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...
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![**Problem Statement:**
Given that \( f(1) = 4 \) and \( f'(x) \ge 1 \) for \( 1 \le x \le 4 \), determine the minimum possible value of \( f(4) \).
**Solution Explanation:**
To find the minimum possible value of \( f(4) \), we need to integrate the information provided about the derivative \( f'(x) \).
- We are given that \( f(1) = 4 \).
- We are also given that the derivative \( f'(x) \), which represents the rate of change of \( f(x) \), is always greater than or equal to 1 in the interval from \( x = 1 \) to \( x = 4 \).
Since \( f'(x) \ge 1 \), the function \( f(x) \) is increasing at a rate of at least 1 unit per unit change in \( x \).
To find how much the function \( f(x) \) can increase from \( x = 1 \) to \( x = 4 \), we can integrate the minimum rate of change over this interval:
\[ f(4) - f(1) \ge \int_{1}^{4} 1 \, dx \]
Calculating the integral:
\[ \int_{1}^{4} 1 \, dx = 4 - 1 = 3 \]
Adding this to the initial value \( f(1) \):
\[ f(4) \ge f(1) + 3 \]
\[ f(4) \ge 4 + 3 \]
\[ f(4) \ge 7 \]
**Conclusion:**
The smallest possible value that \( f(4) \) can be is 7. Therefore, the minimum possible value for \( f(4) \) is \( \boxed{7} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa80fd0a7-9ae8-4bc5-b926-d69bfc10c0d7%2F3c22cf56-6c8b-4cf6-9116-a2b3b84bcf10%2Fvc2j3fg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Given that \( f(1) = 4 \) and \( f'(x) \ge 1 \) for \( 1 \le x \le 4 \), determine the minimum possible value of \( f(4) \).
**Solution Explanation:**
To find the minimum possible value of \( f(4) \), we need to integrate the information provided about the derivative \( f'(x) \).
- We are given that \( f(1) = 4 \).
- We are also given that the derivative \( f'(x) \), which represents the rate of change of \( f(x) \), is always greater than or equal to 1 in the interval from \( x = 1 \) to \( x = 4 \).
Since \( f'(x) \ge 1 \), the function \( f(x) \) is increasing at a rate of at least 1 unit per unit change in \( x \).
To find how much the function \( f(x) \) can increase from \( x = 1 \) to \( x = 4 \), we can integrate the minimum rate of change over this interval:
\[ f(4) - f(1) \ge \int_{1}^{4} 1 \, dx \]
Calculating the integral:
\[ \int_{1}^{4} 1 \, dx = 4 - 1 = 3 \]
Adding this to the initial value \( f(1) \):
\[ f(4) \ge f(1) + 3 \]
\[ f(4) \ge 4 + 3 \]
\[ f(4) \ge 7 \]
**Conclusion:**
The smallest possible value that \( f(4) \) can be is 7. Therefore, the minimum possible value for \( f(4) \) is \( \boxed{7} \).
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