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: 8

Answer

The equation of the tangent line $l$ of the curve at point $A$ is $$(l): y=\frac{1}{3}x+\frac{2}{3}$$

Work Step by Step

$$y=f(x)=\frac{2x+1}{x+2}$$ Given point $A (1,1)$ According to definition, the slope of the tangent line $l$ at the given point $A$ is $$m_l=\lim\limits_{x\to1}\frac{f(x)-f(1)}{x-1}$$$$m_l=\lim\limits_{x\to1}\frac{\frac{2x+1}{x+2} -1}{x-1}$$$$m_l=\lim\limits_{x\to1}\frac{\frac{x-1}{x+2}}{x-1}$$$$m_l=\lim\limits_{x\to1}\frac{x-1}{(x+2)(x-1)}$$$$m_l=\lim\limits_{x\to1}\frac{1}{x+2}$$$$m_l=\frac{1}{1+2}=\frac{1}{3}$$ So, the tangent line $l$ of the given function at the given point $A$ would have the form $$(l): y=\frac{1}{3}x+b$$ Since the tangent line $l$ passes through point $A(1,1)$, we have $$\frac{1}{3}\times1+b=1$$$$\frac{1}{3}+b=1$$$$b=\frac{2}{3}$$ Therefore, the equation of the tangent line $l$ of the curve at point $A$ is $$(l): y=\frac{1}{3}x+\frac{2}{3}$$
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