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 - Page 133: 20


$a.$ The slope is $m_{tan}=-1$. $b.$ The equation is $y=-x+2.$

Work Step by Step

$a.$ Using the formula from definition (2) with $a=1$ and $f(a)=1$ (coordinates of the point $P(1,1)$) we have $$m_{tan}=\lim_{h\to0}\frac{f(1+h)-f(1)}{h}=\lim_{h\to0}\frac{\frac{1}{1+h}-1}{h}=\lim_{h\to0}\frac{\frac{1-(1+h)}{1+h}}{h}=\lim_{h\to0}\frac{-h}{h(1+h)}\lim_{h\to0}\frac{-1}{(1+h)}=\frac{-1}{1+0}=-1.$$ $b.$ Using the formula $y-f(a)=m_{tan}(x-a)$ with the same values for $a$ and $f(a)$ as in part $a$ and the calculated value $m_{tan} = -1$ we get $$y-1=-1\cdot(x-1)\Rightarrow y-1=-x+1$$ which gives $$y=-x+2.$$
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