https://file.aiquickdraw.com/mathbot/file/1726001935942ycacwfm3.jpg

Solve this problem with all the written work./

Let's go through the problem you posted step by step. The given function is piecewise:

\[
f(x) = \begin{cases} 
3 - x & \text{if } x < 2 \\
\frac{x}{2} + 1 & \text{if } x > 2
\end{cases}
\]

### 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.

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 }.

Learn how to solve a piecewise-defined function involving limits at x = 2. This problem demonstrates key concepts in limits, continuity, and solving piecewise functions. Ideal for high school students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation