Math Problem Statement
Tentukan nilai a agar f(x) kontiniu di x = 2 dengan f(x) = \frac{1 - 2a}{x} jika x \leq 2 dan f(x) = a^2x^2 jika x > 2.
Solution
The problem asks to find the value of such that the function is continuous at .
The piecewise function is given as:
\begin{cases} \frac{1 - 2a}{x}, & \text{for } x \leq 2 \\ a^2x^2, & \text{for } x > 2 \end{cases}$$ ### Step-by-step solution: To ensure that the function is continuous at $$x = 2$$, the following condition must be satisfied: $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)$$ #### 1. Left-hand limit $$\lim_{x \to 2^-} f(x)$$: For $$x \leq 2$$, $$f(x) = \frac{1 - 2a}{x}$$. Substituting $$x = 2$$: $$\lim_{x \to 2^-} f(x) = \frac{1 - 2a}{2}$$ #### 2. Right-hand limit $$\lim_{x \to 2^+} f(x)$$: For $$x > 2$$, $$f(x) = a^2x^2$$. Substituting $$x = 2$$: $$\lim_{x \to 2^+} f(x) = a^2(2^2) = 4a^2$$ #### 3. Continuity at $$x = 2$$: For continuity, the left-hand limit must equal the right-hand limit: $$\frac{1 - 2a}{2} = 4a^2$$ #### 4. Solve for $$a$$: Multiply both sides of the equation by 2 to eliminate the denominator on the left side: $$1 - 2a = 8a^2$$ Rearrange the equation: $$8a^2 + 2a - 1 = 0$$ This is a quadratic equation in $$a$$. Solve using the quadratic formula: $$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Where $$A = 8$$, $$B = 2$$, and $$C = -1$$. Substituting these values: $$a = \frac{-2 \pm \sqrt{2^2 - 4(8)(-1)}}{2(8)} = \frac{-2 \pm \sqrt{4 + 32}}{16} = \frac{-2 \pm \sqrt{36}}{16}$$ $$a = \frac{-2 \pm 6}{16}$$ Thus, the two possible solutions are: $$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$$ are $$\frac{1}{4}$$ or $$\frac{-1}{2}$$. --- Would you like further details or explanations? Here are five related questions you might find useful: 1. What are the conditions for a function to be continuous at a point? 2. How does the quadratic formula help in solving non-linear equations? 3. Can you have a function that is continuous but not differentiable at a point? 4. What are piecewise functions and how are their limits calculated? 5. How do you check for continuity for different types of functions? **Tip:** When solving for continuity at a point, always check both the left-hand and right-hand limits.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity
Piecewise Functions
Limits
Quadratic Equations
Formulas
Limit: \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)
Quadratic equation: ax^2 + bx + c = 0
Left-hand limit: \frac{1 - 2a}{2}
Right-hand limit: a^2(2^2)
Theorems
Continuity Theorem
Quadratic Formula
Suitable Grade Level
Grade 11-12 or early college level