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: 22

Answer

$$[0,2]$$

Work Step by Step

Let $-x^2 +2x \geq 0$ be a function $f$. Find the $x$-intercepts by solving $-x^2 +2x=0$ Factor: $-x^2 +x = -x(x-2)$ $-x=0$ or $x-2=0$ $x = 0$ or $x =2$ 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: $(-∞, 0)(0,2)(2, +∞)$ Choose one test value within each interval and evaluate $f$ at that number. $(-∞, 0)$ Test Value = $-1$ $f=-x(x-2)$ $f=-(-1)(-1-2)$ $f=-3$ Conclusion: $f (x) < 0$ for all $x$ in $(-∞, 0)$. $(0, 2)$ Test Value = $1$ $f=-x(x-2)$ $f=-1(1-2)$ $f=1$ Conclusion: $f (x) > 0$ for all $x$ in $(0, 2)$. $(2,+∞)$ Test Value = $3$ $f=-x(x-2)$ $f=-3(3-2)$ $f=-3$ Conclusion: $f (x) < 0$ for all $x$ in $(2,+∞)$. Write the solution set, selecting the interval or intervals that satisfy the given inequality. We are interested in solving $f (x) \geq 0$, where $f (x) = -x^2+2x$. Based on the solution above, $f(x)\gt0$ for all $x$ in $(0,2)$. However, because the inequality involves $\geq$ (greater than or equal to), we must also include the solutions of $-x^2+2x = 0$, namely $0$ and $2$, in the solution set. Thus, the solution set of the given inequality is: $$[0,2]$$
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