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: 32


$a.$ The value of the derivative is $f'(1)=-2$. $b.$ The equation is $y=-2x+3$.

Work Step by Step

$a.$ By definition of a derivative at point $a=1$ we have $$f'(1)=\lim_{h\to0}\frac{f(1+h)-f(1)}{h}=\lim_{h\to0}\frac{\frac{1}{(1+h)^2}-\frac{1}{1^2}}{h}=\lim_{h\to0}\frac{\frac{1-(1+h)^2}{(1+h)^2}}{h}=\lim_{h\to0}\frac{1-(1+h^2+2h)}{h(1+h)^2}=\lim_{h\to0}\frac{1-1-h^2-2h}{h(1+h)^2}=\lim_{h\to0}\frac{-h^2-2h}{h(1+h)^2}=\lim_{h\to0}\frac{-h(h+2)}{h(h+1)^2}=\lim_{h\to0}\frac{-(h+2)}{(h+1)^2}=\frac{-(0+2)}{(0+1)^2}=\frac{-2}{1}=-2.$$ $b.$ We know that the equation of the tangent at point $(a,f(a))$ is given by $y-f(a)=f'(a)(x-a)$. Using $a=1$, $f(a)=1/1^2=1$ and $f'(a)=-2$ we get $$y-1=-2(x-1)\Rightarrow y-1=-2x+2$$ which gives $$y=-2x+3.$$
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