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: 18


$a.$ The slope is $m_{tan}=0.$ $b.$ The equation is $y=8.$

Work Step by Step

$a.$ Using the formula from definition (2) with $a=0$ and $f(a)=8$ (coordinates of the point $P(0,8)$) we have $$m_{tan}=\lim_{h\to0}\frac{f(0+h)-f(0)}{h}=\lim_{h\to0}\frac{8-2(0+h)^2-8}{h}=\lim_{h\to0}\frac{-2h^2}{h}=\lim_{h\to0}-2h=-2\cdot0=0.$$ $b.$ 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 $m_{tan} = 0$ we get $$y-8=0\cdot(x-0)\Rightarrow y-8=0$$ which gives $$y=8.$$
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