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: 13



Work Step by Step

We are given $s(t)=t^3+2t+9$ For $t = 1$ the instantaneous velocity is $\lim\limits_{h \to 0}\frac{s(1+h)-s(1)}{h}$ $s(1+h)=(1+h)^3+2(1+h)+9=1+3h+3h^2+h^3+2+2h+9=h^3+3h^2+5h+12$ $s(1)=12$ The instantaneous rate of change is then $\lim\limits_{h \to 0}\frac{h^3+3h^2+5h+12-12}{h}=\lim\limits_{h \to 0}\frac{h^3+3h^2+5h}{h}=\lim\limits_{h \to 0}(h^2+3h+5)=5$
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