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: 14



Work Step by Step

We are given $s(t)=t^3+2t+9$ For $t = 1$ the instantaneous velocity is $\lim\limits_{h \to 0}\frac{s(4+h)-s(14)}{h}$ $s(4+h)=(4+h)^3+2(4+h)+9=64+48h+12h^2+h^3+8+2h+9=h^3+12h^2+50h+81$ $s(4)=81$ The instantaneous rate of change is then $\lim\limits_{h \to 0}\frac{h^3+12h^2+50h+81-81}{h}=\lim\limits_{h \to 0}\frac{h^3+12h^2+50h}{h}=\lim\limits_{h \to 0}h^2+12h+50=50$
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