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: 21

Answer

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

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 $$m_{tan}=\lim_{h\to0}\frac{f(1+h)-f(1)}{h}=\lim_{h\to0}\frac{(1+h)^3-1}{h}=\lim_{h\to0}\frac{h^3+3h^2+3h+1-1}{h}=\lim_{h\to0}\frac{h^3+3h^2+3h}{h}=\lim_{h\to0}(h^2+3h+3)=0^2+3\cdot0+3=3.$$ $b.$ 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 $m_{tan} = 3$ we get $$y-1=3(x-1)\Rightarrow y-1=3x-3$$ which gives $$y=3x-2.$$
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