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: 20



Work Step by Step

Let $3x^2 - 5x \leq 0$ be a function $f$. Find the $x$-intercepts by solving $3x^2 - 5x=0$ Factor: $3x^2 - 5x = x(3x-5)$ $x=0$ or $3x-5=0$ $x = 0$ or $x = \frac{5}{3}$ 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: $(-∞, \frac{5}{3})(\frac{5}{3},0)(0, +∞)$ Choose one test value within each interval and evaluate $f$ at that number. $(-∞, 0)$ Test Value = $-1$ $f=x(3x-5)$ $f=(-1)(3(-1)-5)$ $f=8$ Conclusion: $f (x) > 0$ for all $x$ in $(-∞, 0)$. $(0,\frac{5}{3})$ Test Value = $1$ $f=x(3x-5)$ $f=(1)(3(1)-5)$ $f=-2$ Conclusion: $f (x) < 0$ for all $x$ in $(\frac{5}{3},0)$. $(\frac{5}{3},+∞)$ Test Value = $2$ $f=x(3x-5)$ $f=(2)(3(2)-5)$ $f=2$ Conclusion: $f (x) > 0$ for all $x$ in $(0,+∞)$. Write the solution set, selecting the interval or intervals that satisfy the given inequality. We are interested in solving $f (x) \leq 0$, where $f (x) = 3x^2-5x$. Based on the solution above, $f(x)\lt0$ for all $x$ in $(\frac{5}{3},0)$. However, because the inequality involves $\leq$ (less than or equal to), we must also include the solutions of $3x^2-5x = 0$, namely $0$ and $\frac{5}{3}$, in the solution set. Thus, the solution set of the given inequality is: $$[0,\frac{5}{3}]$$
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