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...
Related questions
Question
Find the linear approximation of f(x)=lnx at x=7 and use it to approximate ln(7.05).
L(x)=? x + ?
ln(7.05) ≈ ?
![### Linear Approximation of Logarithmic Functions
**Problem Statement:**
Find the linear approximation of \( f(x) = \ln x \) at \( x = 7 \) and use it to approximate \(\ln(7.05)\).
**Solution:**
The linear approximation \( L(x) \) of a function \( f(x) \) at a point \( x = a \) is given by:
\[
L(x) = f(a) + f'(a)(x - a)
\]
In this case,
- \( f(x) = \ln x \)
- \( a = 7 \)
First, compute \( f'(x) \):
\[
f'(x) = \frac{1}{x}
\]
Now, apply the values:
- \( f(a) = \ln 7 \)
- \( f'(a) = \frac{1}{7} \)
The linear approximation is:
\[
L(x) = \ln 7 + \frac{1}{7}(x - 7)
\]
**Approximation of \(\ln(7.05)\):**
Using the linear approximation for \( x = 7.05 \):
\[
L(7.05) = \ln 7 + \frac{1}{7}(7.05 - 7)
\]
This simplifies to:
\[
\ln(7.05) \approx \ln 7 + \frac{0.05}{7}
\]
**Conclusion:**
The approximate value of \(\ln(7.05)\) is:
\[
\ln(7.05) \approx 1.95123
\]
This linear approximation method provides a quick and easy way to estimate logarithmic values near a known point.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc9b69680-5ad4-41fa-bd27-ae6a7c1e734f%2F2e60ac88-cf2e-43cf-b24a-60a965775eb6%2F3q5fam_processed.png&w=3840&q=75)
Transcribed Image Text:### Linear Approximation of Logarithmic Functions
**Problem Statement:**
Find the linear approximation of \( f(x) = \ln x \) at \( x = 7 \) and use it to approximate \(\ln(7.05)\).
**Solution:**
The linear approximation \( L(x) \) of a function \( f(x) \) at a point \( x = a \) is given by:
\[
L(x) = f(a) + f'(a)(x - a)
\]
In this case,
- \( f(x) = \ln x \)
- \( a = 7 \)
First, compute \( f'(x) \):
\[
f'(x) = \frac{1}{x}
\]
Now, apply the values:
- \( f(a) = \ln 7 \)
- \( f'(a) = \frac{1}{7} \)
The linear approximation is:
\[
L(x) = \ln 7 + \frac{1}{7}(x - 7)
\]
**Approximation of \(\ln(7.05)\):**
Using the linear approximation for \( x = 7.05 \):
\[
L(7.05) = \ln 7 + \frac{1}{7}(7.05 - 7)
\]
This simplifies to:
\[
\ln(7.05) \approx \ln 7 + \frac{0.05}{7}
\]
**Conclusion:**
The approximate value of \(\ln(7.05)\) is:
\[
\ln(7.05) \approx 1.95123
\]
This linear approximation method provides a quick and easy way to estimate logarithmic values near a known point.
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