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: 29

Answer

$a.$ The value of the derivative is $f'(-2)=-14$. $b.$ The equation is $y=-14x-16.$

Work Step by Step

$a.$ By definition of a derivative at point $a=-2$ we have $$f'(-2)=\lim_{h\to0}\frac{f(-2+h)-f(-2)}{h}=\lim_{h\to0}\frac{4(-2+h)^2+2(-2+h)-(4(-2)^2+2(-2))}{h}=\lim_{h\to0}\frac{4(h^2+4-4h)+2h-4-12}{h}=\lim_{h\to0}\frac{4h^2+16-16h+2h-4-12}{h}=\lim_{h\to0}\frac{4h^2-14h}{h}.=\lim_{h\to0}(4h-14)=4\cdot0-14=-14$$ $b.$ We know that the equation of the tangent at point $(a,f(a))$ is given by $y-f(a)=f'(a)(x-a)$. Using $a=-2$, $f(a)=4(-2)^2+2(-2)=12$ and $f'(a)=-14$ we get $$y-12=-14(x-(-2))\Rightarrow y-12=-14x-28$$ which gives $$y=-14x-16.$$
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