## Precalculus: Mathematics for Calculus, 7th Edition

$[-2, -1) ∪ (0, 1]$
$$1+\frac{2}{x+1}\leq \frac{2}{x}$$ Subtract $\frac{2}{x}$ from both sides $$1+\frac{2}{x+1}-\frac{2}{x}\leq0$$ And simplify $$\frac{x(x+1)+2x-2(x+1)}{x(x+1)}\leq0$$ $$\frac{x^2+x+2x-2x-2}{x(x+1)}\leq0$$ $$\frac{x^2+x-2}{x(x+1)}\leq0$$ Factor out in numerator $$\frac{(x-1)(x+2)}{x(x+1)}\leq0$$ We have key points at: $x-1=0$ ; $x=1$ $x+2=0$ ; $x=-2$ $x=0$ $x+1=0$ ; $x=-1$ Which gives us intervals: $(-\infty, -2]$ - Positive $[-2, -1)$ - Negative $(-1, 0)$ - Positive $(0, 1]$ - Negative $[1, \infty)$ - Positive We need closed intervals of negative numbers, so the answer is: $[-2, -1) ∪ (0, 1]$