Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.5 Inequalities - R.5 Exercises - Page R-21: 49

Answer

The solution is $[2,3)$

Work Step by Step

$\frac{2k}{k-3} \leq \frac{4}{k-3}$ $\frac{2k}{k-3} - \frac{4}{k-3}\leq0$ First solve the corresponding equation $\frac{2k}{k-3} - \frac{4}{k-3}=0$ $2k-4=0$ $k=2$ For $(-\infty,2)$, choose $0$: $\frac{2(0)}{0-3} - \frac{4}{0-3}=\frac{4}{3}\gt0$ For $(2,3)$, choose $2.5$: $\frac{2(2.5)}{2.5-3} - \frac{4}{2.5-3}=-2\lt0$ For $(3,\infty)$, choose $4:$, $\frac{2(4)}{4-3} - \frac{4}{4-3}=4\gt$ The solution is $[2,3)$
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