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: 34


$a.$ The value of the derivative is $f'(12)=1/4.$ $b.$ The equation is $y=\frac{1}{4}x+3$.

Work Step by Step

$a.$ By definition of a derivative at point $a=12$ we have $$f'(12)=\lim_{h\to0}\frac{f(12+h)-f(12)}{h}=\lim_{h\to0}\frac{\sqrt{3(12+h)}-\sqrt{3\cdot12}}{h}=\lim_{h\to0}\frac{\sqrt{36+3h}-6}{h}=\lim_{h\to0}\frac{\sqrt{36+3h}-6}{h}\cdot\frac{\sqrt{36+3h}+6}{\sqrt{36+3h}+6}=\lim_{h\to0}\frac{\sqrt{36+3h}^2-6^2}{h(\sqrt{36+3h}+6)}=\lim_{h\to0}\frac{36+3h-36}{h(\sqrt{36+3h}+6)}=\lim_{h\to0}\frac{3h}{h(\sqrt{36+3h}+6)}=\lim_{h\to0}\frac{3}{\sqrt{36+3h}+6}=\frac{3}{\sqrt{36+3\cdot0}+6}=\frac{1}{4}.$$ $b.$ We know that the equation of the tangent at point $(a,f(a))$ is given by $y-f(a)=f'(a)(x-a)$. Using $a=12$, $f(a)=\sqrt{3\cdot12}=6$ and $f'(a)=1/4$ we get $$y-6=\frac{1}{4}(x-12)\Rightarrow y-6=\frac{1}{4}x-3$$ which gives $$y=\frac{1}{4}x+3.$$
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