f(x)是定义于R上的函数,满足两个条件f(x+y)=f(x)f(1-y)+f(1-x)f(y)...f(x)是定义于R上的函数,满足两个条件:(1)f(x+y)=f(x)f(1-y)+f(1-x)f(y)(2)f(x)在[0,1]上单调递增;问:(1)f(1)=1;(2)f(x)的奇偶性(3
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/12 02:39:16
![f(x)是定义于R上的函数,满足两个条件f(x+y)=f(x)f(1-y)+f(1-x)f(y)...f(x)是定义于R上的函数,满足两个条件:(1)f(x+y)=f(x)f(1-y)+f(1-x)f(y)(2)f(x)在[0,1]上单调递增;问:(1)f(1)=1;(2)f(x)的奇偶性(3](/uploads/image/z/4107271-31-1.jpg?t=f%28x%29%E6%98%AF%E5%AE%9A%E4%B9%89%E4%BA%8ER%E4%B8%8A%E7%9A%84%E5%87%BD%E6%95%B0%2C%E6%BB%A1%E8%B6%B3%E4%B8%A4%E4%B8%AA%E6%9D%A1%E4%BB%B6f%28x%2By%29%3Df%28x%29f%281-y%29%2Bf%281-x%29f%28y%29...f%28x%29%E6%98%AF%E5%AE%9A%E4%B9%89%E4%BA%8ER%E4%B8%8A%E7%9A%84%E5%87%BD%E6%95%B0%2C%E6%BB%A1%E8%B6%B3%E4%B8%A4%E4%B8%AA%E6%9D%A1%E4%BB%B6%EF%BC%9A%EF%BC%881%EF%BC%89f%28x%2By%29%3Df%28x%29f%281-y%29%2Bf%281-x%29f%28y%29%EF%BC%882%EF%BC%89f%28x%29%E5%9C%A8%5B0%2C1%5D%E4%B8%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%EF%BC%9B%E9%97%AE%EF%BC%9A%EF%BC%881%EF%BC%89f%281%29%3D1%3B%EF%BC%882%EF%BC%89f%28x%29%E7%9A%84%E5%A5%87%E5%81%B6%E6%80%A7%EF%BC%883)
f(x)是定义于R上的函数,满足两个条件f(x+y)=f(x)f(1-y)+f(1-x)f(y)...f(x)是定义于R上的函数,满足两个条件:(1)f(x+y)=f(x)f(1-y)+f(1-x)f(y)(2)f(x)在[0,1]上单调递增;问:(1)f(1)=1;(2)f(x)的奇偶性(3
f(x)是定义于R上的函数,满足两个条件f(x+y)=f(x)f(1-y)+f(1-x)f(y)...
f(x)是定义于R上的函数,满足两个条件:
(1)f(x+y)=f(x)f(1-y)+f(1-x)f(y)
(2)f(x)在[0,1]上单调递增;
问:
(1)f(1)=1;
(2)f(x)的奇偶性
(3)f(2x-1)≥1/2的解集
f(x)是定义于R上的函数,满足两个条件f(x+y)=f(x)f(1-y)+f(1-x)f(y)...f(x)是定义于R上的函数,满足两个条件:(1)f(x+y)=f(x)f(1-y)+f(1-x)f(y)(2)f(x)在[0,1]上单调递增;问:(1)f(1)=1;(2)f(x)的奇偶性(3
f(x+y)=f(x)f(1-y)+f(1-x)f(y) ,
设x=y=1/3,
f(1/3+1/3)=f(1/3)f(1-1/3)+f(1-1/3)f(1/3) ,
f(2/3)=f(1/3)f(2/3)+f(2/3)f(1/3) ,
f(2/3)=2f(1/3)f(2/3) ,
1/2=f(1/3)
[0,1]上单调递增,
f(1/3)=1/2,
f(2x-1)≥1/2=f(1/3),
设y=1, f(1+x)=f(1-x), [关于x=1对称,f(1/3)=f(5/3),]
即f(x)=f(2-x),是奇函数 ,
-f(x)=f(-x)=f(2+x),
f(4+x)=f[2+(2+x)]=-f(2+x)=f(x),
周期为4,
(0,2),f(x)>0;(2,4),f(x)<0,
(0,2),f(x)>1/2,1/3
1/3 太巧了
mgtjqm 太强了