Math Problem Statement
Tentukan nilai a agar f(x) kontinu di x = 2 dengan f(x) = \begin{cases} \frac{1 - 2a}{x}, & jika \ x \leq 2 \ \ a^2x^2, & jika \ x > 2 \end{cases}
Solution
To solve the problem and determine the value of such that the function is continuous at , we need to ensure that the left-hand limit and right-hand limit of at are equal, and that they also equal .
The function is given as:
\begin{cases} \frac{1 - 2a}{x}, & \text{if } x \leq 2 \\ a^2x^2, & \text{if } x > 2 \end{cases}$$ ### Continuity condition: To be continuous at $$x = 2$$, we require: $$\lim_{x \to 2^-} f(x) = f(2) = \lim_{x \to 2^+} f(x)$$ 1. **Left-hand limit (as $$x \to 2^-$$):** $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{1 - 2a}{x} = \frac{1 - 2a}{2}$$ 2. **Right-hand limit (as $$x \to 2^+$$):** $$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} a^2x^2 = a^2(2)^2 = 4a^2$$ 3. **Function value at $$x = 2$$:** From the first piece of the function for $$x \leq 2$$, we have: $$f(2) = \frac{1 - 2a}{2}$$ ### Setting the limits equal: For continuity at $$x = 2$$, we equate the left-hand and right-hand limits: $$\frac{1 - 2a}{2} = 4a^2$$ ### Solving for $$a$$: 1. Multiply both sides by 2 to eliminate the denominator: $$1 - 2a = 8a^2$$ 2. Rearrange the equation: $$8a^2 + 2a - 1 = 0$$ 3. Solve the quadratic equation using the quadratic formula: $$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $$a = 8$$, $$b = 2$$, and $$c = -1$$. Substituting the values: $$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} = \frac{-2 \pm 6}{16}$$ 4. This gives two solutions: $$a_1 = \frac{-2 + 6}{16} = \frac{4}{16} = \frac{1}{4}$$ $$a_2 = \frac{-2 - 6}{16} = \frac{-8}{16} = -\frac{1}{2}$$ Thus, the values of $$a$$ that make the function continuous at $$x = 2$$ are $$a = \frac{1}{4}$$ and $$a = -\frac{1}{2}$$. Would you like to go deeper into any part of this explanation or have any questions? Here are 5 follow-up questions for you: 1. Why is the condition for continuity important in functions like this? 2. How does the quadratic formula help in solving equations with higher degrees? 3. What is the significance of having two possible values for $$a$$? 4. Can a function be continuous but not differentiable at a point? Why or why not? 5. How would you graph the function $$f(x)$$ for different values of $$a$$? **Tip:** When solving quadratic equations, always simplify the equation first, and double-check your calculations when using the quadratic formula.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Calculus
Continuity
Limits
Quadratic Equations
Formulas
\lim_{x \to c^-} f(x) = f(c) = \lim_{x \to c^+} f(x)
Quadratic equation formula: ax^2 + bx + c = 0
Theorems
Continuity theorem
Quadratic formula
Suitable Grade Level
Grades 10-12