Answer
$$[0,1]$$
Work Step by Step
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:
$$[0,1]$$
