College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 2, Functions - Section 2.1 - Functions - 2.1 Exercises - Page 192: 24


$h(-1)=-2$ $h(2)=\frac{5}{2}$ $h(\frac{1}{2})=\frac{5}{2}$ $h(x-1)=x-1+ \frac{1}{x-1}$ $h(\frac{1}{x})=\frac{1}{x}+x$

Work Step by Step

We are given: $h(x)=x+ \frac{1}{x}$ We evaluate: $h(-1)=(-1)+\frac{1}{-1}=-1-1=-2$ $h(2)=2+\frac{1}{2}=\frac{5}{2}$ $h(\frac{1}{2})=\frac{1}{2}+\frac{1}{\frac{1}{2}}=\frac{1}{2}+2=\frac{1}{2}+\frac{2*2}{2}=\frac{5}{2}$ $h(x-1)=x-1+ \frac{1}{x-1}$ $h(\frac{1}{x})=\frac{1}{x}+\frac{1}{\frac{1}{x}}=\frac{1}{x}+x$
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