Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 8 - Section 8.4 - Nonlinear Inequalities in One Variable - Exercise Set - Page 511: 65

Answer

$(-∞, -1)$ U $(0, 1)$

Work Step by Step

$x-1/x <0$ $x(x-1/x) <0*x$ $x^2-1 < 0$ $(x+1)(x-1) < 0$ $x+1=0$ $x+1-1=0-1$ $x=-1$ $x-1=0$ $x-1+1=0+1$ $x=1$ The denominator of the original equation is zero when $x=0$ We have four ranges, and we need to test one value from each range to see if the range is part of the solution set. These ranges are $(-∞, -1)$, $(-1, 0)$, $(0, 1)$, and $(1,∞)$. Since we don't have a "less than or equal to" sign or a "greater than or equal to" sign, we exclude the endpoints and use parentheses and not brackets. Let $x=-2$, $x=-.5$, $x=.5$, $x=2$ $x=-2$ $x-1/x <0$ $-2-1/-2 <0$ $-2+1/2 <0$ $-3/2 < 0$ (true) $x=-.5$ $x-1/x <0$ $-.5-1/-.5 <0$ $-.5+2 <0$ $1.5 < 0$ (false) $x=.5$ $x-1/x <0$ $.5-1/.5 <0$ $.5-2 <0$ $-1.5 < 0$ (true) $x=2$ $x-1/x <0$ $2-1/2 <0$ $3/2 < 0$ (false)
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