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: 42


a) 0.4641 b) 0.4 c) 0.0641

Work Step by Step

We are given: $f(x)=x^4$ $\delta f=f(1+0.1)-f(1)$ or $\delta f=f(1.1)-f(1)=1.1^4-1=1.4641-(1)=0.4641$ or $\delta f=0.4641$ differentiation of f(x): $f'(x)=4x^3$ $df=f'(1)\times0.1$ $df=(4)\times0.1=0.4$ $df=0.4$ so $|\delta f-df|=|0.4641-0.4|=|0.0641|=0.0641$ The final answer is: a)0.4641, b)0.4, c)0.0641
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.