Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 1 - Section 1.8 - Inequalities - 1.8 Exercises - Page 89: 55


The solution is $(-\infty,-5]\cup[2,\infty)$ The graph is:

Work Step by Step

$(x+3)^{2}(x-2)(x+5)\ge0$ Begin immediately by finding the intervals, because all nonzero terms are on one side of the inequality and the nonzero side is given in factored form. The factors are $(x+3)^{2}$ $,$ $x-2$ and $x+5$. Set them equal to $0$ and solve for $x$: $(x+3)^{2}=0$ $x+3=0$ $x=-3$ $x-2=0$ $x=2$ $x+5=0$ $x=-5$ The factors are zero when $x=-3,2,-5$. These three numbers divide the real line into the following intervals: $(-\infty,-5)$ $,$ $(-5,-3)$ $,$ $(-3,2)$ $,$ $(2,\infty)$ Elaborate a diagram, using test point to determine the sign of each factor in each interval: (refer to the attached image below) It can be seen from the diagram that the inequality is satisfied in the intervals $(-\infty,-5)$ and $(2,\infty)$. Also, the inequality involves $\ge$ so the endpoints also satisfy the inequality. The solution is $(-\infty,-5]\cup[2,\infty)$
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