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: 19

Answer

$a.$ The slope is $m_{tan}=4.$ $b.$ The equation is $y=4x-8.$

Work Step by Step

$a.$ Using the formula from definition (2) with $a=2$ and $f(a)=0$ (coordinates of the point $P(2,0)$) we have $$m_{tan}=\lim_{h\to0}\frac{f(2+h)-f(2)}{h}=\lim_{h\to0}\frac{(2+h)^2-4-0}{h}=\lim_{h\to0}\frac{4+h^2+4h-4}{h}=\lim_{h\to0}\frac{h^2+4h}{h}=\lim_{h\to0}(h+4)=0+4=4.$$ $b.$ 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 $m_{tan} = 4$ we get $$y-0=4(x-2)\Rightarrow y=4x-8.$$
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