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: 15

Answer

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

Work Step by Step

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