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: 28

Answer

$g'(1) = 4$ Equation of the tangent line to the curve $y = x^4 - 2$ at the point $(1, -1): y=4x-5$

Work Step by Step

Given $g(x)=x^4-2$ \begin{gather*} g'(1) = \lim\limits_{x \to 1} \dfrac{g(x)-g(1)}{x-1} = \lim\limits_{x \to 1} \dfrac{x^4-2 - (1^4-2)}{x-1} = \lim\limits_{x \to 1} \dfrac{x^4-2 - (-1}{x-1} = \lim\limits_{x \to 1} \dfrac{x^4-2 +1}{x-1} = \lim\limits_{x \to 1} \dfrac{x^4-1}{x-1} = \lim\limits_{x \to 1} \dfrac{(x^2-1)(x^2+1)}{x-1} = \lim\limits_{x \to 1} \dfrac{(x-1)(x+1)(x^2+1)}{x-1} = \lim\limits_{x \to 1} [(x+1)(x^2+1)] = (1+1)(1^2+1) = \\ (2)(2) = 4 \longrightarrow g'(1) = 4 \end{gather*} From the point-slope equation, we have: \begin{gather*} y-g(1)=g'(1)(x-1) \longrightarrow y-(1^4-2)=4(x-1) \longrightarrow y+1=4x-4 \longrightarrow y= 4x-5 \end{gather*}
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