Question

求函数y=(sinx-2)/(sinx-1)的值域

Answer 1

求函数y=(sinx-2)/(sinx-1)的值域

解：ysinx-y=sinx-2，故得sinx=(y-2)/(y-1)；
(y-2)/(y-1)≦1...............(1)
(y-2)/(y-1)≧-1..............(2)
由(1)得(y-2)/(y-1)-1=(-1)/(y-1)≦0，即有1/(y-1)≧0，故得y>1...........①
由(2)得(y-2)/(y-1)+1=(2y-3)/(y-1)≧0，故得y<1或y≧3/2...................②
.①∩②={y∣y≧3/2}，这就是该函数y=(sinx-2)/(sinx-1)的值域。

Answer 2

值域：[1,3/2]
∵y=(sinx-2)/(sinx-1)
∴y=(sinx-1-1)/(sinx-1)
=1-1/(sinx-1)
∵sinx∈[-1,1]
∴sinx-1∈[-2,0]
1/(sinx-1)∈[-1/2,0]
- 1/(sinx-1)∈[0,1/2]
1-1/(sinx-1)∈[1,3/2]

应该没出错，若有问题请追问，谢谢~~

Answer 3

y=[(sinx-1)-1)]/(sinx-1)
=1-1/(sinx-1)
因为sinx-1∈[-2,0]
所以1/(sinx-1)∈(-∞，-1/2]
- 1/(sinx-1)∈(0,1/2]
y=1-1/(sinx-1)∈(1,3/2]