a. Answer The linearization of the function is L ( x ) = 2 Explanation Given information: The given function is f ( x ) = x + 1 x , x = 1 Since f ( 1 ) = 1 + 1 = 2 the point of tangency is ( 1 , 2 ) .the slope of the tangent line is f ′ ( 1 ) = 1 − 1 1 = 0 thus, L ( x ) = f ( a ) + f ′ ( a ) ( x − a ) L ( x ) = 2 + 0 ( x − 1 ) L ( x ) = 2 b. To determine To check the accuracy of approximation L ( 1 + 0.1 ) ≈ f ( 1 + 0.1 ) Answer The approximation error is | 2 − 2.009 | ≈ 0.009 = 9 × 10 − 3 , the error is less than 10 − 2 Explanation Given information: The given value of a is 1 To approximate ( 1 + 0.1 ) + 1 1 + 0.1 , we use x = 1 + 0.1 : ( 1 + 0.1 ) + 1 1 + 0.1 = f ( 1 + 0.1 ) ≈ L ( 1 + 0.1 ) = 2 The calculator gives ( 1 + 0.1 ) + 1 1 + 0.1 = 2.009 , so the approximation error is | 2 − 2.009 | ≈ 0.009 = 9 × 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 3E

To determine

To find the linearization L(x)of f(x) at x=1

Expert Solution

Answer to Problem 3E

The linearization of the function is L(x)=2

Explanation of Solution

Given information:

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

Since f(1)=1+1=2 the point of tangency is (1,2) .the slope of the tangent line is f′(1)=1−11=0 thus,

L(x)=f(a)+f′(a)(x−a)L(x)=2+0(x−1)L(x)=2

To determine

To check the accuracy of approximation L(1+0.1)≈f(1+0.1)

Expert Solution

Answer to Problem 3E

The approximation error is |2−2.009|≈0.009=9×10−3 , the error is less than 10−2

Explanation of Solution

Given information:

The given value of a is 1

To approximate (1+0.1)+11+0.1 , we use x=1+0.1 :

(1+0.1)+11+0.1=f(1+0.1)≈L(1+0.1)=2

The calculator gives (1+0.1)+11+0.1=2.009 , so the approximation error is

|2−2.009|≈0.009=9×10−3 , the error is less than 10−2

Chapter 5.5, Problem 2E

Chapter 5.5, Problem 4E

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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