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: 24


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

Work Step by Step

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