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: 20

Answer

$(-3, -5/2] U [5/2, 3) $

Work Step by Step

$\frac{4x^2 - 25}{x^2 - 9} \leq 0$ Find the zeros of the expressions in the numerator AND the denominator $(2x + 5) (2x - 5) = 0$; $(x-3)(x+3) = 0$ $x = -5/2, 5/2, 3, -3$ Test numbers in between those zero values to determine if the function is negative or positive (-∞, -3) $\frac{(-)(-)}{(-)(-)} = (+)$ (-3, -5/2] $\frac{(-)(-)}{(-)(+)} = (-)$ [-5/2, 5/2] $\frac{(+)(-)}{(-)(+)} = (+)$ [5/2, 3) $\frac{(+)(+)}{(-)(+)} = (-)$ (3, ∞) $\frac{(+)(+)}{(+)(+)} = (+)$ Thus the solution is $(-3, -5/2] U [5/2, 3) $
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