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: 27


$a.$ The value of the derivative is $f'(-3)=8.$ $b.$ The equation is $y=8x$.

Work Step by Step

$a.$ By definition of a derivative at point $a=-3$ we have $$f'(-3)=\lim_{h\to0}\frac{f(-3+h)-f(-3)}{h}=\lim_{h\to0}\frac{8(-3+h)-8\cdot(-3)}{h}=\lim_{h\to0}\frac{-24+8h+24}{h}=\lim_{h\to0}\frac{8h}{h}=\lim_{h\to0}8=8.$$ $b.$ We know that the equation of the tangent at point $(a,f(a))$ is given by $y-f(a)=f'(a)(x-a)$. Using $a=-3$, $f(a)=8\cdot(-3)=-24$ and $f'(a)=8$ we get $$y-(-24)=8(x-(-3))\Rightarrow y+24=8x+24$$ which gives $$y=8x.$$
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