Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.9 Linear Approximations and Differentials - 2.9 Exercises - Page 193: 30


The answer is perfectly reasonable by a linear approximation shown below...

Work Step by Step

The slope of $\sqrt {2x}$ near $x = 2$ is about $1/2$ as $$f'(x) = 1/\sqrt{2x}$$ Thus, if x were to increase by $0.01$, then $y$ would increase by $0.005$ $\sqrt{2*2}+0.005= 2.005$
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