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: 28

Answer

$a.$ The value of the derivative is $f'(3)=6$. $b.$ The equation is $y=6x-9$.

Work Step by Step

$a.$ By definition of a derivative at point $a=3$ we have $$f'(3)=\lim_{h\to0}\frac{f(3+h)-f(3)}{h}=\lim_{h\to0}\frac{(3+h)^2-3^2}{h}=\lim_{h\to0}\frac{9+h^2+6h-9}{h}=\lim_{h\to0}\frac{h^2+6h}{h}=\lim_{h\to0}(h+6)=0+6=6.$$ $b.$ We know that the equation of the tangent at point $(a,f(a))$ is given by $y-f(a)=f'(a)(x-a)$. Using $a=3$, $f(a)=3^2=9$ and $f'(a)=6$ we get $$y-9=6(x-3)\Rightarrow y-9=6x-18$$ which gives $$y=6x-9.$$
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