Math Problem Statement
Find the two values of b for which f(x) is a continuous function at -1, given f(x) = {b - 2x if x < -1, 6/(x - b) if x ≥ -1}.
Solution
The problem asks to find the two values of for which the function is continuous at . The function is defined as:
\begin{cases} b - 2x & \text{if } x < -1 \\ \frac{6}{x - b} & \text{if } x \geq -1 \end{cases}$$ To ensure continuity at $$x = -1$$, the left-hand limit and right-hand limit at $$x = -1$$ must equal the function value at that point. Let's calculate these limits and set them equal to each other. ### 1. Left-hand limit ($$x \to -1^{-}$$): For $$x < -1$$, the function is given by $$f(x) = b - 2x$$. Therefore, the left-hand limit as $$x \to -1$$ is: $$\lim_{x \to -1^{-}} f(x) = b - 2(-1) = b + 2$$ ### 2. Right-hand limit ($$x \to -1^{+}$$): For $$x \geq -1$$, the function is given by $$f(x) = \frac{6}{x - b}$$. Therefore, the right-hand limit as $$x \to -1$$ is: $$\lim_{x \to -1^{+}} f(x) = \frac{6}{-1 - b} = \frac{6}{-1 - b}$$ ### 3. Continuity condition: For $$f(x)$$ to be continuous at $$x = -1$$, the left-hand limit must equal the right-hand limit. Thus, we set: $$b + 2 = \frac{6}{-1 - b}$$ ### 4. Solving for $$b$$: Now, let's solve this equation for $$b$$. Multiply both sides by $$-1 - b$$ to get rid of the denominator: $$(b + 2)(-1 - b) = 6$$ Expand both sides: $$-(b + 2)(b + 1) = 6$$ Expanding the left-hand side: $$-(b^2 + 3b + 2) = 6$$ Simplifying: $$-b^2 - 3b - 2 = 6$$ Move everything to one side: $$b^2 + 3b + 8 = 0$$ ### 5. Solving the quadratic equation: To solve $$b^2 + 3b + 8 = 0$$, we use the quadratic formula: $$b = \frac{-3 \pm \sqrt{3^2 - 4(1)(8)}}{2(1)}$$ Simplifying: $$b = \frac{-3 \pm \sqrt{9 - 32}}{2}$$ $$b = \frac{-3 \pm \sqrt{-23}}{2}$$ Since the discriminant is negative, the solutions are complex: $$b = \frac{-3 \pm i\sqrt{23}}{2}$$ Thus, the two values of $$b$$ are: $$b = \frac{-3 + i\sqrt{23}}{2} \quad \text{and} \quad b = \frac{-3 - i\sqrt{23}}{2}$$ Among these two values, the one with the greater absolute value is typically the one with the positive imaginary part, i.e., $$b = \frac{-3 + i\sqrt{23}}{2}$$. --- Do you want more details, or have any further questions? Here are five related questions to extend your learning: 1. How do you compute the modulus (absolute value) of a complex number? 2. What does it mean for a function to be continuous at a point? 3. Can you solve this quadratic equation by factoring? 4. What happens if a function is discontinuous at a point? How does it affect graph behavior? 5. How can we generalize continuity conditions for piecewise functions? **Tip:** When solving quadratic equations, always check the discriminant ($$b^2 - 4ac$$) to determine if the roots are real or complex.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity
Limits
Quadratic Equations
Piecewise Functions
Formulas
Continuity condition: left-hand limit = right-hand limit
Quadratic equation: ax^2 + bx + c = 0
Theorems
Quadratic formula
Continuity of piecewise functions
Suitable Grade Level
Grades 11-12 (or first-year calculus students)
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