University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Practice Exercises - Page 279: 130


Absolute maximum value is at $f(e)=10e$ Absolute minimum value is at $f(e^2)=0$

Work Step by Step

Let us consider $y=f(x)=10x (2-\ln x)=20 x-10 \ln x$ Now, $f'(x)=10(1-\ln x)$ Now, $f'(e)= 10(1-\ln e)=10(1-1)0$ Thus, at $x=e; f'(x)=0$ This shows a critical point. Now, $f(e)= 10e (2-\ln e)=10e$ So, absolute maximum value is at $f(e)=10e$ and $f(e^2)= 10e^2(2-\ln e^2)=10e^2(2-2)=0$ So, absolute minimum value is at $f(e^2)=0$
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