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: 13

Answer

$a.$ The slope is $m_{tan} = -1$ $b.$ The equatin is $y=-x-2$ $c.$ The graph is on the figure below
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Work Step by Step

$a.$ For the slope of the tangent line at point $P(a,f(a))$ we have by definition (1) with $a=-1$ and $f(a) = -1$: $$m_{tan} = \lim_{x\to-1}\frac{f(x)-f(-1)}{x-(-1)}= \lim_{x\to-1}\frac{\frac{1}{x}+1}{x+1} = \lim_{x\to-1}\frac{\frac{1+x}{x}}{x+1} = \lim_{x\to-1}\frac{x+1}{x(x+1)}= \lim_{x\to-1}\frac{1}{x}=-1.$$ $b.$ From definition (1) we have $y-f(a) = m_{tan} (x-a)$. Using this formula with $a=-1$, $f(a)=-1$ and $m_{tan}=-1$: $$y-(-1)=-1(x-(-1))\Rightarrow y+1=-x-1$$ which gives $$y=-x-2.$$ $c.$ The graph is on the figure below. The function is solid and the tangent is dashed.
Small 1510687697
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