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: 44

Answer

a) 1.061, b) 1.0, c) 0.061

Work Step by Step

We are given: $f(x)=x^3-2x+3$ for $\delta f$ at 2: $\delta f=f(2+0.1)-f(2)$ or $\delta f=(2.1)^3-2\times2.1+3-(2)^3-2\times2+3=8.061-7=1.061$ or $\delta f=1.061$ differentiation of f(x): $f'(x)=3x^2-2$ $df=f'(2)\times0.1$ $df=(3\times2^2-2)\times0.1=1.0$ $df=1.0$ so $|\delta f-df|=|1.061-1.0|=|0.061|=0.061$ The final answer is: a) 1.061, b) 1.0, c) 0.061
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