Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.3 Introduction To Techniques Of Differentiation - Exercises Set 2.3 - Page 140: 27

Answer

putting the value of $f'(x)=0$

Work Step by Step

$f(x)=y=\frac{x^2+1}{x}$ Now using Quotient rule $f'(x)=\frac{dy}{dx}=\frac{d}{dx}(\frac{x^2+1}{x})$ $=\frac{x.\frac{d}{dx}x^2+1-\frac{d}{dx} x . (x^2+1)}{x^2}$ $=\frac{x.2x-1.(x^2+1)}{x^2}$ $=\frac{2x^2-x^2-1}{x^2}$ $=\frac{x^2(2-1)-1}{x^2}$ $=1-\frac{1}{x^2}$ putting the value of $f'(x)=1-\frac{1}{1^2}$ putting the value of $f'(x)=1-1$ putting the value of $f'(x)=0$
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