已知函数f(x)=2cos²x+2根号3sinxcosx.求 (1)函数f(x)的周期。 (2)函数f(x)的单调递减区间。
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2011-09-27 · 知道合伙人教育行家
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f(x)=2cos²x+2根号3sinxcosx
= (cos2x+1) + √3sin2x
= √3sin2x + cos2x + 1
= 2(sin2xcosπ/6+cos2xsinπ/6) +1
= 2sin(2x+π/6) + 1
最小正周期 = 2π/2 = π
2x+π/6∈(2kπ+π/2,2kπ+3π/2),其中k∈Z时单调减
所以单调减区间:x∈(kπ+π/6,kπ+2π/3),其中k∈Z
x∈[0,π/2]时:
2x+π/6∈【π/3,4π/3】
其中,2x+π/6∈【π/3,π/2)时单调增,2x+π/6∈【π/2,4π/3)时单调减
∴单调增区间:x∈∈【π/12,π/6),单调减区间x∈【π/6,7π/12)
2x+π/6=π/2时,最大值f(x)max = 2+1 = 3
2x+π/6=4π/3时,最小值f(x)min = 2*(-√3/2)+1 = 1-√3
= (cos2x+1) + √3sin2x
= √3sin2x + cos2x + 1
= 2(sin2xcosπ/6+cos2xsinπ/6) +1
= 2sin(2x+π/6) + 1
最小正周期 = 2π/2 = π
2x+π/6∈(2kπ+π/2,2kπ+3π/2),其中k∈Z时单调减
所以单调减区间:x∈(kπ+π/6,kπ+2π/3),其中k∈Z
x∈[0,π/2]时:
2x+π/6∈【π/3,4π/3】
其中,2x+π/6∈【π/3,π/2)时单调增,2x+π/6∈【π/2,4π/3)时单调减
∴单调增区间:x∈∈【π/12,π/6),单调减区间x∈【π/6,7π/12)
2x+π/6=π/2时,最大值f(x)max = 2+1 = 3
2x+π/6=4π/3时,最小值f(x)min = 2*(-√3/2)+1 = 1-√3
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