Math Problem Statement
Find the values of a and b that make k(x) continuous everywhere for the given piecewise function:
k(x) =
(x^2 - 4) / (x - 2) for x < 2
ax^2 + bx + 3 for 2 ≤ x ≤ 3
2x - a + b for x > 3
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