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: 33


$a.$ The value of the derivative is $f'(4)=1/3$. $b.$ The equation is $y=\frac{1}{3}x+\frac{5}{3}.$

Work Step by Step

$a.$ By definition of a derivative at point $a=4$ we have $$f'(4)=\lim_{h\to0}\frac{f(4+h)-f(4)}{h}=\lim_{h\to0}\frac{\sqrt{2(4+h)+1}-\sqrt{2\cdot 4+1}}{h}=\lim_{h\to0}\frac{\sqrt{9+2h}-3}{h}=\lim_{h\to0}\frac{\sqrt{9+2h}-3}{h}\cdot\frac{\sqrt{9+2h}+3}{\sqrt{9+2h}+3}=\lim_{h\to0}\frac{\sqrt{9+2h}^2-3^2}{h(\sqrt{9+2h}+3)}=\lim_{h\to0}\frac{9+2h-9}{h(\sqrt{9+2h}+3)}=\lim_{h\to0}\frac{2h}{h(\sqrt{9+2h}+3)}=\lim_{h\to0}\frac{2}{\sqrt{9+2h}+3}=\frac{2}{\sqrt{9+2\cdot0}+3}=\frac{1}{3}.$$ $b.$ We know that the equation of the tangent at point $(a,f(a))$ is given by $y-f(a)=f'(a)(x-a)$. Using $a=4$, $f(a)=\sqrt{2\cdot4+1}=3$ and $f'(a)=1/3$ we get $$y-3=\frac{1}{3}(x-4)\Rightarrow y-3=\frac{1}{3}x-\frac{4}{3}$$ which gives $$y=\frac{1}{3}x-\frac{4}{3}+3=\frac{1}{3}x+\frac{5}{3}.$$
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