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
![**Problem 1: Estimation Using Linear Approximation**
Estimate the difference between the fractions \( \frac{3}{19.2} \) and \( \frac{3}{19} \) using linear approximation.
**Solution Explanation:**
Linear approximation, often used in calculus, is a method of approximating the value of a function near a given point using the tangent line at that point.
Here, the problem involves estimating the change in the function \( f(x) = \frac{3}{x} \) as \( x \) changes from 19 to 19.2.
Using the linear approximation formula:
\[ f(x + \Delta x) \approx f(x) + f'(x) \cdot \Delta x \]
- **\( x = 19 \)**
- **\( \Delta x = 0.2 \)**
1. First, find the derivative of \( f(x) = \frac{3}{x} \):
\[
f'(x) = -\frac{3}{x^2}
\]
2. Calculate \( f'(19) \):
\[
f'(19) = -\frac{3}{19^2}
\]
3. Apply the linear approximation to estimate \( f(19.2) \):
\[
f(19.2) \approx f(19) + f'(19) \cdot 0.2
\]
The aim is to find the approximate value of the difference \( \frac{3}{19.2} - \frac{3}{19} \) using this method.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdd27f7e0-834e-40ed-ac12-9ff71783ae4a%2F7f8d080d-1c46-4ba9-90e9-82a52a091f61%2Fwtd21j_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 1: Estimation Using Linear Approximation**
Estimate the difference between the fractions \( \frac{3}{19.2} \) and \( \frac{3}{19} \) using linear approximation.
**Solution Explanation:**
Linear approximation, often used in calculus, is a method of approximating the value of a function near a given point using the tangent line at that point.
Here, the problem involves estimating the change in the function \( f(x) = \frac{3}{x} \) as \( x \) changes from 19 to 19.2.
Using the linear approximation formula:
\[ f(x + \Delta x) \approx f(x) + f'(x) \cdot \Delta x \]
- **\( x = 19 \)**
- **\( \Delta x = 0.2 \)**
1. First, find the derivative of \( f(x) = \frac{3}{x} \):
\[
f'(x) = -\frac{3}{x^2}
\]
2. Calculate \( f'(19) \):
\[
f'(19) = -\frac{3}{19^2}
\]
3. Apply the linear approximation to estimate \( f(19.2) \):
\[
f(19.2) \approx f(19) + f'(19) \cdot 0.2
\]
The aim is to find the approximate value of the difference \( \frac{3}{19.2} - \frac{3}{19} \) using this method.
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