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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Question
![**Title: Linearization of Functions**
**Objective:**
Understand the process of linearizing a function \( f(x) = (1 + x)^k \) at \( x = 0 \) and demonstrate that the linearization \( L(x) \) is \( 1 + kx \).
---
**Linearization Concept:**
For a differentiable function at \( x = a \), the linear approximation is given by:
\[ L(x) = f(a) + f'(a)(x - a) \]
This linearization provides an approximation of the function near \( x = a \).
---
**Procedure:**
1. **Determine \( f(0) \):**
\( f(0) = (1 + 0)^k = 1 \)
*(Simplified result is 1.)*
2. **Determine \( f'(x) \) (the derivative of \( f(x) \)):**
Choose the correct expression for the derivative:
- \( \text{A. } f'(x) = (1 + x)^k \)
- \( \text{B. } f'(x) = k(1 + x)^{k-1} \)
- \( \text{C. } f'(x) = (1 + x)^{k-1} \)
- \( \text{D. } f'(x) = k(1 + x)^k \)
*Correct choice: B. \( f'(x) = k(1 + x)^{k-1} \)*
3. **Determine \( f'(0) \):**
Substitute \( x = 0 \) into the derivative:
\( f'(0) = k(1 + 0)^{k-1} = k \)
*(Simplified result is k.)*
4. **Determine the Linearization \( L(x) \):**
Using the values \( f(0) \) and \( f'(0) \) calculated previously, find \( L(x) \):
- \( \text{A. } L(x) = f(0) + f'(0)(x - 0) = 0 + k(x - 0) = 1 + kx \)
- \( \text{B. } L(x) = f(0) + f'(0)(x - 0) = 1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6f5a197-bd68-4fd1-8232-f49c284d6121%2F8cd53ea0-b5f9-43da-a386-a503f3f358bd%2Fmf261ac_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Linearization of Functions**
**Objective:**
Understand the process of linearizing a function \( f(x) = (1 + x)^k \) at \( x = 0 \) and demonstrate that the linearization \( L(x) \) is \( 1 + kx \).
---
**Linearization Concept:**
For a differentiable function at \( x = a \), the linear approximation is given by:
\[ L(x) = f(a) + f'(a)(x - a) \]
This linearization provides an approximation of the function near \( x = a \).
---
**Procedure:**
1. **Determine \( f(0) \):**
\( f(0) = (1 + 0)^k = 1 \)
*(Simplified result is 1.)*
2. **Determine \( f'(x) \) (the derivative of \( f(x) \)):**
Choose the correct expression for the derivative:
- \( \text{A. } f'(x) = (1 + x)^k \)
- \( \text{B. } f'(x) = k(1 + x)^{k-1} \)
- \( \text{C. } f'(x) = (1 + x)^{k-1} \)
- \( \text{D. } f'(x) = k(1 + x)^k \)
*Correct choice: B. \( f'(x) = k(1 + x)^{k-1} \)*
3. **Determine \( f'(0) \):**
Substitute \( x = 0 \) into the derivative:
\( f'(0) = k(1 + 0)^{k-1} = k \)
*(Simplified result is k.)*
4. **Determine the Linearization \( L(x) \):**
Using the values \( f(0) \) and \( f'(0) \) calculated previously, find \( L(x) \):
- \( \text{A. } L(x) = f(0) + f'(0)(x - 0) = 0 + k(x - 0) = 1 + kx \)
- \( \text{B. } L(x) = f(0) + f'(0)(x - 0) = 1
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