x^2 + x - 1 & \text{for } -1 \leq x \leq 1 \end{cases}$$ Would you like a detailed explanation of the steps or have any questions? Here are 5 related questions for further exploration: 1. How do we check if the function is differentiable at $$x = -1$$ and $$x = 1$$? 2. What is the graphical interpretation of a piecewise function's continuity and differentiability? 3. How can we generalize the process of finding constants $$a$$ and $$b$$ for other piecewise functions? 4. How would the solution change if the intervals of the piecewise function were different? 5. Can we explore the second derivative for the interval $$-1 \leq x \leq 1$$ and its implications? **Tip**: Always ensure the boundary points of a piecewise function are evaluated carefully for both continuity and differentiability.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Continuity
Solving Linear Equations
Formulas
Continuity condition: f(a-) = f(a+)
Solving linear equations
Theorems
Continuity at a point
Suitable Grade Level
Grades 10-12