已知函数fx=-根号2sin（2x+π/4）+6sinxcosx-2cos²x+1 x属于R求fx最小正周期及单调递增区间求fx在0-π/2上最大值最小值
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已知函数fx=-根号2sin（2x+π/4）+6sinxcosx-2cos²x+1 x属于R求fx最小正周期及单调递增区间求fx在0-π/2上最大值最小值
已知函数fx=-根号2sin（2x+π/4）+6sinxcosx-2cos²x+1 x属于R
求fx最小正周期及单调递增区间
求fx在0-π/2上最大值最小值
已知函数fx=-根号2sin（2x+π/4）+6sinxcosx-2cos²x+1 x属于R求fx最小正周期及单调递增区间求fx在0-π/2上最大值最小值
f（x）=-sin2x-cos2x+3sin2x-cos2x
=2sin2x-2cos2x
=2根号2sin(2x-π/4)
T=2π/2=π
-π/2+2kπ≤2x-π/4≤π/2+2kπ k属于Z
-π/8+kπ≤x≤3π/8+kπ
递增区间[-π/8+kπ,3π/8+kπ]
在0-π/2上最大值为2根号2,x=3π/8最小值为-2 x=0
fx=2根号2sin(2x-π/4)
周期是π，单调增区间（-π/8+kπ，3π/8+kπ）
最大值为2根号2，最小值为-2
f(x)=-√2sin(2x+π/4)+6sinxcosx-2cos²x+1
=-√2(sin2xcosπ/4+cos2xsinπ/4)+3sin2x-2×(1+cos2x)/2+1
=-√2(√2/2·sin2x+√2/2·cos2x)+3sin2x-cos2x
=-sin2x-cos2x+...
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f(x)=-√2sin(2x+π/4)+6sinxcosx-2cos²x+1
=-√2(sin2xcosπ/4+cos2xsinπ/4)+3sin2x-2×(1+cos2x)/2+1
=-√2(√2/2·sin2x+√2/2·cos2x)+3sin2x-cos2x
=-sin2x-cos2x+3sin2x-cos2x
=2sin2x-2cos2x
设tanψ=-2/2=-1,ψ=-π/4
则f(x)=√[2²+(-2)²]sin(2x-π/4)=2√2sin(2x-π/4)
1)最小正周期=2π/丨2丨=π
2)易知f(x)在区间[kπ-π/8,kπ+3π/8)上单调递增，在区间[kπ+3π/8,kπ+7π/8)上单调递减，k∈Z
∴在区间[0,3π/8)上单调递增，在区间[3π/8,π/2]上单调递减
∴f(x)在区间[0,π/2]上最大值为f(3π/8)=2√2,最小值为f(0)=-2
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