Find the linearization L(x) of the function g(x) = /a · f(/E) at z = 100 given the following information. f(10) = 3 f(100) = -7 f'(10) = –4 f'(100) = –10 Answer: L(x) : %3D
Find the linearization L(x) of the function g(x) = /a · f(/E) at z = 100 given the following information. f(10) = 3 f(100) = -7 f'(10) = –4 f'(100) = –10 Answer: L(x) : %3D
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: Linearization of a Function**
Find the linearization \( L(x) \) of the function \( g(x) = \sqrt{x} \cdot f(\sqrt{x}) \) at \( x = 100 \) given the following information.
\[
f(10) = 3 \quad f(100) = -7 \quad f'(10) = -4 \quad f'(100) = -10
\]
**Solution:**
To find the linearization \( L(x) \) at \( x = 100 \), we consider the formula for linearization:
\[
L(x) = g(100) + g'(100) \cdot (x - 100)
\]
**Calculations:**
1. **Evaluate \( g(100) \)**:
\[
g(100) = \sqrt{100} \cdot f(\sqrt{100}) = 10 \cdot f(10) = 10 \cdot 3 = 30
\]
2. **Differentiate \( g(x) \) and evaluate \( g'(100) \)**:
Using the product rule:
\[
g(x) = \sqrt{x} \cdot f(\sqrt{x}) \Rightarrow g'(x) = \frac{1}{2\sqrt{x}} \cdot f(\sqrt{x}) + \sqrt{x} \cdot \frac{f'(\sqrt{x})}{2\sqrt{x}}
\]
Simplifying:
\[
g'(x) = \frac{f(\sqrt{x})}{2\sqrt{x}} + \frac{\sqrt{x} f'(\sqrt{x})}{2x}
\]
Evaluate at \( x = 100 \):
\[
g'(100) = \frac{f(10)}{20} + \frac{10 \cdot f'(10)}{200}
\]
\[
= \frac{3}{20} + \frac{10 \cdot (-4)}{200} = \frac{3}{20} - \frac{2}{20} = \frac{1}{20}
\]
3. **Linearization formula**:
\[
L(x) = 30 + \frac{1}{20} \cdot (x - 100)
\]
**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F341515a7-9f2b-49dc-a023-366552e2ebc5%2F9b4240e9-3664-4cf6-9896-b14605819f2b%2Fv0gh0b_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem: Linearization of a Function**
Find the linearization \( L(x) \) of the function \( g(x) = \sqrt{x} \cdot f(\sqrt{x}) \) at \( x = 100 \) given the following information.
\[
f(10) = 3 \quad f(100) = -7 \quad f'(10) = -4 \quad f'(100) = -10
\]
**Solution:**
To find the linearization \( L(x) \) at \( x = 100 \), we consider the formula for linearization:
\[
L(x) = g(100) + g'(100) \cdot (x - 100)
\]
**Calculations:**
1. **Evaluate \( g(100) \)**:
\[
g(100) = \sqrt{100} \cdot f(\sqrt{100}) = 10 \cdot f(10) = 10 \cdot 3 = 30
\]
2. **Differentiate \( g(x) \) and evaluate \( g'(100) \)**:
Using the product rule:
\[
g(x) = \sqrt{x} \cdot f(\sqrt{x}) \Rightarrow g'(x) = \frac{1}{2\sqrt{x}} \cdot f(\sqrt{x}) + \sqrt{x} \cdot \frac{f'(\sqrt{x})}{2\sqrt{x}}
\]
Simplifying:
\[
g'(x) = \frac{f(\sqrt{x})}{2\sqrt{x}} + \frac{\sqrt{x} f'(\sqrt{x})}{2x}
\]
Evaluate at \( x = 100 \):
\[
g'(100) = \frac{f(10)}{20} + \frac{10 \cdot f'(10)}{200}
\]
\[
= \frac{3}{20} + \frac{10 \cdot (-4)}{200} = \frac{3}{20} - \frac{2}{20} = \frac{1}{20}
\]
3. **Linearization formula**:
\[
L(x) = 30 + \frac{1}{20} \cdot (x - 100)
\]
**
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