cosx−2 |
cosx−1 |
1 |
cosx−1 |
∵-1≤cosx≤1,∴-2≤cosx-1≤0,
∴cosx-1是分母,∴-2≤cosx<0,
∴当cosx-1=-2时,函数y=
cosx−2 |
cosx−1 |
1 |
cosx−1 |
1 |
−2 |
3 |
2 |
当cosx-1→0时,函数y=
cosx−2 |
cosx−1 |
1 |
cosx−1 |
∴函数y=
cosx−2 |
cosx−1 |
3 |
2 |
故答案为:[
3 |
2 |
cosx−2 |
cosx−1 |
cosx−2 |
cosx−1 |
1 |
cosx−1 |
cosx−2 |
cosx−1 |
1 |
cosx−1 |
1 |
−2 |
3 |
2 |
cosx−2 |
cosx−1 |
1 |
cosx−1 |
cosx−2 |
cosx−1 |
3 |
2 |
3 |
2 |