Precalculus: Mathematics for Calculus, 7th Edition

SOLUTION $\frac{-x(3x+2)}{x+1}\gt0$ CORRECT INTERVALS $(-\infty,-1) (\frac{-2}{3},0)$ GRAPH In the real number line there should be three numbers -1, $\frac{-2}{3}$ and 0. You shpuld make a n open circle in the three of them. In -1 you are foing to do a line pointing to $-\infty$. Then yo are going to connect $\frac{-2}{3}$ and 0.
SOLUTION $\frac{x}{x+1}\gt3x$ $\frac{x}{x+1}-3x\gt0$ $\frac{x}{x+1}-\frac{3x(x+1)}{x+1}\gt0$ $\frac{-3x^{2}-2x}{x+1}\gt0$ $\frac{-x(3x+2)}{x+1}\gt0$ INTERVALS To create the intervals you should look at the solutions offer by the inequality, which means the solution from the numerator and the denominator. This solution should not be included in the interval so we use parentheses. NUMERATOR: x=0 and x=$\frac{-2}{3}$ DENOMINATOR: x=-1 We use these key number to create the intervals. $(-\infty, -1) (-1,\frac{-2}{3}) (\frac{-2}{3}, 0) (0, \infty)$ you have to test the valies in between to know which one is a solution. $(-\infty,-1) y(-2)=\frac{2(3(-2)+2)}{(-2+1)}\gt0 y(-2)=8$ it is a solution because it is greater than 0 $(-1,\frac{-2}{3}) y(-0.8)=\frac{0.8(3(-0.8)+2)}{-0.8+1}\gt0 =y(-0.8)=-1.6$ It isn’t a solution $\frac{-2}{3},0). Y(-0.3)=\frac{0.3(3(-0.3)+2}{-0.3+1}\gt0= Y(-0.3)=0.47$ It is a solution $(0,\infty) y(5)=\frac{-5(3(5)+2)}{5+1}\gt0 =y(5)=-14.17$ it isn’t a solution CORRECT INTERVALS $(-\infty,-1) (\frac{-2}{3},0)$ GRAPH In the real number line there should be three numbers -1, $\frac{-2}{3}$ and 0. You shpuld make a n open circle in the three of them. In -1 you are foing to do a line pointing to $-\infty$. Then yo are going to connect $\frac{-2}{3}$ and 0.