Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.7 - Derivatives and Rates of Change - 2.7 Exercises - Page 149: 28


$g'(1)=4$ The equation of the tangent line is $$(l):y=4x-5$$

Work Step by Step

$$g(x)=x^4-2$$ 1) The derivative of $g(x)$ at a number $a$ is calculated as follows: $$g'(a)=\lim\limits_{x\to a}\frac{g(x)-g(a)}{x-a}$$ $$g'(a)=\lim\limits_{x\to a}\frac{(x^4-2)-(a^4-2)}{x-a}$$ $$g'(a)=\lim\limits_{x\to a}\frac{x^4-a^4}{x-a}$$ $$g'(a)=\lim\limits_{x\to a}\frac{(x^2-a^2)(x^2+a^2)}{x-a}$$ $$g'(a)=\lim\limits_{x\to a}\frac{(x-a)(x+a)(x^2+a^2)}{x-a}$$ $$g'(a)=\lim\limits_{x\to a}(x+a)(x^2+a^2)$$ $$g'(a)=(a+a)(a^2+a^2)$$ $$g'(a)=2a\times2a^2=4a^3$$ So $g'(1)=4\times1^3=4$ 2) The equation of the tangent line $l$ of the curve $y$ at point $(1,-1)$ would have the following form $$(l):y=g'(1)x+b$$$$(l):y=4x+b$$ Since point $(1-1)$ lies in the tangent line $l$, we can find $b$ by applying the point into the equation of the tangent line $l$: $$4\times1+b=-1$$$$4+b=-1$$$$b=-5$$ Overall, the equation of tangent line $l$ is $$(l):y=4x-5$$
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