Question

求函数y=-sin²x+sinx+1(-π/4≤x≤π/4)的值域。

Answer 1

-π/4≤x≤π/4
-√2/2≤sinx≤√2/2

y=-sin²x+sinx+1
=-sin²x+sinx-1/4+1/4+1
=-(sin²x-sinx+1/4)+5/4
=-(sinx-1/2)²+5/4

0<=(sinx-1/2)²<=(-√2/2-1/2)²=3/4+√2/2
-3/4-√2/2<=-(sinx-1/2)²<=0
-3/4-√2/2+5/4<=-(sinx-1/2)²+5/4<=5/4
1/2-√2/2<=-(sinx-1/2)²+5/4<=5/4

所以函数y=-sin²x+sinx+1(-π/4≤x≤π/4)的值域为：[1/2-√2/2,5/4]

Answer 2

解：y=-sin²x+sinx+1=-(sinx-1/2)²+5/4
∵-π/4≤x≤π/4
∴-√2/2≤sinx≤√2/2
则sinx=1/2∈[-√2/2，√2/2]
则存在最大值为sinx=1/2时，y=5/4
最小值为sinx=-√2/2时，y=(1-√2)/2
即y的值域为[(1-√2)/2，5/4]