Math Problem Statement
Evaluate the limits and find the value of c for which the function f(x) is continuous at x = 1. The piecewise function is defined as f(x) = 4x + 2c for x < 1, and f(x) = cx^2 + 1 for x ≥ 1.
Solution
Let's analyze the mathematical problem provided in the image.
Given Function:
\begin{cases} 4x + 2c & \text{untuk } x < 1 \\ cx^2 + 1 & \text{untuk } x \geq 1 \end{cases}$$ The question asks to evaluate the limit of $$f(x)$$ as $$x \to 1$$ and other conditions related to $$c$$. ### Key Concept: In order for $$f(x)$$ to have a limit at $$x = 1$$, both the left-hand limit (as $$x \to 1^-$$) and the right-hand limit (as $$x \to 1^+$$) must exist and be equal. 1. **Left-hand limit as $$x \to 1^-$$**: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (4x + 2c) = 4(1) + 2c = 4 + 2c$$ 2. **Right-hand limit as $$x \to 1^+$$**: $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (cx^2 + 1) = c(1^2) + 1 = c + 1$$ To have the function continuous and a limit at $$x = 1$$, we require: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)$$ This gives the equation: $$4 + 2c = c + 1$$ Solving for $$c$$: $$4 + 2c = c + 1 \\ 2c - c = 1 - 4 \\ c = -3$$ ### Now let's analyze the statements: - **a.** $$\lim_{x \to 1^+} f(x) = 10$$ for $$c$$ positive: - If $$c$$ is positive, the right-hand limit would not be $$10$$ (as this limit depends on the value of $$c$$). - **False.** - **b.** $$\lim_{x \to 1^+} f(x) = 2$$ for $$c$$ positive: - From $$\lim_{x \to 1^+} f(x) = c + 1$$, if $$c$$ is positive, the limit cannot be 2. - **False.** - **c.** $$\lim_{x \to 1^+} f(x) = -5$$ for $$c$$ negative: - If $$c = -3$$, the right-hand limit is $$\lim_{x \to 1^+} f(x) = -3 + 1 = -2$$, not $$-5$$. - **False.** - **d.** $$\lim_{x \to 1^-} f(x) = 5$$ for $$c$$ positive: - The left-hand limit depends on $$4 + 2c$$. If $$c = -3$$, the left-hand limit would not be 5. - **False.** - **e.** $$c^2 + 2c - 3 = 0$$: - Solving this quadratic equation: $$c^2 + 2c - 3 = 0 \implies (c - 1)(c + 3) = 0$$ So, $$c = 1$$ or $$c = -3$$. - **True.** - **f.** $$c^2 - 2c - 3 = 0$$: - This quadratic doesn't hold for $$c = -3$$. - **False.** - **g.** $$\lim_{x \to 1^+} f(x) = 2$$ for $$c$$ negative: - For $$c = -3$$, the right-hand limit is $$-2$$, not 2. - **False.** - **h.** $$\lim_{x \to 1^+} f(x) = -2$$ for $$c$$ negative: - For $$c = -3$$, the right-hand limit is $$-2$$. - **True.** ### Conclusion: The correct statements are: - **e.** $$c^2 + 2c - 3 = 0$$ - **h.** $$\lim_{x \to 1^+} f(x) = -2$$ for $$c$$ negative Would you like further details on the solution or clarification on any point? --- ### Follow-up Questions: 1. What are the conditions for a function to be continuous at a point? 2. Can you explain how limits from the left-hand and right-hand side impact continuity? 3. How do you solve quadratic equations like $$c^2 + 2c - 3 = 0$$? 4. How would the limits change if $$c = 1$$ instead of $$-3$$? 5. Why is it necessary for the limits to match at $$x = 1$$ for continuity? ### Tip: When solving limit problems with piecewise functions, always check both sides of the limit to ensure consistency for continuity.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Limits
Continuity
Piecewise Functions
Quadratic Equations
Formulas
Left-hand limit: lim (x -> 1^-) f(x) = 4x + 2c
Right-hand limit: lim (x -> 1^+) f(x) = cx^2 + 1
Continuity condition: lim (x -> 1^-) f(x) = lim (x -> 1^+) f(x)
Quadratic equation: c^2 + 2c - 3 = 0
Theorems
Limit Theorem for Piecewise Functions
Continuity Conditions
Suitable Grade Level
Grades 11-12