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: 32


$ f'(1)=24$ Equation of the tangent line: $ y=24x-16$

Work Step by Step

Recall that $ f'(a)=\lim\limits_{h \to 0}\frac{f(a+h)-f(a)}{h}$ Given that $ a=1$. Therefore, $ f'(1)=\lim\limits_{h \to 0}\frac{f(1+h)-f(1)}{h}=\lim\limits_{h \to 0}\frac{[8(1+h)^{3}]-(8\times1^{3})}{h}$ $=\lim\limits_{h \to 0}\frac{8(1^{3}+3h+3h^{2}+h^{3}-1)}{h}=\lim\limits_{h \to 0}\frac{8h^{3}+24h^{2}+24h}{h}=\lim\limits_{h \to 0}(8h^{2}+24h+24)=24$ Equation of the tangent line is of the form $ y-f(a)=f'(a)(x-a)$ Knowing that $ f(a)=f(1)=8, f'(a)=f'(1)=24$ and $ a=1$, we obtain the equation of the tangent line through $ a $ as below. $ y-8=24(x-1)$ Or $ y=24x-16$
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