Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.1 Derivatives and Rates of Change - 2.1 Exercises - Page 114: 31


$f'(a) = 6a-4$

Work Step by Step

Given $f(x) = 3x^2-4x+1$ $f'(a) = \lim\limits_{x \to a} \dfrac{f(x)-f(a)}{x-a} = \lim\limits_{x \to a} \dfrac{3x^2-4x+1-(3a^2-4a+1)}{x-a} = \lim\limits_{x \to a} \dfrac{3x^2-4x+1-3a^2+4a-1}{x-a} = \lim\limits_{x \to a} \dfrac{3x^2-3a^2-4x+4a}{x-a} = \lim\limits_{x \to a} \dfrac{3(x^2-a^2)-4(x-a)}{x-a} =\lim\limits_{x \to a} \dfrac{3(x+a)(x-a)-4(x-a)}{x-a} = \lim\limits_{x \to a} \dfrac{(x-a)[3(x+a)-4]}{x-a} = \lim\limits_{x \to a} [3(x+a)-4] = \\ 3(a+a)-4 = 3(2a)-4 = 6a-4$
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