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: 41


a) 0.231 b) 0.2 c) 0.031

Work Step by Step

We are given: $f(x)=x^3-x$ $\delta f=f(1+0.1)-f(1)$ or $\delta f=f(1.1)-f(1)=0.231-0=0.231$ or $\delta f=0.231$ differentiation of f(x): $f'(x)=3x^2-1$ $df=f'(1)\times0.1$ $df=(3-1)\times0.1=0.2$ $df=0.2$ so $|\delta f-df|=|0.231-0.2|=|0.031|=0.031$ Thus the final answer is: a)0.231, b)0.2, c)0.031
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