函数f(x)=log1
f(x)=log[√2sin(x-π/4)]
1。
定义域为√2sin(x-π/4)＞0
===> sin(x-π/4)＞0
===> x-π/4∈(2kπ,2kπ+π)
===> x∈(2kπ+π/4,2kπ+5π/4)（k∈Z)
因为sin(x-π/4)∈(0,1]
所以，√2sin(x-π/4)∈(0,√2]
所以，值域y=log[√2sin(x-π/4)]∈[-1/2,+∞)
2。
f(x)=log[√2sin(x-π/4)]
=log{√2*[sinx*cos(π/4)-cosx*sin(π/4)]}
=log[√2*(√2/2)*(sinx-cosx)]
=log(sinx...全部
f(x)=log[√2sin(x-π/4)]
1。
定义域为√2sin(x-π/4)＞0
===> sin(x-π/4)＞0
===> x-π/4∈(2kπ,2kπ+π)
===> x∈(2kπ+π/4,2kπ+5π/4)（k∈Z)
因为sin(x-π/4)∈(0,1]
所以，√2sin(x-π/4)∈(0,√2]
所以，值域y=log[√2sin(x-π/4)]∈[-1/2,+∞)
2。
f(x)=log[√2sin(x-π/4)]
=log{√2*[sinx*cos(π/4)-cosx*sin(π/4)]}
=log[√2*(√2/2)*(sinx-cosx)]
=log(sinx-cosx)
所以，f(-x)=log(-sinx-cosx)
所以，它是非奇非偶函数
3。
由(1)知，其定义域为x∈(2kπ+π/4,2kπ+5π/4)（k∈Z）
①当x-π/4∈(2kπ,2kπ+π/2)，即x∈(2kπ+π/4,2kπ+3π/4)（k∈Z）时：
√2*sin(x-π/4)＞0，且单调递增
那么，y=log[√2*sin(x-π/4)]就单调递减；
②当x-π/4∈[2kπ+π/2,2kπ+π)，即x∈(2kπ+3π/4,2kπ+5π/4)（k∈Z）时：
√2*sin(x-π/4)＞0，且单调递减
那么，y=log[√2*sin(x-π/4)]就单调递增；。
收起
