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
Find the linearization of the function f(x) = x^8 at x = 1
![### Problem Statement
Find the linearization of the function \( f(x) = x^8 \) at \( x = 1 \).
\[ L(x) = \]
Using this linearization, \( f(1.02) \) is approximately \(\_\_\_\_\_\_\_\_\_\_\).
### Solution Steps
#### 1. Understanding Linearization
Linearization of a function at a point provides an approximation of the function using the tangent line at that specific point. The linearization \( L(x) \) of \( f(x) \) at \( x = a \) is given by:
\[
L(x) = f(a) + f'(a)(x - a)
\]
#### 2. Identifying the Function and Point of Interest
Given function:
\[
f(x) = x^8
\]
Point of interest:
\[
x = 1
\]
#### 3. Calculate \( f(a) \) and \( f'(a) \)
**Value of the function at \( x = 1 \):**
\[
f(1) = 1^8 = 1
\]
**First derivative of \( f(x) = x^8 \):**
\[
f'(x) = 8x^7
\]
**Value of the derivative at \( x = 1 \):**
\[
f'(1) = 8(1)^7 = 8
\]
#### 4. Formulating the Linearization
Using the formula \( L(x) = f(a) + f'(a)(x - a) \):
\[
L(x) = f(1) + f'(1)(x - 1)
\]
Substituting the values:
\[
L(x) = 1 + 8(x - 1)
\]
The linearization of \( f(x) = x^8 \) at \( x = 1 \) is:
\[
L(x) = 8x - 7
\]
#### 5. Approximating \( f(1.02) \)
Using the linear approximation:
\[
f(1.02) \approx L(1.02)
\]
Substituting \( x = 1.02 \) into the linearization:
\[
L(1.02) = 8(1.02) - 7 = 0.16 +](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff3e1f0c0-c921-4249-b733-3c4d539fb1d9%2F1d73aa42-24d8-4281-87af-63a037746f61%2Ftaebagp_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Find the linearization of the function \( f(x) = x^8 \) at \( x = 1 \).
\[ L(x) = \]
Using this linearization, \( f(1.02) \) is approximately \(\_\_\_\_\_\_\_\_\_\_\).
### Solution Steps
#### 1. Understanding Linearization
Linearization of a function at a point provides an approximation of the function using the tangent line at that specific point. The linearization \( L(x) \) of \( f(x) \) at \( x = a \) is given by:
\[
L(x) = f(a) + f'(a)(x - a)
\]
#### 2. Identifying the Function and Point of Interest
Given function:
\[
f(x) = x^8
\]
Point of interest:
\[
x = 1
\]
#### 3. Calculate \( f(a) \) and \( f'(a) \)
**Value of the function at \( x = 1 \):**
\[
f(1) = 1^8 = 1
\]
**First derivative of \( f(x) = x^8 \):**
\[
f'(x) = 8x^7
\]
**Value of the derivative at \( x = 1 \):**
\[
f'(1) = 8(1)^7 = 8
\]
#### 4. Formulating the Linearization
Using the formula \( L(x) = f(a) + f'(a)(x - a) \):
\[
L(x) = f(1) + f'(1)(x - 1)
\]
Substituting the values:
\[
L(x) = 1 + 8(x - 1)
\]
The linearization of \( f(x) = x^8 \) at \( x = 1 \) is:
\[
L(x) = 8x - 7
\]
#### 5. Approximating \( f(1.02) \)
Using the linear approximation:
\[
f(1.02) \approx L(1.02)
\]
Substituting \( x = 1.02 \) into the linearization:
\[
L(1.02) = 8(1.02) - 7 = 0.16 +
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