Question

y=（x+2）/（x-3）.求y的值域

Answer 1

（一）

y=(x-3)/(x+2)=（x+2-5)/(x+2)=1-5/(x+2) x≠-2

x∈（-∞,-2)时,-∞＜1/(x+2)＜0,0＜-1/(x+2)＜+∞,

1＜1-5/(x+2)＜+∞

x∈（-2,+∞)时,0＜1/(x+2)＜+∞,-∞＜-1/(x+2)＜0,

-∞＜1-5/(x+2)＜1

值域：（-∞,1）,（1,+∞）