Question

已知函数f(x)=-√3sin²x+sinxcosx

求函数在fx∈[ 0,π／2 ]的值域

Answer 1

f(x)=-√3sin²x+sinxcosx
=-√3(1-cos2x)/2+sin2x/2
=-√3/2+√3/2cos2x+sin2x/2
=-√3/2+sin(2x+π/3)
∵x∈[0,π/2]
∴2x+π/3∈[π/3,4π/3]
∴sin(2x+π/3)∈[-√3/2,1]
∴值域为[-√3,(2-√3)/2]

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Answer 2

解
f(x)=1/2(2sinxcosx)+√3/2(1-2sin²x)-√3/2
=1/2sin2x+√3/2cos2x-√3/2
=sin2xcosπ/3+cos2xsinπ/3-√3/2
=sin(2x+π/3)-√3/2
∵x∈[0,π/2]
∴2x+π/3∈[π/3,4π/3]
∴sin(2x+π/3)∈[-√3/2,1]
∴f(x)∈[-√3,1-√3/2]
值域为[-√3.1-√3/2]

Answer 3

f(x)=-√3sin²x+sinxcosx
=-√3(1-cos2x)/2+sin2x/2
=-√3/2+√3/2cos2x+sin2x/2
=sin(2x+π/3)-√3/2
∵x∈[0,π/2]
∴2x+π/3∈[π/3,4π/3]
画出图像。最高点可以取到。最低点在4π/3处取得
所以值域是[-√3.1-√3/2]