## Precalculus: Mathematics for Calculus, 7th Edition

The zeroes are (x-2)— > x=2 (x+1)— > x=-1 In this case the inquality has to be less than or equal to zero, therefore we know that x=2 can’t work because a 0 in the denominator will make it undefined. In this case the real line will be divided into: $(-\infty, -1] [-1, 2) (2, \infty)$ After that you have to choose one value between the value and test it — — — — — — — — - $(-\infty, -1]— — — — [-1, 2)— —— (2, \infty)$ Sign of x+1— — — (-2+1)= -1 (-)— — — — (1+1)=2 (+)— —(3+1)=4 (+) Sign of x-2 — — —(-2-2)=-4 (-)— — — — (1-2)=-1 (-)— — (3-2)=1 (+) (x+1)(x+2)— — — — —(+)— — — — — — — — -(—)— — — — — — — (+) The intervals that satisfy the inequality are [-1,2)