Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Review Exercises - Page 1179: 58


a. $20.5\ in^3/in$, $20.05\ in^3/in$ b. $20\ in^3/in$

Work Step by Step

a. Given $f(x)=5x^2$, for the average rate of change from 2 to 2.1 inches, we need to evaluate the difference quotient $\frac{f(x+h)-f(x)}{h}=\frac{f(2.1)-f(2)}{2.1-2}=\frac{5(2.1)^2-5(2)^2}{0.1}=20.5\ in^3/in$. Similarly, from 2 to 2.01 inches, we have $\frac{f(2.01)-f(2)}{2.01-2}=\frac{5(2.01)^2-5(2)^2}{0.01}=20.05\ in^3/in$ b. For the instantaneous rate of change at $x=2$, we have $\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{f(2+h)-f(2)}{h}=\lim_{h\to0}\frac{5(2+h)^2-5(2)^2}{h}=\lim_{h\to0}\frac{20+20h+5h^2-20}{h}=\lim_{h\to0}(20+5h)=20\ in^3/in$ That is, $f'(2)=20\ in^3/in$
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