Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.1 Introducing the Derivative - 3.1 Execises: 57

Answer

This expression represents the slope of the tangent of the function $f(x)=\frac{1}{x+1}$ at the point $a=2$ and it has the value of $m_{tan}=-\frac{1}{9}$.

Work Step by Step

By the definition of the slope we have that $$m_{tan}=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}.$$ We see that if we take $a=2$ and $f(x)=\frac{1}{x+1}$ (then also $f(a)=f(2)=\frac{1}{2+1}=\frac{1}{3}$) we get the expression for the slope of the function $f(x)=\frac{1}{x+1}$ at $a=2$. Now lets find the slope: $$m_{tan}=\lim_{x\to2}\frac{\frac{1}{x+1}-\frac{1}{3}}{x-2}=\lim_{x\to2}\frac{\frac{3-(x+1)}{3(x+1)}}{x-2}=\lim_{x\to2}\frac{2-x}{3(x+1)(x-2)}=\lim_{x\to2}\frac{-(x-2)}{3(x+1)(x-2)}=\lim_{x\to2}\frac{-1}{3(x+1)}=\frac{-1}{3(2+1)}=-\frac{1}{9}.$$
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