Math Problem Statement
Tentukan nilai a agar f(x) kontinu di x = 2 dengan f(x) = { (1 - 2a)/x jika x ≤ 2; a^2x^2 jika x > 2 }.
Solution
To solve this problem, we need to ensure that the function is continuous at . For a function to be continuous at a point, the following conditions must be met:
- must exist.
- The left-hand limit as must equal the right-hand limit as .
- The left-hand and right-hand limits must equal .
The function is defined as:
\begin{cases} \frac{1 - 2a}{x}, & \text{jika } x \leq 2 \\ a^2 x^2, & \text{jika } x > 2 \end{cases}$$ #### Step 1: Calculate $$f(2)$$. Since $$f(x) = \frac{1 - 2a}{x}$$ when $$x \leq 2$$, we can substitute $$x = 2$$ directly into this expression: $$f(2) = \frac{1 - 2a}{2}$$ #### Step 2: Calculate the left-hand limit as $$x \to 2^-$$. The left-hand limit is based on $$\frac{1 - 2a}{x}$$, so: $$\lim_{x \to 2^-} f(x) = \frac{1 - 2a}{2}$$ #### Step 3: Calculate the right-hand limit as $$x \to 2^+$$. For $$x > 2$$, we use $$f(x) = a^2 x^2$$, so the right-hand limit is: $$\lim_{x \to 2^+} f(x) = a^2 (2^2) = 4a^2$$ #### Step 4: Set the left-hand limit equal to the right-hand limit. To ensure continuity at $$x = 2$$, the left-hand and right-hand limits must be equal: $$\frac{1 - 2a}{2} = 4a^2$$ #### Step 5: Solve the equation. Now, solve for $$a$$ from the equation $$\frac{1 - 2a}{2} = 4a^2$$: $$1 - 2a = 8a^2$$ $$8a^2 + 2a - 1 = 0$$ This is a quadratic equation in $$a$$. We can solve it using the quadratic formula: $$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$b = 2$$, $$a = 8$$, and $$c = -1$$. Substitute these values into the quadratic formula: $$a = \frac{-2 \pm \sqrt{(2)^2 - 4(8)(-1)}}{2(8)}$$ $$a = \frac{-2 \pm \sqrt{4 + 32}}{16}$$ $$a = \frac{-2 \pm \sqrt{36}}{16}$$ $$a = \frac{-2 \pm 6}{16}$$ Thus, we have two solutions: $$a = \frac{-2 + 6}{16} = \frac{4}{16} = \frac{1}{4}$$ or $$a = \frac{-2 - 6}{16} = \frac{-8}{16} = \frac{-1}{2}$$ ### Conclusion: The values of $$a$$ that make $$f(x)$$ continuous at $$x = 2$$ are $$a = \frac{1}{4}$$ or $$a = \frac{-1}{2}$$. Would you like a more detailed explanation of any of these steps? ### Follow-up Questions: 1. What are the conditions for a function to be continuous at a point? 2. How is the quadratic formula applied to solve quadratic equations? 3. What is the significance of left-hand and right-hand limits in continuity? 4. Can a function be discontinuous even if it is defined at a specific point? 5. How would the solution change if the function were defined differently for $$x > 2$$? ### Tip: When solving quadratic equations, always check both roots, as both may be valid depending on the context of the problem.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity
Limits
Quadratic Equations
Formulas
Limit definition for continuity
Quadratic equation formula: ax^2 + bx + c = 0
Theorems
Quadratic formula
Definition of continuity at a point
Suitable Grade Level
Grades 11-12