Math Problem Statement
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.
Solution
The piecewise function you have provided is:
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.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise functions
Continuity
Limits
Formulas
Left-hand limit at x = -2: f(-2^-) = ax + b + a
Right-hand limit at x = -2: f(-2^+) = (a - b)x - 2a
Left-hand limit at x = 1: f(1^-) = (a - b)x - 2a
Right-hand limit at x = 1: f(1^+) = (a - 1)x + a + 4
Theorems
Continuity theorem
Suitable Grade Level
Undergraduate calculus