Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.1 Definition of the Derivative - Exercises - Page 103: 35

Answer

$ f'(8)=-\frac{1}{64}$ Equation of the tangent line: $ y=-\frac{1}{64}x+\frac{1}{4}$

Work Step by Step

Recall that $ f'(a)=\lim\limits_{h \to 0}\frac{f(a+h)-f(a)}{h}$ Given that $ a=8$. Therefore, $ f'(8)=\lim\limits_{h \to 0}\frac{f(8+h)-f(8)}{h}=\lim\limits_{h \to 0}\frac{(8+h)^{-1}-(8^{-1})}{h}$ $=\lim\limits_{h \to 0}\frac{\frac{-h}{8(8+h)}}{h}=\lim\limits_{h \to 0}\frac{-1}{8(8+h)}=-\frac{1}{64}$ Equation of the tangent line is of the form $ y-f(a)=f'(a)(x-a)$ Knowing that $ f(a)=f(8)=\frac{1}{8}, f'(a)=f'(8)=-\frac{1}{64}$ and $ a=8$, we obtain the equation of the tangent line through $ a $ as below. $ y-\frac{1}{8}=-\frac{1}{64}(x-8)$ Or $ y=-\frac{1}{64}x+\frac{1}{4}$
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