Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Review Exercises - Page 432: 83



Work Step by Step

We are given the function $ T(x)=\dfrac{4}{x+3}+\dfrac{2}{x}$ This implies that $ T(x)=\dfrac{4}{x+3}+\dfrac{2}{x}=\dfrac{4x+2(x+3)}{x(x+3)}$ Therefore, $ T(x)=\dfrac{4x+2x+6}{x(x+3)}=\dfrac{6x+6}{x(x+3)}$
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