*23) Gketch the graph of f by hand and use r sketch to find the absolute and local maximdm and winimum valucs of f. your 2xナ 1 L4-2x f 1Ex <3 1f osX f(x) =

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
Question
**Exercise 28:**

Sketch the graph of \( f \) by hand and use your sketch to find the absolute and local maximum and minimum values of \( f \).

\[ f(x) = 
  \begin{cases} 
   2x + 1 & \text{if } 0 \leq x < 1 \\
   4 - 2x & \text{if } 1 \leq x \leq 3 
  \end{cases}
\]

**Explanation of Graph:**

- The graph consists of two linear pieces defined within different intervals.
- For \( 0 \leq x < 1 \), the function is \( f(x) = 2x + 1 \). This is a line with a slope of 2 and a y-intercept at 1.
- For \( 1 \leq x \leq 3 \), the function is \( f(x) = 4 - 2x \). This is a line with a slope of -2 and a y-intercept at 4.

**Diagram Explanation:**

The graph starts with an upward sloping line segment from \( (0, 1) \) to just before \( (1, 3) \). At \( x = 1 \), the graph switches to a downward sloping line starting from \( (1, 2) \) to \( (3, -2) \). The intersection of these two segments at \( x = 1 \) represents a transition from the first linear expression to the second one.
Transcribed Image Text:**Exercise 28:** Sketch the graph of \( f \) by hand and use your sketch to find the absolute and local maximum and minimum values of \( f \). \[ f(x) = \begin{cases} 2x + 1 & \text{if } 0 \leq x < 1 \\ 4 - 2x & \text{if } 1 \leq x \leq 3 \end{cases} \] **Explanation of Graph:** - The graph consists of two linear pieces defined within different intervals. - For \( 0 \leq x < 1 \), the function is \( f(x) = 2x + 1 \). This is a line with a slope of 2 and a y-intercept at 1. - For \( 1 \leq x \leq 3 \), the function is \( f(x) = 4 - 2x \). This is a line with a slope of -2 and a y-intercept at 4. **Diagram Explanation:** The graph starts with an upward sloping line segment from \( (0, 1) \) to just before \( (1, 3) \). At \( x = 1 \), the graph switches to a downward sloping line starting from \( (1, 2) \) to \( (3, -2) \). The intersection of these two segments at \( x = 1 \) represents a transition from the first linear expression to the second one.
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