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: 56


a) $ f′(x)=\dfrac{-1}{x^2}$ b) $\dfrac{-1}{4}, \dfrac{-1}{4}$

Work Step by Step

a) Now, $ f′(x)=\lim_\limits{ h\to 0} \dfrac{f(x+h)-f(x)}{h}$ or, $=\lim_\limits{ h\to 0} \dfrac{\dfrac{1}{x+h}-1/x}{h}$ or, $=\lim_\limits{ h\to 0} \dfrac{-1}{x(x+h)}$ or, $=6x^2+6x(0)+2(0)^2-1$ or, $ f′(x)=\dfrac{-1}{x^2}$ b) $ f′(-2)=\dfrac{-1}{(-2)^2}=\dfrac{-1}{4}$ and $ f′(2)=\dfrac{-1}{(2)^2}=\dfrac{-1}{4}$
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