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: 31


$ f'(3)=-11$ Equation of the tangent line: $ y=-11t+18$

Work Step by Step

Recall that $ f'(a)=\lim\limits_{h \to 0}\frac{f(a+h)-f(a)}{h}$ Given that $ a=3$. Therefore, $ f'(3)=\lim\limits_{h \to 0}\frac{f(3+h)-f(3)}{h}=\lim\limits_{h \to 0}\frac{[(3+h)-2(3+h)^{2}]-[3-2(3)^{2}]}{h}$ $=\lim\limits_{h \to 0}\frac{(3+h-2\times3^{2}-2h^{2}-2\times6h)-(-15)}{h}=\lim\limits_{h \to 0}\frac{-2h^{2}-11h}{h}=\lim\limits_{h \to 0}(-2h-11)=-11$ Equation of the tangent line is of the form $ y-f(a)=f'(a)(t-a)$ Knowing that $ f(a)=f(3)=-15,f'(a)=f'(3)=-11$ and $ a=3$, we obtain the equation of the tangent line through $ a $ as below. $ y-(-15)=-11(t-3)$ Or $ y=-11t+18$
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