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: 16

Answer

$a.$ The slope is $m_{tan}=2.$ $b.$ The equation is $y=2x-3.$

Work Step by Step

$a.$ Using the formula from definition (2) with $a=1$ and $f(a)=-1$ (coordinates of the point $P(1,-1)$) we have $$m_{tan}=\lim_{h\to0}\frac{f(1+h)-f(1)}{h}=\lim_{h\to0}\frac{3(1+h)^2-4(1+h)-(-1)}{h}=\lim_{h\to0}\frac{3(1+h^2+2h)-4(1+h)+1}{h}=\lim_{h\to0}\frac{3+3h^2+6h-4-4h+1}{h}=\lim_{h\to0}\frac{3h^2+2h}{h}=\lim_{h\to0}(3h+2)=3\cdot 0+2=2.$$ $b.$ 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 $$y-(-1)=2(x-1)\Rightarrow y+1=2x-2$$ which gives $$y=2x-3.$$
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