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: 23


$a.$ The slope is $m_{tan}=\frac{2}{25}$. $b.$ The equation is $y=\frac{2}{25}x+\frac{7}{25}$.

Work Step by Step

$a.$ Using the formula from definition (2) with $a=-1$ and $f(a)=1/5$ (coordinates of the point $P(-1,1/5)$) we have $$m_{tan}=\lim_{h\to0}\frac{f(-1+h)-f(-1)}{h}=\lim_{h\to0}\frac{\frac{1}{3-2(-1+h)}-\frac{1}{5}}{h}=\lim_{h\to0}\frac{\frac{1}{5-2h}-\frac{1}{5}}{h}=\lim_{h\to0}\frac{\frac{5-(5-2h)}{5(5-h)}}{h}=\lim_{h\to0}\frac{2h}{5h(5-h)}=\lim_{h\to0}\frac{2}{5(5-h)} = \frac{2}{5(5-0)} = \frac{2}{25}.$$ $b.$ 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 $m_{tan} = 2/25$ we get $$y-\frac{1}{5}=\frac{2}{25}(x-(-1))\Rightarrow y- \frac{1}{5}=\frac{2}{25}x+\frac{2}{25}$$ which gives $$y=\frac{2}{25}x+\frac{2}{25}+\frac{1}{5}=\frac{2}{25}x+\frac{7}{25}.$$
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