University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.11 - Linearization and Differentials - Exercises - Page 199: 40


a) 0.02 b) 0 c) 0.02

Work Step by Step

We are given: $f(x)=2x^2+4x-3$ for $\delta f$ at 0.1: $\delta f=f(-1+0.1)-f(-1)$ or $\delta f=2(-0.9)^2+4\times0.9-3-2\times(- 1)+4(-1)-3=-4.98-(-5)=0.02$ or$\delta f=0.02$ differentiation of f(x): $f'(x)=4x+4$ $df=f'(-1)\times0.1$ $df=(-4+4)\times0.1=0.0$ $df=0.0$ so $|\delta f-df|=|0.02-0.0|=|0.02|=0.02$ The final answer is: a)0.02, b)0, c)0.02
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