Math Problem Statement
Certainly! To construct a function that satisfies all the given conditions, we can use piecewise definitions and appropriate functions that fit the behavior described. Hereβs a formula for a function π ( π₯ ) f(x) that meets all the conditions:
Vertical asymptote at π₯
β 2 x=β2 with a limit approaching 1. Value at π₯
0 x=0 should be 2. Value at π₯
3 x=3 should be 0. As π₯ x approaches 0 β 0 β , π ( π₯ ) f(x) approaches 2. As π₯ x approaches 0 + 0 + , π ( π₯ ) f(x) approaches infinity. Suggested Formula: \frac{1}{x + 2} + 1 & \text{for } x < 0 \text{ (approaching } -2 \text{ from the left side)} \ \frac{2}{x} + 2 & \text{for } x > 0 \text{ (approaching } 0 \text{ from the right side)} \ 0 & \text{for } x = 3 \text{ (discrete point)} \ 2 & \text{for } x = 0 \text{ (discrete point)} \end{cases} ] ### Explanation: 1. For ( x < -2 ): - Use ( \frac{1}{x + 2} + 1 ). This term approaches 1 as ( x ) approaches -2 from the left side. 2. For ( x > 0 ): - Use ( \frac{2}{x} + 2 ). As ( x ) approaches 0 from the right, this term goes to infinity, matching the behavior described. 3. At ( x = 0 ): - Set the function explicitly to 2 to satisfy the condition ( f(0) = 2 ). 4. At ( x = 3 ): - Set the function explicitly to 0 to meet the condition ( f(3) = 0 ). ### Piecewise Construction: - For ( x \to -2 ) from the left, ( \frac{1}{x + 2} + 1 ) approaches 1. - For ( x \to 0^- ), the function ( \frac{2}{x} + 2 ) approaches 2, satisfying ( \lim_{x \to 0^-} f(x) = 2 ). - For ( x \to 0^+ ), the function ( \frac{2}{x} + 2 ) goes to infinity, satisfying ( \lim_{x \to 0^+} f(x) = \infty ).
Solution
To clarify, the function that you're proposing seems to be structured as a piecewise function to satisfy various conditions, such as asymptotic behavior, specific values at certain points, and limits as approaches specific values.
However, there is a slight mix-up in the provided format. Let's clean up the construction of the function and present it in a clear and concise manner.
Given Conditions:
- Vertical asymptote at with .
- Value at should be 2.
- Value at should be 0.
- As approaches , approaches 2.
- As approaches , approaches infinity.
Suggested Piecewise Function:
\begin{cases} \frac{1}{x + 2} + 1 & \text{for } x < 0 \text{ (handles asymptote at } x = -2\text{ and limit as } x \to -2^-)\\ \frac{2}{x} + 2 & \text{for } 0 < x < 3 \text{ (approaches } \infty \text{ as } x \to 0^+ \text{ and matches } \lim_{x \to 0^-} f(x) = 2) \\ 0 & \text{for } x = 3 \text{ (specific value at } x = 3\text{)} \\ 2 & \text{for } x = 0 \text{ (specific value at } x = 0\text{)} \end{cases}$$ ### Explanation: 1. **For $$x < 0$$**: - The term $$\frac{1}{x + 2} + 1$$ is designed to create a vertical asymptote at $$x = -2$$. As $$x$$ approaches $$-2$$ from the left, $$\frac{1}{x + 2}$$ becomes very large (negative), and adding 1 shifts the limit so that the overall expression approaches 1. This fits the condition $$\lim_{x \to -2^-} f(x) = 1$$. 2. **For $$0 < x < 3$$**: - The term $$\frac{2}{x} + 2$$ handles the behavior as $$x$$ approaches 0 from the right. As $$x$$ gets closer to 0 from the right, $$\frac{2}{x}$$ becomes very large, making $$f(x)$$ approach infinity. This also matches the condition $$\lim_{x \to 0^-} f(x) = 2$$. 3. **At $$x = 0$$**: - The function is defined to be exactly 2 at $$x = 0$$ to meet the requirement $$f(0) = 2$$. 4. **At $$x = 3$$**: - The function is defined to be exactly 0 at $$x = 3$$ to meet the condition $$f(3) = 0$$. ### Summary: This piecewise function meets all the specified criteria and behaves as expected for each interval and at the points of interest. --- Do you want further details on any specific part of this function? Here are some related questions that could deepen your understanding: 1. How do piecewise functions work in general, and how are they evaluated? 2. What exactly happens at a vertical asymptote, and why does it affect the limit? 3. What is the difference between the left-hand and right-hand limits, and why are they important in this context? 4. How can we ensure continuity in a piecewise function, and why isn't this function continuous at all points? 5. How do limits and asymptotic behavior help in defining the shape and behavior of a function? **Tip:** When dealing with piecewise functions, carefully evaluate each interval separately, considering the specific behavior of the function within that region.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Limits
Asymptotes
Formulas
Piecewise function definition
Theorems
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Suitable Grade Level
Advanced High School