a. Answer The linearization of the function is L ( x ) = x Explanation Given information: The given function is f ( x ) = ln ( x + 1 ) , x = 0 Since f ( 0 ) = ln ( 1 + 0 ) = 0 the point of tangency is ( 0 , 0 ) .the slope of the tangent line is f ′ ( 0 ) = 1 1 + 0 = 1 thus, L ( x ) = f ( a ) + f ′ ( a ) ( x − a ) L ( x ) = 0 + 1 ( x − 0 ) L ( x ) = x b. To determine To check the accuracy of approximation L ( 0 + 0.1 ) ≈ f ( 0 + 0.1 ) Answer The approximation error is | 0.1 − 0.0953 | ≈ 0.004 = 4 × 10 − 3 , the error is less than 10 − 2 Explanation Given information: The given value of a is 0 To approximate ln ( ( 0 + 0.1 ) + 1 ) , we use x = 0 + 0.1 : ln ( ( 0 + 0.1 ) + 1 ) = f ( 0 + 0.1 ) ≈ L ( 0 + 0.1 ) = 0 + 0.1 = 0.1 The calculator gives ln ( ( 0 + 0.1 ) + 1 ) = 0.0953 , so the approximation error is | 0.1 − 0.0953 | ≈ 0.004 = 4 × 10 − 3 , the error is less than 10 − 2

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Question

Chapter 5.5, Problem 4E

To determine

To find the linearization L(x)of f(x) at x=0

Expert Solution

Answer to Problem 4E

The linearization of the function is L(x)=x

Explanation of Solution

Given information:

The given function is f(x)=ln(x+1), x=0

Since f(0)=ln(1+0)=0 the point of tangency is (0,0) .the slope of the tangent line is f′(0)=11+0=1 thus,

L(x)=f(a)+f′(a)(x−a)L(x)=0+1(x−0)L(x)=x

To determine

To check the accuracy of approximation L(0+0.1)≈f(0+0.1)

Expert Solution

Answer to Problem 4E

The approximation error is |0.1−0.0953|≈0.004=4×10−3 , the error is less than 10−2

Explanation of Solution

Given information:

The given value of a is 0

To approximate ln((0+0.1)+1) , we use x=0+0.1 :

ln((0+0.1)+1)=f(0+0.1)≈L(0+0.1)=0+0.1=0.1

The calculator gives ln((0+0.1)+1)=0.0953 , so the approximation error is

|0.1−0.0953|≈0.004=4×10−3 , the error is less than 10−2

Chapter 5.5, Problem 3E

Chapter 5.5, Problem 5E

Chapter 5 Solutions

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

Ch. 5.1 - Prob. 1QR Ch. 5.1 - Prob. 2QR Ch. 5.1 - Prob. 3QR Ch. 5.1 - Prob. 4QR Ch. 5.1 - Prob. 5QR Ch. 5.1 - Prob. 6QR Ch. 5.1 - Prob. 7QR Ch. 5.1 - Prob. 8QR Ch. 5.1 - Prob. 9QR Ch. 5.1 - Prob. 10QR

Ch. 5.1 - Prob. 11QR Ch. 5.1 - Prob. 12QR Ch. 5.1 - Prob. 1E Ch. 5.1 - Prob. 2E Ch. 5.1 - Prob. 3E Ch. 5.1 - Prob. 4E Ch. 5.1 - Prob. 5E Ch. 5.1 - Prob. 6E Ch. 5.1 - Prob. 7E Ch. 5.1 - Prob. 8E Ch. 5.1 - Prob. 9E Ch. 5.1 - Prob. 10E Ch. 5.1 - Prob. 11E Ch. 5.1 - Prob. 12E Ch. 5.1 - Prob. 13E Ch. 5.1 - Prob. 14E Ch. 5.1 - Prob. 15E Ch. 5.1 - Prob. 16E Ch. 5.1 - Prob. 17E Ch. 5.1 - Prob. 18E Ch. 5.1 - Prob. 19E Ch. 5.1 - Prob. 20E Ch. 5.1 - Prob. 21E Ch. 5.1 - Prob. 22E Ch. 5.1 - Prob. 23E Ch. 5.1 - Prob. 24E Ch. 5.1 - Prob. 25E Ch. 5.1 - Prob. 26E Ch. 5.1 - Prob. 27E Ch. 5.1 - Prob. 28E Ch. 5.1 - Prob. 29E Ch. 5.1 - Prob. 30E Ch. 5.1 - Prob. 31E Ch. 5.1 - Prob. 32E Ch. 5.1 - Prob. 33E Ch. 5.1 - Prob. 34E Ch. 5.1 - Prob. 35E Ch. 5.1 - Prob. 36E Ch. 5.1 - Prob. 37E Ch. 5.1 - Prob. 38E Ch. 5.1 - Prob. 39E Ch. 5.1 - Prob. 40E Ch. 5.1 - Prob. 41E Ch. 5.1 - Prob. 42E Ch. 5.1 - Prob. 43E Ch. 5.1 - Prob. 44E Ch. 5.1 - Prob. 45E Ch. 5.1 - Prob. 46E Ch. 5.1 - Prob. 47E Ch. 5.1 - Prob. 48E Ch. 5.1 - 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