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:**
Find the \( c \) value that satisfies the Mean Value Theorem of the function \( f(x) = x^3 - x \) on the interval \([0, 2]\).
**Answer Box:**
\( c = \) [ ]
---
**Explanation:**
To solve this, you will apply the Mean Value Theorem which states that for a function \( f \) that is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), there exists a number \( c \) in \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
For the given function \( f(x) = x^3 - x \):
1. **Find the derivative \( f'(x) \):**
\[ f'(x) = 3x^2 - 1 \]
2. **Calculate \( f(a) \) and \( f(b) \):**
- \( f(0) = 0^3 - 0 = 0 \)
- \( f(2) = 2^3 - 2 = 8 - 2 = 6 \)
3. **Evaluate \(\frac{f(b) - f(a)}{b - a}\):**
\[ \frac{6 - 0}{2 - 0} = \frac{6}{2} = 3 \]
4. **Set the derivative equal to the average rate of change and solve for \( c \):**
\[ 3c^2 - 1 = 3 \]
\[ 3c^2 = 4 \]
\[ c^2 = \frac{4}{3} \]
\[ c = \pm\sqrt{\frac{4}{3}} \]
However, since \( c \) must be in the interval \((0, 2)\), we only take the positive value:
\[ c = \sqrt{\frac{4}{3}} \]
---
Thus, the value of \( c \) that satisfies the Mean Value Theorem is \( c = \sqrt{\frac{4}{3}} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3a34a1e4-7c25-4c64-ab7e-d5dd0d192579%2F89ca455f-1af8-4fff-add2-5595217ac51e%2Fbmusqfm_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the \( c \) value that satisfies the Mean Value Theorem of the function \( f(x) = x^3 - x \) on the interval \([0, 2]\).
**Answer Box:**
\( c = \) [ ]
---
**Explanation:**
To solve this, you will apply the Mean Value Theorem which states that for a function \( f \) that is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), there exists a number \( c \) in \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
For the given function \( f(x) = x^3 - x \):
1. **Find the derivative \( f'(x) \):**
\[ f'(x) = 3x^2 - 1 \]
2. **Calculate \( f(a) \) and \( f(b) \):**
- \( f(0) = 0^3 - 0 = 0 \)
- \( f(2) = 2^3 - 2 = 8 - 2 = 6 \)
3. **Evaluate \(\frac{f(b) - f(a)}{b - a}\):**
\[ \frac{6 - 0}{2 - 0} = \frac{6}{2} = 3 \]
4. **Set the derivative equal to the average rate of change and solve for \( c \):**
\[ 3c^2 - 1 = 3 \]
\[ 3c^2 = 4 \]
\[ c^2 = \frac{4}{3} \]
\[ c = \pm\sqrt{\frac{4}{3}} \]
However, since \( c \) must be in the interval \((0, 2)\), we only take the positive value:
\[ c = \sqrt{\frac{4}{3}} \]
---
Thus, the value of \( c \) that satisfies the Mean Value Theorem is \( c = \sqrt{\frac{4}{3}} \).
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