Math Problem Statement
Solution
The piecewise function in the image is written as and asks to find the values of and that make continuous everywhere. Here's the breakdown of the function:
\begin{cases} \frac{x^2 - 4}{x - 2} & \text{for } x < 2 \\ ax^2 + bx + 3 & \text{for } 2 \leq x \leq 3 \\ 2x - a + b & \text{for } x > 3 \end{cases}$$ ### Continuity Conditions For $$k(x)$$ to be continuous everywhere, it must be continuous at the points where the piecewise function changes, which are at $$x = 2$$ and $$x = 3$$. #### 1. Continuity at $$x = 2$$ - The first part is $$\frac{x^2 - 4}{x - 2}$$, which simplifies using factorization: $$\frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2} = x + 2 \quad \text{for} \quad x \neq 2$$ So, as $$x \to 2$$, the limit of the first part is $$2 + 2 = 4$$. For continuity at $$x = 2$$, we need $$k(2)$$ from the second piece (since the second piece applies at $$x = 2$$) to equal 4. Therefore, substitute $$x = 2$$ into $$ax^2 + bx + 3$$: $$a(2^2) + b(2) + 3 = 4$$ Simplifying: $$4a + 2b + 3 = 4$$ $$4a + 2b = 1 \quad \text{(Equation 1)}$$ #### 2. Continuity at $$x = 3$$ For continuity at $$x = 3$$, the values of the second and third parts must match at $$x = 3$$. - Substituting $$x = 3$$ into the second part $$ax^2 + bx + 3$$: $$a(3^2) + b(3) + 3 = 9a + 3b + 3$$ - The third part is $$2x - a + b$$. Substituting $$x = 3$$: $$2(3) - a + b = 6 - a + b$$ For continuity at $$x = 3$$, these two expressions must be equal: $$9a + 3b + 3 = 6 - a + b$$ Simplifying: $$9a + 3b + 3 = 6 - a + b$$ $$10a + 2b = 3 \quad \text{(Equation 2)}$$ ### Solving the System of Equations Now, solve the system of equations: 1. $$4a + 2b = 1$$ 2. $$10a + 2b = 3$$ Subtract Equation 1 from Equation 2: $$(10a + 2b) - (4a + 2b) = 3 - 1$$ $$6a = 2$$ $$a = \frac{1}{3}$$ Substitute $$a = \frac{1}{3}$$ into Equation 1: $$4\left(\frac{1}{3}\right) + 2b = 1$$ $$\frac{4}{3} + 2b = 1$$ $$2b = 1 - \frac{4}{3} = \frac{3}{3} - \frac{4}{3} = -\frac{1}{3}$$ $$b = -\frac{1}{6}$$ ### Conclusion The values of $$a$$ and $$b$$ that make $$k(x)$$ continuous everywhere are: $$a = \frac{1}{3}, \quad b = -\frac{1}{6}$$ Would you like a detailed explanation of the next steps or any additional clarification? Here are 5 related questions for further practice: 1. How do you check if a function is continuous at a point? 2. What is the difference between a piecewise function and a regular function? 3. Can a piecewise function have a discontinuity at a point where the pieces meet? 4. What are the conditions for a function to be differentiable? 5. How would the continuity conditions change if the piecewise function had different intervals? **Tip:** When solving for continuity in piecewise functions, always check the behavior of the function at the boundaries between different expressions.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity
Piecewise Functions
Solving Systems of Equations
Formulas
Limit of a function: lim_{x -> c} f(x) = L
Simplified limit of (x^2 - 4) / (x - 2): x + 2
Quadratic equation: ax^2 + bx + 3
Linear equation: 2x - a + b
Theorems
Definition of Continuity: A function is continuous at x = c if lim_{x -> c^-} f(x) = lim_{x -> c^+} f(x) = f(c)
Suitable Grade Level
Grades 11-12
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