Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 3 - Review - Test - Page 324: 12

Answer

$(-\infty,-1]\cup (\frac{5}{2},3]$

Work Step by Step

Rewrite the inequality as $x-\frac{6-x}{2x-5}\leq0$ which gives $\frac{(x+1)(x-3)}{2x-5}\leq0$ Test signs in intervals $(-\infty,-1), (-1,5/2), (5/2,3), (3,\infty)$, we get $-,+,-,+$ for the function. Test end points and we get the solution as $x\in(-\infty,-1]\cup (\frac{5}{2},3]$
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