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: 22

Answer

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

Work Step by Step

$a.$ Using the formula from definition (2) with $a=0$ and $f(a)=1$ (coordinates of the point $P(0,1)$) we have $$m_{tan}=\lim_{h\to0}\frac{f(0+h)-f(0)}{h}=\lim_{h\to0}\frac{\frac{1}{2(0+h)+1}-1}{h}=\lim_{h\to0}\frac{\frac{1}{2h+1}-1}{h}=\lim_{h\to0}\frac{\frac{1-(2h+1)}{2h+1}}{h}=\lim_{h\to0}\frac{-2h}{h(2h+1)}=\lim_{h\to0}\frac{-2}{2h+1}=\frac{-2}{2\cdot0+1}=-2.$$ $b.$ 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 $m_{tan} = -2$ we get $$y-1=-2(x-0)\Rightarrow y-1=-2x$$ which gives $$y=-2x+1.$$
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