Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![### Sketching the Function \( f(x) = \lfloor x + 2 \rfloor - 1 \) on the Interval \([-2, 2]\)
**Problem Statement:**
Sketch \( f(x) = \lfloor x + 2 \rfloor - 1 \) on the interval \([-2, 2]\).
**Function Breakdown:**
This function involves the floor function, denoted as \( \lfloor \cdot \rfloor \), which rounds down to the nearest integer. Specifically, \( f(x) \) can be described in two parts:
1. **Transformation of the Input:**
The expression \( x + 2 \) adjusts the input \( x \) by adding 2. So, every value of \( x \) is increased by 2 before any other operations are applied.
2. **Applying the Floor Function:**
The floor function, \( \lfloor x + 2 \rfloor \), takes the adjusted input and rounds it down to the nearest integer.
3. **Final Adjustment:**
After the floor function has been applied, 1 is subtracted from the result.
**Interval Analysis:**
We consider the function on the range \([-2, 2]\). Here are the steps to generate the sketch:
- **For \( x \in [-2, -1) \):**
\( x + 2 \) ranges from \( 0 \) to \( 1 \). So, \( \lfloor x + 2 \rfloor = \lfloor 0 \leq x + 2 < 1 \rfloor = 0 \).
Therefore, \( f(x) = 0 - 1 = -1 \).
- **For \( x \in [-1, 0) \):**
\( x + 2 \) ranges from \( 1 \) to \( 2 \). So, \( \lfloor 1 \leq x + 2 < 2 \rfloor = 1 \).
Therefore, \( f(x) = 1 - 1 = 0 \).
- **For \( x \in [0, 1) \):**
\( x + 2 \) ranges from \( 2 \) to \( 3 \). So, \( \lfloor 2 \leq x + 2 < 3 \rfloor](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F70fe4f99-9e69-4bef-afe6-80d54e8903ee%2Fcbfa86a2-33cf-4b52-b308-578fd28bef8e%2F84j49ku_processed.png&w=3840&q=75)
Transcribed Image Text:### Sketching the Function \( f(x) = \lfloor x + 2 \rfloor - 1 \) on the Interval \([-2, 2]\)
**Problem Statement:**
Sketch \( f(x) = \lfloor x + 2 \rfloor - 1 \) on the interval \([-2, 2]\).
**Function Breakdown:**
This function involves the floor function, denoted as \( \lfloor \cdot \rfloor \), which rounds down to the nearest integer. Specifically, \( f(x) \) can be described in two parts:
1. **Transformation of the Input:**
The expression \( x + 2 \) adjusts the input \( x \) by adding 2. So, every value of \( x \) is increased by 2 before any other operations are applied.
2. **Applying the Floor Function:**
The floor function, \( \lfloor x + 2 \rfloor \), takes the adjusted input and rounds it down to the nearest integer.
3. **Final Adjustment:**
After the floor function has been applied, 1 is subtracted from the result.
**Interval Analysis:**
We consider the function on the range \([-2, 2]\). Here are the steps to generate the sketch:
- **For \( x \in [-2, -1) \):**
\( x + 2 \) ranges from \( 0 \) to \( 1 \). So, \( \lfloor x + 2 \rfloor = \lfloor 0 \leq x + 2 < 1 \rfloor = 0 \).
Therefore, \( f(x) = 0 - 1 = -1 \).
- **For \( x \in [-1, 0) \):**
\( x + 2 \) ranges from \( 1 \) to \( 2 \). So, \( \lfloor 1 \leq x + 2 < 2 \rfloor = 1 \).
Therefore, \( f(x) = 1 - 1 = 0 \).
- **For \( x \in [0, 1) \):**
\( x + 2 \) ranges from \( 2 \) to \( 3 \). So, \( \lfloor 2 \leq x + 2 < 3 \rfloor
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