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: 34


$ f'(4)=100$ Equation of the tangent line: $ y=100t-256$

Work Step by Step

Recall that $ f'(a)=\lim\limits_{h \to 0}\frac{f(a+h)-f(a)}{h}$ Given that $ a=4$. Therefore, $ f'(4)=\lim\limits_{h \to 0}\frac{f(4+h)-f(4)}{h}=\lim\limits_{h \to 0}\frac{[2(4+h)^{3}+4(4+h)]-(2\times4^{3}+4\times4)}{h}$ $=\lim\limits_{h \to 0}\frac{[2(4^{3}+3\times4^{2}\times h+3\times4\times h^{2}+h^{3})+16+4h]-144}{h}=\lim\limits_{h \to 0}\frac{2h^{3}+24h^{2}+100h}{h}=\lim\limits_{h \to 0}(2h^{2}+24h+100)=100$ Equation of the tangent line is of the form $ y-f(a)=f'(a)(t-a)$ Knowing that $ f(a)=f(4)=144, f'(a)=f'(4)=100$ and $ a=4$, we obtain the equation of the tangent line through $ a $ as below. $ y-144=100(t-4)$ Or $ y=100t-256$
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