Math Problem Statement
Tentukan nilai a dan b agar f(x) kontinu pada domainnya. The piecewise function f(x) is defined as follows:
- f(x) = (x^2 - 4)/(x - 2), for x < 2
- f(x) = ax^2 - bx + 3, for 2 ≤ x < 3
- f(x) = 2x - a + b, for x ≥ 3
Solution
The question asks for the values of and such that the function is continuous on its domain. The function is defined piecewise as:
\begin{cases} \frac{x^2 - 4}{x - 2} & ; x < 2 \\ ax^2 - bx + 3 & ; 2 \leq x < 3 \\ 2x - a + b & ; x \geq 3 \end{cases}$$ ### Step 1: Simplify $$f(x)$$ for $$x < 2$$ The first piece is $$f(x) = \frac{x^2 - 4}{x - 2}$$ for $$x < 2$$. Notice that: $$x^2 - 4 = (x - 2)(x + 2)$$ Thus, $$f(x)$$ for $$x < 2$$ simplifies to: $$f(x) = x + 2 \quad \text{for} \quad x < 2$$ since the factor $$x - 2$$ cancels out. ### Step 2: Continuity at $$x = 2$$ To ensure the function is continuous at $$x = 2$$, the value of the left-hand limit and right-hand limit must be equal at $$x = 2$$. - Left-hand limit as $$x \to 2^-$$ (from the first piece): $$\lim_{x \to 2^-} f(x) = 2 + 2 = 4$$ - Right-hand limit as $$x \to 2^+$$ (from the second piece): $$\lim_{x \to 2^+} f(x) = a(2)^2 - b(2) + 3 = 4a - 2b + 3$$ Setting these equal for continuity at $$x = 2$$: $$4 = 4a - 2b + 3$$ $$4a - 2b = 1 \quad \text{(Equation 1)}$$ ### Step 3: Continuity at $$x = 3$$ Similarly, for the function to be continuous at $$x = 3$$, the values of the second and third pieces must match at $$x = 3$$. - Left-hand limit as $$x \to 3^-$$ (from the second piece): $$\lim_{x \to 3^-} f(x) = a(3)^2 - b(3) + 3 = 9a - 3b + 3$$ - Right-hand limit as $$x \to 3^+$$ (from the third piece): $$\lim_{x \to 3^+} f(x) = 2(3) - a + b = 6 - a + b$$ Setting these equal for continuity at $$x = 3$$: $$9a - 3b + 3 = 6 - a + b$$ $$10a - 4b = 3 \quad \text{(Equation 2)}$$ ### Step 4: Solve the system of equations We now solve the system of equations: 1. $$4a - 2b = 1$$ 2. $$10a - 4b = 3$$ We can multiply the first equation by 2 to align it with the second equation: $$8a - 4b = 2$$ Now subtract the second equation from this: $$(8a - 4b) - (10a - 4b) = 2 - 3$$ $$-2a = -1$$ $$a = \frac{1}{2}$$ Substitute $$a = \frac{1}{2}$$ into Equation 1: $$4\left(\frac{1}{2}\right) - 2b = 1$$ $$2 - 2b = 1$$ $$-2b = -1$$ $$b = \frac{1}{2}$$ ### Final Answer: The values of $$a$$ and $$b$$ such that $$f(x)$$ is continuous are: $$a = \frac{1}{2}, \quad b = \frac{1}{2}$$ Would you like more details on any part of this solution? Here are 5 related questions to expand on this: 1. How do you determine continuity for piecewise functions? 2. Why is factoring useful in simplifying rational expressions? 3. What is the significance of left-hand and right-hand limits? 4. How do we verify the values of $$a$$ and $$b$$ ensure differentiability at $$x = 2$$ and $$x = 3$$? 5. Can piecewise functions have discontinuities in their domain? **Tip**: Always simplify piecewise expressions as much as possible before working with limits. This can make solving continuity conditions much easier!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity of Functions
Piecewise Functions
Limits
Algebraic Equations
Formulas
f(x) = (x^2 - 4) / (x - 2)
f(x) = ax^2 - bx + 3
f(x) = 2x - a + b
Continuity: lim(x -> c^-) = lim(x -> c^+)
Theorems
Limit Theorem
Continuity of Piecewise Functions
Suitable Grade Level
Grade 10-12 (High School)
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