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: 9

Answer

$a.$ The slope is $m_{tan} = 6$. $b.$ The equation is $y=6x-14.$ $c.$ The graph is on the figure below.
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Work Step by Step

By definition (1) we have $a.$ The slope is given by $$m_{tan}=\lim_{x\to 3}\frac{f(x)-f(3)}{x-3} =\lim_{x\to 3}\frac{x^2-5-(3^2-5)}{x-3}=\lim_{x\to 3}\frac{x^2-9}{x-3} = \lim_{x\to 3}\frac{(x-3)(x+3)}{x-3} = \lim_{x\to 3}(x+3) = 6 $$ where in the last step we used the substitution to evaluate the limit. $b.$ Applying formula for tangent line $y-f(a) = m_{tan}(x-a)$ with $a=3$ and $m_{tan} = 6$ we have $$y-(3^2-5)=6(x-3)\Rightarrow y-4=6x-18$$ which gives $$y=6x-14.$$ $c.$ The graph is on the figure below. The function is solid and the tangent is dashed
Small 1510684259
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