Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 8 - Section 8.5 - Polynomial and Rational Inequalities - Exercise Set - Page 649: 31

Answer

$$[1,2]∪[3+∞)$$

Work Step by Step

Let $(x - 1)(x - 2)(x - 3) \geq 0$ be a function $f$. Find the $x$-intercepts by solving $(x - 1)(x - 2)(x - 3)=0$ $x-1=0$ or $x-2=0$ or $(x-3)$ $x = 1$ or $x =2$ or $x=3$ These $x$-intercepts serve as boundary points that separate the number line into intervals. The boundary points of this equation therefore divide the number line into three intervals: $(-∞, 1)(1,2),(2,3),(3, +∞)$ Choose one test value within each interval and evaluate $f$ at that number. $(-∞, 1)$ Test Value = $0$ $f=(x - 1)(x - 2)(x - 3)$ $f=(0 - 1)(0 - 2)(0- 3)$ $f=-6$ Conclusion: $f (x) < 0$ for all $x$ in $(-∞, 1)$. $(1, 2)$ Test Value = $\frac{3}{2}$ $f=(x - 1)(x - 2)(x - 3)$ $f=(\frac{3}{2} - 1)(\frac{3}{2} - 2)(\frac{3}{2} - 3)$ $f=\frac{3}{8}$ Conclusion: $f (x) > 0$ for all $x$ in $(1, 2)$. $(2, 3)$ Test Value = $\frac{5}{2}$ $f=(x - 1)(x - 2)(x - 3)$ $f=(\frac{5}{2} - 1)(\frac{5}{2} - 2)(\frac{5}{2} - 3)$ $f=-\frac{3}{8}$ Conclusion: $f (x) < 0$ for all $x$ in $(2,3)$. $(3,+∞)$ Test Value = $4$ $f=(x - 1)(x - 2)(x - 3)$ $f=(4 - 1)(4 - 2)(4 - 3)$ $f=6$ Conclusion: $f (x) > 0$ for all $x$ in $(3,+∞)$. Write the solution set, selecting the interval or intervals that satisfy the given inequality. We are interested in solving $f (x) \geq 0$, where $f (x) = (x - 1)(x - 2)(x - 3)$. Based on the solution above, $f(x)\gt0$ for all $x$ in $(1,2)$ or $(3+∞)$. However, because the inequality involves $\geq$ (greater than or equal to), we must also include the solutions of $(x - 1)(x - 2)(x - 3) = 0$, namely $1$, $2$ and $3$, in the solution set. Thus, the solution set of the given inequality is: $$[1,2]∪[3+∞)$$
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