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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![### Piecewise Function Analysis
**Problem Statement:**
Sketch the function \( f(x) \) and provide the domain and range. The function is defined piecewise as follows:
\[ f(x) = \begin{cases}
x^2 + 1 & \text{if } x < -1 \\
x + 1 & \text{if } x \geq -1
\end{cases} \]
**Instructions for Graphing:**
1. **For \( x < -1 \):**
- The function is \( f(x) = x^2 + 1 \).
- This is a parabola that opens upwards with its vertex shifted 1 unit up from the origin.
2. **For \( x \geq -1 \):**
- The function is \( f(x) = x + 1 \).
- This is a linear function with a slope of 1 and y-intercept of 1.
3. **Boundary at \( x = -1 \):**
- Evaluate the function at the boundary point \( x = -1 \).
- For \( x = -1 \), the quadratic segment is \( f(-1) = (-1)^2 + 1 = 2 \).
- Also, for \( x = -1 \), the linear segment is \( f(-1) = -1 + 1 = 0 \).
**Graphical Explanation:**
- Plot the parabola \( f(x) = x^2 + 1 \) for \( x < -1 \). Ensure it does not include the point at \( x = -1 \). The point on this curve just to the left of \( x = -1 \) is approaching \( y = 2 \).
- Plot the line \( f(x) = x + 1 \) for \( x \geq -1 \). Include the point at \( x = -1 \). This means plotting the line starting from \( (-1, 0) \), inclusive, and moving rightward.
- The graph will be a combination of these two segments, with a discontinuity at \( x = -1 \).
**Domain and Range:**
- **Domain:** The domain of \( f(x) \) includes all real numbers because there are no restrictions provided in the piecewise function definition. Therefore, the domain is \( (-\infty, \infty)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F70fe4f99-9e69-4bef-afe6-80d54e8903ee%2F33c490ec-b19a-4344-9bb7-1abb568f6c0b%2Fomides_processed.png&w=3840&q=75)
Transcribed Image Text:### Piecewise Function Analysis
**Problem Statement:**
Sketch the function \( f(x) \) and provide the domain and range. The function is defined piecewise as follows:
\[ f(x) = \begin{cases}
x^2 + 1 & \text{if } x < -1 \\
x + 1 & \text{if } x \geq -1
\end{cases} \]
**Instructions for Graphing:**
1. **For \( x < -1 \):**
- The function is \( f(x) = x^2 + 1 \).
- This is a parabola that opens upwards with its vertex shifted 1 unit up from the origin.
2. **For \( x \geq -1 \):**
- The function is \( f(x) = x + 1 \).
- This is a linear function with a slope of 1 and y-intercept of 1.
3. **Boundary at \( x = -1 \):**
- Evaluate the function at the boundary point \( x = -1 \).
- For \( x = -1 \), the quadratic segment is \( f(-1) = (-1)^2 + 1 = 2 \).
- Also, for \( x = -1 \), the linear segment is \( f(-1) = -1 + 1 = 0 \).
**Graphical Explanation:**
- Plot the parabola \( f(x) = x^2 + 1 \) for \( x < -1 \). Ensure it does not include the point at \( x = -1 \). The point on this curve just to the left of \( x = -1 \) is approaching \( y = 2 \).
- Plot the line \( f(x) = x + 1 \) for \( x \geq -1 \). Include the point at \( x = -1 \). This means plotting the line starting from \( (-1, 0) \), inclusive, and moving rightward.
- The graph will be a combination of these two segments, with a discontinuity at \( x = -1 \).
**Domain and Range:**
- **Domain:** The domain of \( f(x) \) includes all real numbers because there are no restrictions provided in the piecewise function definition. Therefore, the domain is \( (-\infty, \infty)
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