Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 3 - The Derivative - 3.3 Rates of Change - 3.3 Exercises - Page 158: 11



Work Step by Step

We are given $s(t)=5t^2-2t-7$ For $t = 2$ the instantaneous velocity is $\lim\limits_{b \to 2}\frac{s(b)-s(2)}{b-2}$ $s(b)=5b^2-2b-7$ $s(6)=9$ The instantaneous rate of change is then $\lim\limits_{b \to 2}\frac{5b^2-2b-7-9}{b-2}=\lim\limits_{b \to 2}\frac{(b-2)(b+\frac{8}{5})}{b-2}=\lim\limits_{b \to 2}(b+\frac{8}{5})=\frac{18}{5}$
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