Question

求函数y=sinx/(2+sinx)的值域

Answer 1

y=(2+sinx-2)/(2+sinx)
=1-2/(2+sinx)

-1<=sinx<=1
1<=sinx<=3
1/3<=1/(2+sinx)<=1
-2<=-2/(2+sinx)<=-2/3
-1<=1-2/(2+sinx)<=1/3
值域[-1,1/3]

追问

看清楚题目啊是(sinx)/(2+sinx)

追答

不信拉倒

Answer 2

y = sinx/(2+sinx)
= 1- 2/(2+sinx)
y' = [2/(2+sinx)]. cosx
y' =0
[2/(2+sinx)]. cosx = 0
x = π/2(max) , 3π/2 ( min)
max y = 1/3
min y = -1/1 =-1
值域=[-1,1/3]