Explanation Given: The function is f ( x ) = ln ( x 2 + 1 ) and the value of a is 1 and the value of Δ x is 0.1 . Concept used: Error: The formula to calculate the error is as follows: Error = | Δ f − f ′ ( a ) Δ x | Percentage error: The formula to calculate the percentage error is as follows: Percentage error = | Error Actual value | ⋅ 100 Calculation: The given function is expressed as follow: f ( x ) = ln ( x 2 + 1 ) …. (1) The expression Δ f represents the change is the value of f ( x ) as the value of value of x is changed by Δ x . The value of Δ f is calculated as follows: Δ f = f ( a + Δ x ) − f ( a ) = ln ( ( a + Δ x ) 2 + 1 ) − ln ( a 2 + 1 ) = ln ( ( a + Δ x ) 2 + 1 a 2 + 1 ) To obtain the value of Δ f substitute 1 for a and 0.1 for Δ x in the above equation as shown below: Δ f = ln ( ( 1 + 0.1 ) 2 + 1 1 2 + 1 ) = ln ( 2.21 2 ) = ln ( 1.105 ) = 0.099 This implies that the value of Δ f is 0.099 . Differentiate equation (1) with respect to x as shown below: f ′ ( x ) = d ( ln ( x 2 + 1 ) ) d x = 1 x 2 + 1 ⋅ 2 x = 2 x x 2 + 1 Label the above equation as equation (2). f ′ ( x ) = 2 x x 2 + 1 …..(2) The expression to obtain the linear approximation of a function is as follows: f ( a + Δ x ) ≈ f ( a ) + f ′ ( a ) Δ x

CALCULUS EARLY TRANSCENDENTALS W/ACHIEVE

4th Edition

ISBN: 9781319057664

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 4.1, Problem 12E

To determine

To find:

The value of Δf, error and the percentage error for the function f(x)=ln(x2+1) such that the value of a is 1 and the value of Δx is 0.1.

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Chapter 4.1, Problem 11E

Chapter 4.1, Problem 13E

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Chapter 4 Solutions

CALCULUS EARLY TRANSCENDENTALS W/ACHIEVE

Ch. 4.1 - Prob. 1PQ Ch. 4.1 - Prob. 2PQ Ch. 4.1 - Prob. 3PQ Ch. 4.1 - Prob. 4PQ Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E

Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E Ch. 4.1 - Prob. 11E Ch. 4.1 - Prob. 12E Ch. 4.1 - Prob. 13E Ch. 4.1 - Prob. 14E Ch. 4.1 - Prob. 15E Ch. 4.1 - Prob. 16E Ch. 4.1 - Prob. 17E Ch. 4.1 - Prob. 18E Ch. 4.1 - Prob. 19E Ch. 4.1 - Prob. 20E Ch. 4.1 - Prob. 21E Ch. 4.1 - Prob. 22E Ch. 4.1 - Prob. 23E Ch. 4.1 - Prob. 24E Ch. 4.1 - Prob. 25E Ch. 4.1 - Prob. 26E Ch. 4.1 - Prob. 27E Ch. 4.1 - Prob. 28E Ch. 4.1 - Prob. 29E Ch. 4.1 - Prob. 30E Ch. 4.1 - Prob. 31E Ch. 4.1 - Prob. 32E Ch. 4.1 - Prob. 33E Ch. 4.1 - Prob. 34E Ch. 4.1 - Prob. 35E Ch. 4.1 - Prob. 36E Ch. 4.1 - Prob. 37E Ch. 4.1 - Prob. 38E Ch. 4.1 - Prob. 39E Ch. 4.1 - Prob. 40E Ch. 4.1 - Prob. 41E Ch. 4.1 - Prob. 42E Ch. 4.1 - Prob. 43E Ch. 4.1 - Prob. 44E Ch. 4.1 - Prob. 45E Ch. 4.1 - Prob. 46E Ch. 4.1 - Prob. 47E Ch. 4.1 - Prob. 48E Ch. 4.1 - Prob. 49E Ch. 4.1 - Prob. 50E Ch. 4.1 - Prob. 51E Ch. 4.1 - Prob. 52E Ch. 4.1 - Prob. 53E Ch. 4.1 - Prob. 54E Ch. 4.1 - Prob. 55E Ch. 4.1 - Prob. 56E Ch. 4.1 - 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The value of Δ f , error and the percentage error for the function f ( x ) = ln ( x 2 + 1 ) such that the value of a is 1 and the value of Δ x is 0.1 .

Solution Summary: The author calculates the value of Delta f, the error and the percentage error for the function.

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To find:

The value of Δf, error and the percentage error for the function f(x)=ln(x2+1) such that the value of a is 1 and the value of Δx is 0.1.

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Chapter 4 Solutions

The value of Δ f , error and the percentage error for the function f ( x ) = ln ( x 2 + 1 ) such that the value of a is 1 and the value of Δ x is 0.1 .

Solution Summary: The author calculates the value of Delta f, the error and the percentage error for the function.

Videos

To find: The value of Δf, error and the percentage error for the function f(x)=ln(x2+1) such that the value of a is 1 and the value of Δx is 0.1.

Want to see the full answer?

Chapter 4 Solutions

To find:

The value of Δf, error and the percentage error for the function f(x)=ln(x2+1) such that the value of a is 1 and the value of Δx is 0.1.