Question

求下列函数的值域(1)y=sin(2x-π/3),x∈[-π/4,π/4]

(2)y=sin²x-cos+3，x∈[π/6,2/3π]

Answer 1

(1)
x∈[-π/4,π/4]，∴2x-π/3∈[-5π/6，π/6]
∴y=sin(2x-π/3)∈[-1，1/2]
即值域：[-1，1/2]
(2)
y=sin²x-cos+3
=1-cos²x-cosx+3
=-cos²x-cosx+4
∴cosx∈[-1/2,(根号3)/2]
对称轴cosx=-1/2，开口向下
∴最大值在-1/2处取得，最小值在(根号3)/2处取得
∴y最大=-1/4+1/2+4=17/4
y最小=-3/4-(根号3)/2+4=13/4-(根号3)/2
即值域：[13/4-(根号3)/2，17/4]

Answer 2

y=sin(2x-Pai/3)

-Pai/4<=x<=Pai/4
-5Pai/6<=2x-Pai/3<=Pai/6
故有-1<=y<=1/2
即值域是[-1,1/2]
y=sin^2x-cosx+3=1-cos^x+cosx+3=-(cosx-1/2)^2+17/4
x属于[Pai/6,2/3Pai],那么有-1/2<=cosx<=根号3/2
故当cosx=1/2时有最大值是:17/4
当cosx=-1/2时有最小值是:13/4
即值域是[13/4,17/4]