Explanation Given: Δ f = f ( 1 + h ) − f ( 1 ) w h e r e f ( x ) = x − 1 Concept: Suppose that y = f ( x ) is a differentiable curve at x = a . Then the tangent line at x = a has equation y = f ( a ) + f ' ( a ) ( x − a ) We call the above equation the linear approximation or linearization of y = f ( x ) at the point ( a , f ( a ) ) and write f ( x ) ≈ L ( x ) = f ( a ) + f ' ( a ) ( x − a ) We sometimes write L a ( x ) to stress that the approximation is near a . Calculation: Description Step 1: Show the given statement E ≤ 2 h 2 f o r − 1 2 ≤ h ≤ 1 2 Let f ( x ) = x − 1 T h e n Δ f <

4th Edition

ISBN: 9781319545956

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 4.1, Problem 74E

To determine

To show directly E=|Δf−f'(1)h| is equal to h2/(1+h) . Then prove that E≤2h2 for −12≤h≤12.

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Chapter 4.1, Problem 73E

Chapter 4.2, Problem 1PQ

Chapter 4 Solutions

CALCULUS W/ACHIEVE PLATFORM

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To show directly E = | Δ f − f ' ( 1 ) h | i s e q u a l t o h 2 / ( 1 + h ) . Then prove that E ≤ 2 h 2 f o r − 1 2 ≤ h ≤ 1 2 .

Solution Summary: The author demonstrates the linear approximation or linearization of y=f(x) at the point (a,f)

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To show directly E=|Δf−f'(1)h| is equal to h2/(1+h) . Then prove that E≤2h2 for −12≤h≤12.

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