## Calculus 10th Edition

Published by Brooks Cole

# Chapter 3 - Applications of Differentiation - 3.1 Exercises: 27

#### Answer

Over the specified interval, the function has an absolute maximum equal to $-\dfrac{1}{2}$ and an absolute minimum equal to $-1.$

#### Work Step by Step

$h'(s)=-\dfrac{1}{(s-2)^2}$ $h'(s)$ is never equal to $0$ but is undefined for $s=2$ but since that is not in the specified interval. The only possible candidates for absolute extremum are the interval's endpoints. $h(0)=-\dfrac{1}{2}.$ $h(1)=-1.$ Over the specified interval, the function has an absolute maximum equal to $-\dfrac{1}{2}$ and an absolute minimum equal to $-1.$

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