## Precalculus: Mathematics for Calculus, 7th Edition

SOLUTIONS $\frac{2x+3}{(x+1)(x+2)} \leq0$ INTERVAL $(-\infty, -2) [\frac{-3}{2},-1)$ GRAPH In the real number line put $-\infty$, -2, $\frac{-3}{2}$ and -1. Put a close circle in $\frac{-3}{2}$. In the -1 and -2 put an open circle. Connect $\frac{-3}{2}$ and -1. In -2 draw a line pointing towards $-\infty$
$\frac{1}{x+1}+\frac{1}{x+2} \leq0$ $\frac{1(x+1)+1(x+2)}{(x+1)(x+2)} \leq0$ $\frac{x+1+x+2}{(x+1)(x+2)} \leq0$ $\frac{2x+3}{(x+1)(x+2)} \leq0$ You have to find the solution from the numerator and denominator. NUMERATOR x=$\frac{-3}{2}$ DENOMINATOR x=-1 x=-2 INTERVAL $(-\infty, -2) (-2,\frac{-3}{2}] [\frac{-3}{2},-1) (-1, \infty)$ $(-\infty, -2)$ $y(-4)=\frac{2(-4)+3}{(-2+1)(-2+2)} \leq0)= y(-4)=-0.83$ Solution $(-2,\frac{-3}{2}]$ $y(-1.8)=\frac{(2(-1.8)+3}{(-1.8+1)(-1.8+2)} \leq0= y(-1.8)=3.75$ No solution $[\frac{-3}{2},-1)$ $y(-1.3)=\frac{2(-1.3)+3}{(-1.3+1)(-1.3+2)} \leq0= y(-1.3)=-1.90$ Solution $(-1, \infty)$ $y(3)=\frac{2(3)+3}{(3+1)(3+2)} \leq0= y(3)=0.45$ No solution CORRECT $(-\infty, -2) [\frac{-3}{2},-1)$ GRAPH In the real number line put $-\infty$, -2, $\frac{-3}{2}$ and -1. Put a close circle in $\frac{-3}{2}$. In the -1 and -2 put an open circle. Connect $\frac{-3}{2}$ and -1. In -2 draw a line pointing towards $-\infty$