## College Algebra (6th Edition)

Solution set: $[0,2]$ Follow the "Procedure for Solving Polynomial lnequalities",\ p.412: 1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$ where $f$ is a polynomial function. $-x^{2}+2x \geq 0\qquad /\times(-1)$ ... the sign turns ... $x^{2}-2x \leq 0 \quad$...factor the trinomial... $x(x-2) \leq 0$ $f(x)= x(x-2) \leq 0$ 2. Solve the equation $f(x)=0$. The real solutions are the boundary points. $x(x-2)= 0$ $x=0$ or $x=2$ 3. Locate these boundary points on a number line, thereby dividing the number line into intervals. 4. Test each interval's sign of $f(x)$ with a test value, $\begin{array}{llll} Intervals: & (-\infty, 0) & (0,2) & (2,\infty)\\ a=test.val. & -1 & 1 & 3\\ f(a) & (-1)(-3) & (1)(-1) & (3)(1)\\ f(a) \leq 0 ? & F & T & F \end{array}$ 5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points. Solution set: $[0,2]$