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: 30

Answer

$a.$ The value of the derivative is $f'(10)=600.$ $b.$ The equation is $y=600x-4000.$

Work Step by Step

$a.$ By definition of a derivative at point $a=10$ we have $$f'(10)=\lim_{h\to0}\frac{f(10+h)-f(10)}{h}=\lim_{h\to0}\frac{2\cdot(10+h)^3-2\cdot 10^3}{h}=\lim_{h\to0}\frac{2(10^3+3h10^2+3h^2\cdot 10+h^3)-2\cdot10^3}{h}=\lim_{h\to0}\frac{2\cdot10^3+6\cdot10^2h+60h^2+2h^3-2\cdot10^3}{h}=\lim_{h\to0}\frac{2h^3+600h+60h^2}{h}=\lim_{h\to0}(2h^2+600+60h)=2\cdot0^2+600+60\cdot0=600.$$ $b.$ We know that the equation of the tangent at point $(a,f(a))$ is given by $y-f(a)=f'(a)(x-a)$. Using $a=10$, $f(a)=2\cdot10^3=2000$ and $f'(a)=600$ we get $$y-2000=600(x-10)\Rightarrow y-2000=600x-6000$$ which gives $$y=600x-4000.$$
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