Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 11 - Section 11.4 - Nonlinear Inequalities in One Variable - Exercise Set - Page 799: 45

Answer

(-infinity, $-6)$ U $(-1, 0)$ U $(7$, infinity)

Work Step by Step

$\frac {x*(x+6)}{(x-7)(x+1)} \geq0$ The denominator is zero when $x=7$ and $x=-1$ $\frac {x*(x+6)}{(x-7)(x+1)} \geq0$ $(x-7)(x+1)*\frac {x*(x+6)}{(x-7)(x+1)} \geq(x-7)(x+1)*0$ $x*(x+6) \geq 0$ $x*(x+6) = 0$ $x=0$ $x+6=0$ $x=-6$ (-infinity, $-6)$ $(-6, -1)$ $(-1, 0)$ $(0, 7)$ $(7$, infinity) Let $x=-10$, $x=-2$, $x=-1/2$, $x=2$, $x=10$ $x=-10$ $\frac {x*(x+6)}{(x-7)(x+1)} \geq0$ $\frac {-10*(-10+6)}{(-10-7)(-10+1)} \geq0$ $\frac {-10*(-16)}{(-17)(-9)} \geq0$ $\frac {160}{153} \geq0$ (true) $x=-2$ $\frac {x*(x+6)}{(x-7)(x+1)} \geq0$ $\frac {-2*(-2+6)}{(-2-7)(-2+1)} \geq0$ $\frac {-2*(4)}{(-9)(-1)} \geq0$ $\frac {-8}{9} \geq0$ (false) $x=-1/2$ $\frac {x*(x+6)}{(x-7)(x+1)} \geq0$ $\frac {-.5*(-.5+6)}{(-.5-7)(-.5+1)} \geq0$ $\frac {-.5*(5.5)}{(-7.5)(.5)} \geq0$ $\frac {-2.75}{-3.75} \geq0$ $\frac {-2.75*4}{-3.75*4} \geq0$ $\frac {-11}{-15} \geq0$ $11/15 \geq 0$ (true) $x=2$ $\frac {x*(x+6)}{(x-7)(x+1)} \geq0$ $\frac {2*(2+6)}{(2-7)(2+1)} \geq0$ $\frac {2*8}{(-5)(3)} \geq0$ $\frac {16}{-15} \geq0$ (false) $x=10$ $\frac {x*(x+6)}{(x-7)(x+1)} \geq0$ $\frac {10*(10+6)}{(10-7)(10+1)} \geq0$ $\frac {10*(16)}{(10-7)(10+1)} \geq0$ $\frac {160}{3*11} \geq0$ $\frac {160}{33} \geq0$ (true)
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