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已知函数y=sin(12x?π6)(1)求该函数的周期,对称...

3-1=2,-3-1=-4 (2-4)/2 = =-1 把y = -1 代入y = 3sin(2x + π/6) - 1得: sin(2x + π/6) = 0 2x+ π/6 = 2kπ 2x = 2kπ - π/6 x = kπ - π/12 (k为整数) 所以(kπ - π/12 , -1)是对称中心 由y = 3sin(2x + π/6) - 1知: y ' = 6cos(2x + π/6) 令y ' ...

定义域为R 值域为[-3,1] 单调增区间[π/3+kπ,5π/6+kπ] 单调减区间[-π/6+kπ,π/3+kπ] k属于Z 图像的对称轴X=-π/6+1/2kπ k属于Z 对称中心点【π/12-1/2kπ,0】k属于Z

(1) f(x)=sin(2(x-π/12)) x∈[0,π]时,2(x-π/12)∈[-π/6,11π/6] 单调递减区间是2(x-π/12)∈[π/2,3π/2] 即x-π/12∈[π/4,3π/4] 则x∈[π/3,5π/6] (2)x∈[-π/12,π/2]时, 2(x-π/12)∈[-π/3,5π/6] 而当2(x-π/12)∈[-π/3,π/3]时,sin(2(x-π/12))∈[sin(-π/3),...

x+π/6,周期2π, 最大sin(x+π/6)=1,y=2,x+π/6=2kπ+π/2,x=2kπ+π/3; 最小sin(x+π/6)=-1,y=0,x+π/6=2kπ-π/2,x=2kπ-2π/3

(1) 2x+π6 0 π2 π 3π2 2π x ?π12 π6 5π12 2π3 11π12 y=2sin(2x+π6) 0 2 0 -2 0…(2分)…(5分)(2)f(x)的最大值为2;…(7分)此时自变量x取值的集合为{x|x=kπ+π6,k∈Z}…(10分)(3)函数的对称轴方程为 x=kπ2+π6,k∈z…(14分)

(1)振幅A=2,周期T=π,初相为π3.(2)y=2sin(2x+π3),列表如下: X 0 π2 π 3π2 2π x π3 4π3 7π3 10π3 13π3 y 0 2 0 -2 0描点连图(3)将y=sinx图象上各点向左平移π3个单位,得到y=sin(x+π3)的图象,再把y=sin(x+π3)的图象上各点的横坐...

(1)函数的最大值为2,取得最大值时,2x+π6=π2+2kπ,即x=kπ+π6(k∈Z);(2)由2x+π6=kπ+π2,可得函数图象的对称轴为x=kπ2+π6(k∈Z);由2x+π6=kπ,可得函数的对称中心为(kπ2?π12,0)(k∈Z);(3)由2x+π6∈[?π2+2kπ,π2+2kπ],可得该函数...

∵是f(x)=2sin(ωx+φ-π/6)偶函数,∴f(x)=2sin(ωx+φ-π/6)=2Cos(ωx),得出φ-π/6=±π/2, ∵0

(1)f(x)=sin2x+cos(2x?π6)=sin2x+cos2xcosπ6+sin2xsinπ6=32sin2x+32cos2x=3(32sin2x+12cos2x)=3sin(2x+π6)…(4分)∴最小正周期T=2π2=π,…(5分)减区间:2kπ+π2≤2x+π6≤2kπ+3π2,k∈Z解得2kπ+π3≤2x≤2kπ+4π3,k∈Z所以单调减区间为[kπ+π6,kπ...

(1)f(x)=2sin(ωx-π6)sin(ωx+π3)=2sin(ωx-π6)sin[(ωx-π6)+π2]=2sin(ωx-π6)cos(ωx-π6)=sin(2ωx-π3),∵T=π,∴ω=1,∴f(x)=sin(2x-π3),∵x∈[π8,5π12],∴2x-π3∈[-π12,π2],根据正弦函数在此区间单调递增,得到:f(x)min=sin...

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