Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises: 31

Answer

The equation of the tangent line $l$ to the given curve at $(1,0)$ is $$(l): y=\frac{2}{3}x-\frac{2}{3}$$

Work Step by Step

$$y=f(x)=\frac{x^2-1}{x^2+x+1}$$ 1) Find $f'(x)$ $$f'(x)=\frac{(x^2-1)'(x^2+x+1)-(x^2+x+1)'(x^2-1)}{(x^2+x+1)^2}$$ $$f'(x)=\frac{2x(x^2+x+1)-(2x+1)(x^2-1)}{(x^2+x+1)^2}$$ $$f'(x)=\frac{2x^3+2x^2+2x-2x^3+2x-x^2+1}{(x^2+x+1)^2}$$ $$f'(x)=\frac{x^2+4x+1}{(x^2+x+1)^2}$$ 2) The slope of the tangent line $l$ to the given curve at point $A(a,b)$ is $f'(a)$. Therefore, at $A(1,0)$, the slope of tangent line $l$ is $$f'(1)=\frac{1^2+4\times1+1}{(1^2+1+1)^2}$$ $$f'(1)=\frac{6}{9}=\frac{2}{3}$$ The equation of the tangent line $l$ to the given curve at $(1,0)$ is $$(l): (y-0)=f'(1)(x-1)$$ $$(l): y=\frac{2}{3}(x-1)$$ $$(l): y=\frac{2}{3}x-\frac{2}{3}$$
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