Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 3 - Section 3.7 - Polynomial and Rational Inequalities - 3.7 Exercises - Page 316: 36

Answer

$(-\infty,-3-\sqrt 2]\cup(-3,-2)\cup[-3+\sqrt 2,-1)$

Work Step by Step

1. Move all items to one side $f(x)=\frac{1}{x+1}+\frac{1}{x+2}-\frac{1}{x+3} =\frac{x^2+5x+6+x^2+3x+3-x^2-2x-2}{(x+1)(x+2)(x+3)} =\frac{x^2+6x+7}{(x+1)(x+2)(x+3)}$ $f(x)==\frac{(x+3+\sqrt 2)(x+3-\sqrt 2)}{(x+1)(x+2)(x+3)}\leq0$ 2. Cut points $-3-\sqrt 2 (-4.4), -3,-2,-3+\sqrt 2 (-1.6),-1$, make a table as shown. 3. Write the solutions $(-\infty,-3-\sqrt 2]\cup(-3,-2)\cup[-3+\sqrt 2,-1)$
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