Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - 3.1 Maximum and Minimum Values - 3.1 Execises - Page 212: 51


Absolute maximum: $f(0.2)=5.2$ Absolute minimum: $f(1)=2$

Work Step by Step

Using closed interval method, as $f$ in continuous over the given range, let's first calculate the value of $f$ at critical numbers followed by values of $f$ at endpoints of the given range. For critical numbers, $f'(x)=0$ or $f'(x)$ should not exist. Let's differentiate $f(x)=x+\frac1x$ with respect to $x$ keeping $[0.2,4]$ as interval. $$\frac{df}{dx}=1-\frac{1}{x^2}$$ $\frac{df}{dx}$ exists on the given interval. Thus, $\frac{df}{dx}=0$, $$1-\frac{1}{x^2}=0\implies x^2-1=0\implies x=-1, 1.$$ Thus, $f(1)=2$. At endpoints, $f(0.2)=5.2$ & $f(4)=4.25$. Thus, absolute maximum $=f(0.2)=5.2$ and absolute minimum $=f(1)=2$.
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