Math Problem Statement
Solve this problem with all the written work./
In Exercises 57-60, complete parts (a), (b), and (c) for the piecewise-defined function. 57. c = 2, f(x) = { 3 - x, x < 2; x/2 + 1, x > 2 }.
Solution
Let's go through the problem you posted step by step. The given function is piecewise:
3 - x & \text{if } x < 2 \\ \frac{x}{2} + 1 & \text{if } x > 2 \end{cases}$$ ### Part (a): Draw the graph of $$f(x)$$ 1. **For $$x < 2$$**: The function is $$f(x) = 3 - x$$, which is a linear function with a slope of $$-1$$ and a y-intercept of 3. 2. **For $$x > 2$$**: The function is $$f(x) = \frac{x}{2} + 1$$, which is a linear function with a slope of $$\frac{1}{2}$$ and a y-intercept of 1. **Note**: There is no value of $$f(x)$$ defined at $$x = 2$$, so there will be an open circle at $$x = 2$$ on the graph. #### Graph description: - The first part, $$f(x) = 3 - x$$, starts from the left and goes until $$x = 2$$, where you will put an open circle. - The second part, $$f(x) = \frac{x}{2} + 1$$, starts from $$x = 2$$ with another open circle and continues for $$x > 2$$. ### Part (b): Determine $$\lim_{x \to 2^-} f(x)$$ and $$\lim_{x \to 2^+} f(x)$$ - **Left-hand limit** $$\lim_{x \to 2^-} f(x)$$: For $$x < 2$$, the function is $$f(x) = 3 - x$$. So, we evaluate the limit as $$x$$ approaches 2 from the left: $$\lim_{x \to 2^-} f(x) = 3 - 2 = 1$$ - **Right-hand limit** $$\lim_{x \to 2^+} f(x)$$: For $$x > 2$$, the function is $$f(x) = \frac{x}{2} + 1$$. So, we evaluate the limit as $$x$$ approaches 2 from the right: $$\lim_{x \to 2^+} f(x) = \frac{2}{2} + 1 = 1 + 1 = 2$$ ### Part (c): Does $$\lim_{x \to 2} f(x)$$ exist? For the limit $$\lim_{x \to 2} f(x)$$ to exist, the left-hand limit and the right-hand limit must be equal. In this case: - $$\lim_{x \to 2^-} f(x) = 1$$ - $$\lim_{x \to 2^+} f(x) = 2$$ Since these two limits are not equal, the limit $$\lim_{x \to 2} f(x)$$ **does not exist**. ### Summary of answers: - **(a)**: The graph consists of two linear pieces: one for $$x < 2$$ with a slope of $$-1$$ and the other for $$x > 2$$ with a slope of $$\frac{1}{2}$$, with open circles at $$x = 2$$. - **(b)**: $$\lim_{x \to 2^-} f(x) = 1$$ and $$\lim_{x \to 2^+} f(x) = 2$$. - **(c)**: $$\lim_{x \to 2} f(x)$$ does **not exist** because the left-hand and right-hand limits are not equal. Would you like more details or clarification on any part of the solution? Here are five related questions: 1. What conditions must be satisfied for a limit to exist at a point? 2. How can you determine if a piecewise function is continuous at a given point? 3. What is the difference between a left-hand limit and a right-hand limit? 4. How do open circles affect the continuity of a graph? 5. Can a function be discontinuous but still have one-sided limits? **Tip**: When dealing with piecewise functions, always check the behavior of the function on either side of the boundary point to determine limits and continuity.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise functions
Limits
Continuity
Formulas
Left-hand limit: lim_{x→c^-} f(x)
Right-hand limit: lim_{x→c^+} f(x)
Theorems
Limit existence theorem
Continuity condition for limits
Suitable Grade Level
Grades 10-12