Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 3 - Review - Exercises - Page 322: 94

Answer

$(-∞, -2] U [2, 3] $

Work Step by Step

$x^3 - 3x^2 - 4x + 12 \leq 0$ Find the zeros of the expression. $x^2 (x - 3) - 4 (x - 3)$ $(x^2 - 4) (x - 3)$ $(x+2)(x-2) (x-3)$ $x = -2, 2, 3$ Test numbers in between those zero values to determine if the function is negative or positive (-∞, -2] $(-)(-)(-) = (-)$ [-2, 2] $(+)(-)(-) = (+)$ [2, 3] $(+)(+)(-) = (-)$ [3, ∞) $(+)(+)(+)= (+)$ Thus the solution is $(-∞, -2] U [2, 3] $
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