## Precalculus: Mathematics for Calculus, 7th Edition

SOLUTION $\frac{2x}{3-x}\geq0$ INTERVAL [0,3) GRAPH In a real number line write 0 and 3. The 0 should be with a close circle and the 3 with an open circle
SOLUTION $\frac{3+x}{3-x}\geq1$ $\frac{3+x}{3-x}-1\geq0$ $\frac{3+x}{3-x}-\frac{1(3-x)}{3-x}\geq0$ $\frac{3+x}{3-x}-\frac{(3-x)}{3-x}\geq0$ $\frac{3+x-3+x}{3-x}\geq0$ $\frac{2x}{3-x}\geq0$ Solutions from the inequality. For the numerator to be 0 x=0. For the denominator to be 0 it has to be x=3. In this case this number is not allowed because it will make the fraction undefined. But we can use it to construct the intervals. INTERVALS $(-\infty,0] [0,3) (3,\infty)$ In order to know whihc one is the soltion you have to test the values. $(-\infty,0]$ = $y(-2)=\frac{2(-2)}{3-(-2)}\geq0$ = $y(-2)=0.8$ This interval is not the solution because the result is less than 0 and it should be greater than or equal to 0. [0,3) = $y(1)=\frac{2(1)}{3-1}\geq0$ = $y(1)= 1$ This interval is a solution because the result is greater or equal to 0. $(3,\infty)$ = $y(4)\frac{2(4)}{3-4}$ = $y(4)=-8$ This interval is not a solution because the result is less than 0. The correct interval is [0,3) GRAPH In a real number line write 0 and 3. The 0 should be with a close circle and the 3 with an open circle. Then connect them. Indicating that 3 should not be used.