设函数f(x)=sin(πx/3-π/6)-2(cosπx/6)^2.(1)求y=f(x)的最小正周期及单调递增区间;(2)若函数y=g(x)与y=f(x)的图像关于直线x=2对称;求当x∈[0,1]时,函数y=g(x)的最大值.
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![设函数f(x)=sin(πx/3-π/6)-2(cosπx/6)^2.(1)求y=f(x)的最小正周期及单调递增区间;(2)若函数y=g(x)与y=f(x)的图像关于直线x=2对称;求当x∈[0,1]时,函数y=g(x)的最大值.](/uploads/image/z/13446744-24-4.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dsin%28%CF%80x%2F3-%CF%80%2F6%29-2%28cos%CF%80x%2F6%29%5E2.%281%29%E6%B1%82y%3Df%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E5%8F%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4%EF%BC%9B%282%29%E8%8B%A5%E5%87%BD%E6%95%B0y%3Dg%28x%29%E4%B8%8Ey%3Df%28x%29%E7%9A%84%E5%9B%BE%E5%83%8F%E5%85%B3%E4%BA%8E%E7%9B%B4%E7%BA%BFx%3D2%E5%AF%B9%E7%A7%B0%EF%BC%9B%E6%B1%82%E5%BD%93x%E2%88%88%5B0%2C1%5D%E6%97%B6%2C%E5%87%BD%E6%95%B0y%3Dg%28x%29%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC.)
设函数f(x)=sin(πx/3-π/6)-2(cosπx/6)^2.(1)求y=f(x)的最小正周期及单调递增区间;(2)若函数y=g(x)与y=f(x)的图像关于直线x=2对称;求当x∈[0,1]时,函数y=g(x)的最大值.
设函数f(x)=sin(πx/3-π/6)-2(cosπx/6)^2.
(1)求y=f(x)的最小正周期及单调递增区间;
(2)若函数y=g(x)与y=f(x)的图像关于直线x=2对称;求当x∈[0,1]时,函数y=g(x)的最大值.
设函数f(x)=sin(πx/3-π/6)-2(cosπx/6)^2.(1)求y=f(x)的最小正周期及单调递增区间;(2)若函数y=g(x)与y=f(x)的图像关于直线x=2对称;求当x∈[0,1]时,函数y=g(x)的最大值.
f(x)=sin(πx/3-π/6)-[cos(πx/3)+1]
=sin(πx/3)cos(π/6)-cos(πx/3)sin(π/6)-cos(πx/3)-1
=(√3/2)sin(πx/3)-(3/2)cos(πx/3)-1
=√3sin(πx/3-π/3)-1
1、函数f(x)的最小正周期是2π/(π/3)=6,增区间:2kπ-π/2≤πx/3-π/3≤2kπ+π/2
6k-1/2≤x≤6k+5/2
即增区间是[6k-1/2,6k+5/2],其中k是整数.
2、y=g(x)与y=f(x)关于x=2对称,则g(x)=f(4-x)=√3sin(πx/3)-1,其中x∈[0,1],则:
(πx/3)∈[0,π/3],则g(x)的最大值是g(1)=√3sin(π/3)-1=1/2
(1) y=f(x)=sin(πx/3-π/6)-1-cos(πx/3)
=sin(πx/3-π/6)-sin(π/2-πx/3)-1
=2cos(π/6)sin(πx/3-π/3)-1
=√3sin(πx/3-π/3)-1
最小正周期T=2π/(π/3)=6
单增区间πx/3-π/3∈[2kπ-π/2, 2kπ+π/2]
x∈[6k-1, 6k+2...
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(1) y=f(x)=sin(πx/3-π/6)-1-cos(πx/3)
=sin(πx/3-π/6)-sin(π/2-πx/3)-1
=2cos(π/6)sin(πx/3-π/3)-1
=√3sin(πx/3-π/3)-1
最小正周期T=2π/(π/3)=6
单增区间πx/3-π/3∈[2kπ-π/2, 2kπ+π/2]
x∈[6k-1, 6k+2] k∈Z
(2) 设g(x)有一点(x, y)
对应f(x) 的点为(4-x, y)
代入f(x):y=√3sin[π(4-x)/3-π/3]-1
所以g(x)=√3sin(π-πx/3)-1=√3sin(πx/3)-1
当x∈[0,1]时,g(x)最大=g(1)=√3sin(π/3)-1=3/2-1=1/2
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由f(x)=sinπx/6 ∵f(1)=-f(7)=1/2 f(2)=-f(8)=√3/2 f(3)=-f(9)=1 f(4)=-f(10)=√3/2 f(5)=-f(11