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: 36


$a.$ The value of the derivative is $f'(2)=-3/25$. $b.$ The equation is $y=-\frac{3}{25}x+\frac{11}{25}.$

Work Step by Step

$a.$ By definition of a derivative at point $a=2$ we have $$f'(2)=\lim_{h\to0}\frac{f(2+h)-f(2)}{h}=\lim_{h\to0}\frac{\frac{1}{3(2+h)-1}-\frac{1}{3\cdot2-1}}{h}=\lim_{h\to0}\frac{\frac{1}{5+3h}-\frac{1}{5}}{h}=\lim_{h\to0}\frac{\frac{5-(5+3h)}{5(5+3h)}}{h}=\lim_{h\to0}\frac{-3h}{5h(5+3h)}=\lim_{h\to0}\frac{-3}{5(5+3h)}=\frac{-3}{5(5+3\cdot0)}=-\frac{3}{25}.$$ $b.$ We know that the equation of the tangent at point $(a,f(a))$ is given by $y-f(a)=f'(a)(x-a)$. Using $a=2$, $f(a)=1/(3\cdot2-1)=1/5$ and $f'(a)=-3/25$ we get $$y-\frac{1}{5}=-\frac{3}{25}(x-2)\Rightarrow y-\frac{1}{5}=-\frac{3}{25}x+\frac{6}{25}$$ which gives $$y=-\frac{3}{25}x+\frac{6}{25}+\frac{1}{5}=-\frac{3}{25}x+\frac{11}{25}.$$
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