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: 25


$a.$ The slope is $m_{tan} = \frac{1}{4}$. $b.$ The equation is $y=\frac{1}{4}x+\frac{7}{4}.$

Work Step by Step

$a.$ Using the formula for the slope of the tangent at point $P(a,f(a))$ from definition (2) with $a=1$ and $f(a)=2$ we have $$m_{tan}=\lim_{h\to0}\frac{f(1+h)-f(1)}{h}=\lim_{h\to0}\frac{\sqrt{(1+h)+3}-2}{h}=\lim_{h\to0}\frac{\sqrt{h+4}-2}{h}\cdot\frac{\sqrt{h+4}+2}{\sqrt{h+4}+2}=\lim_{h\to0}\frac{\sqrt{h+4}^2-2^2}{h(\sqrt{h+4}+2)}=\lim_{h\to0}\frac{h+4-4}{h(\sqrt{h+4}+2)}=\lim_{h\to0}\frac{h}{h(\sqrt{h+4}+2)}=\lim_{h\to0}\frac{1}{\sqrt{h+4}+2}=\frac{1}{\sqrt{0+4}+2}=\frac{1}{4}.$$ $b.$ Using the formula from definition (1) for the equation of the slope $y-f(a)=m_{tan}(x-a)$ with the same values for $a$ and $f(a)$ as in the part $a.$ and the calculated value $m_tan=\frac{1}{4}$ we get $$y-2=\frac{1}{4}(x-1)\Rightarrow y-2=\frac{1}{4}x-\frac{1}{4},$$ which gives $$y=\frac{1}{4}x-\frac{1}{4}+2=\frac{1}{4}x+\frac{7}{4}.$$
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