2. (6371 #3 p. 227) ƒ(5) — ƒ(0) f'(c) = 5-0 x +4 Show that there is no value of c with 0 < c < 5 such that x 4 Why does this not contradict the Mean Value Theorem for Derivatives? Let f(x) =
2. (6371 #3 p. 227) ƒ(5) — ƒ(0) f'(c) = 5-0 x +4 Show that there is no value of c with 0 < c < 5 such that x 4 Why does this not contradict the Mean Value Theorem for Derivatives? Let f(x) =
Calculus: Early Transcendentals
8th Edition
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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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![### Educational Content: Mean Value Theorem Exploration
**Problem 2. (6371 #3 p. 227)**
**Objective**: Analyze the function and explore the application of the Mean Value Theorem for derivatives.
**Given Function**:
\[ f(x) = \frac{x + 4}{x - 4} \]
**Task**: Demonstrate that there is no value of \( c \) within the interval \( 0 < c < 5 \) such that:
\[ f'(c) = \frac{f(5) - f(0)}{5 - 0} \]
**Question**: Why does this finding not contradict the Mean Value Theorem for Derivatives?
**Explanation**:
1. **Define the Function**: The function \( f(x) = \frac{x + 4}{x - 4} \) involves a rational expression where the denominator becomes zero at \( x = 4 \). This point is essential as it implies the function is not continuous across \( [0, 5] \).
2. **Mean Value Theorem (MVT) for Derivatives**: The MVT states that if a function is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one number \( c \) in \((a, b)\) such that:
\[
f'(c) = \frac{f(b) - f(a)}{b - a}
\]
3. **Application to the Given Function**:
- **Continuity Issue**: Since \( f(x) \) is not continuous at \( x = 4 \), and \( x = 4 \) lies within the interval \( (0, 5) \), the MVT cannot be directly applied.
- **Differentiability**: For similar reasons, the function cannot be differentiable at \( x = 4 \).
4. **Conclusion**: The absence of a value \( c \) satisfying the MVT condition does not contradict the theorem itself; instead, it highlights the preconditions (continuity and differentiability) required for the MVT are not met over the interval \( [0, 5] \).
This exercise is a practical evaluation of the conditions under which the Mean Value Theorem holds and illustrates why checking the prerequisites of mathematical theorems is](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffaefa6c0-2df9-403e-a900-0259c11910a5%2F91401fd3-ebf4-4630-8c98-70dc0d4c14db%2F3qoyf6q_processed.png&w=3840&q=75)
Transcribed Image Text:### Educational Content: Mean Value Theorem Exploration
**Problem 2. (6371 #3 p. 227)**
**Objective**: Analyze the function and explore the application of the Mean Value Theorem for derivatives.
**Given Function**:
\[ f(x) = \frac{x + 4}{x - 4} \]
**Task**: Demonstrate that there is no value of \( c \) within the interval \( 0 < c < 5 \) such that:
\[ f'(c) = \frac{f(5) - f(0)}{5 - 0} \]
**Question**: Why does this finding not contradict the Mean Value Theorem for Derivatives?
**Explanation**:
1. **Define the Function**: The function \( f(x) = \frac{x + 4}{x - 4} \) involves a rational expression where the denominator becomes zero at \( x = 4 \). This point is essential as it implies the function is not continuous across \( [0, 5] \).
2. **Mean Value Theorem (MVT) for Derivatives**: The MVT states that if a function is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one number \( c \) in \((a, b)\) such that:
\[
f'(c) = \frac{f(b) - f(a)}{b - a}
\]
3. **Application to the Given Function**:
- **Continuity Issue**: Since \( f(x) \) is not continuous at \( x = 4 \), and \( x = 4 \) lies within the interval \( (0, 5) \), the MVT cannot be directly applied.
- **Differentiability**: For similar reasons, the function cannot be differentiable at \( x = 4 \).
4. **Conclusion**: The absence of a value \( c \) satisfying the MVT condition does not contradict the theorem itself; instead, it highlights the preconditions (continuity and differentiability) required for the MVT are not met over the interval \( [0, 5] \).
This exercise is a practical evaluation of the conditions under which the Mean Value Theorem holds and illustrates why checking the prerequisites of mathematical theorems is
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