# Chapter 8 - Section 8.5 - Polynomial and Rational Inequalities - Exercise Set - Page 649: 21

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