定义在R上的函数f(x)满足f(x+y)+f(x-y)=2f(x)f(y)且f(1/2)=0 f(0)≠0(1)求证:f(x)是偶函数(2)求证:f(x)是周期函数(3)若f(x)在[0,1]内是单调函数,求f(1/3)与f(1/6)的值高中数学 会的说下 谢谢!
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![定义在R上的函数f(x)满足f(x+y)+f(x-y)=2f(x)f(y)且f(1/2)=0 f(0)≠0(1)求证:f(x)是偶函数(2)求证:f(x)是周期函数(3)若f(x)在[0,1]内是单调函数,求f(1/3)与f(1/6)的值高中数学 会的说下 谢谢!](/uploads/image/z/5079478-22-8.jpg?t=%E5%AE%9A%E4%B9%89%E5%9C%A8R%E4%B8%8A%E7%9A%84%E5%87%BD%E6%95%B0f%28x%29%E6%BB%A1%E8%B6%B3f%28x%2By%29%2Bf%28x-y%29%3D2f%28x%29f%28y%29%E4%B8%94f%281%2F2%29%3D0+f%280%29%E2%89%A00%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9Af%28x%29%E6%98%AF%E5%81%B6%E5%87%BD%E6%95%B0%EF%BC%882%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9Af%28x%29%E6%98%AF%E5%91%A8%E6%9C%9F%E5%87%BD%E6%95%B0%EF%BC%883%EF%BC%89%E8%8B%A5f%28x%29%E5%9C%A8%5B0%2C1%5D%E5%86%85%E6%98%AF%E5%8D%95%E8%B0%83%E5%87%BD%E6%95%B0%2C%E6%B1%82f%281%2F3%29%E4%B8%8Ef%281%2F6%29%E7%9A%84%E5%80%BC%E9%AB%98%E4%B8%AD%E6%95%B0%E5%AD%A6+%E4%BC%9A%E7%9A%84%E8%AF%B4%E4%B8%8B+%E8%B0%A2%E8%B0%A2%EF%BC%81)
定义在R上的函数f(x)满足f(x+y)+f(x-y)=2f(x)f(y)且f(1/2)=0 f(0)≠0(1)求证:f(x)是偶函数(2)求证:f(x)是周期函数(3)若f(x)在[0,1]内是单调函数,求f(1/3)与f(1/6)的值高中数学 会的说下 谢谢!
定义在R上的函数f(x)满足f(x+y)+f(x-y)=2f(x)f(y)且f(1/2)=0 f(0)≠0
(1)求证:f(x)是偶函数
(2)求证:f(x)是周期函数
(3)若f(x)在[0,1]内是单调函数,求f(1/3)与f(1/6)的值
高中数学 会的说下 谢谢!
定义在R上的函数f(x)满足f(x+y)+f(x-y)=2f(x)f(y)且f(1/2)=0 f(0)≠0(1)求证:f(x)是偶函数(2)求证:f(x)是周期函数(3)若f(x)在[0,1]内是单调函数,求f(1/3)与f(1/6)的值高中数学 会的说下 谢谢!
(1)将x=y=0代入 f(x+y)+f(x-y)=2f(x)f(y)得,f(0)+f(0)=2f(0)f(0),即
2f(0)=2f(0)f(0),由f(0)≠0得,f(0)=1,
再将x=0代入f(0+y)+f(0-y)=2f(0)f(y)得,
f(y)+f(-y)=2f(y),f(-y)=f(y),故f(x)是偶函数.
(2)将y=1/2代入f(x+y)+f(x-y)=2f(x)f(y)得
f(x+1/2)+f(x-1/2)=2f(x)f(1/2)=0.由f(1/2)=0 得 f(x+1/2)+f(x-1/2) =0.
f(x+1/2)=-f(x-1/2) ,同理f(x-1/2)= f(x-1+1/2)=- f(x-1-1/2)= - f(x-3/2),故f(x+1/2)= -f(x-1/2) =f(x-3/2),
设y= x-3/2,则x=y+3/2,x+1/2=y+2,代入上式得f(y+2)=f(y),故f(x)是周期为2的周期函数.
(3)设x=y代入f(x+y)+f(x-y)=2f(x)f(y)得f(2x)+f(0)=2f(x)f(x),
f(2x) =2f(x)f(x)-1,将x=1/2代入得f(1) =2f(1/2)f(1/2)-1=-1,即
f(1) =-1,
设y=2x代入f(x+y)+f(x-y)=2f(x)f(y)得,f(3x)+f(-x)=2f(x)f(2x),故
f(3x)=2f(x)f(2x)-f(x)=f(x)( 2f(2x)-1)
=f(x)( 4f(x)f(x)-3)=4f(x)f(x)f(x)-3f(x)
将x=1/3代入上式得
f(1)=4 f(1/3) f(1/3) f(1/3)-3 f(1/3),设a=f(1/3)得
4a^3-3a+1=0,解得a=1/2,即f(1/3)=1/2,
再x=1/6代入f(2x) =2f(x)f(x)-1得f(1/3) =2f(1/6)f(1/6)-1,设b=f(1/3)得2bb-1= f(1/3)=1/2,4bb-3=0,b=√3/2.f(x)在[0,1]内单调t得f(1/6)=√3/2.
(1)令y=0,等式变为:f(x)+f(x)=2f(x)f(0),所以f(0)=1;
再x=0,等式化为:f(y)+f(-y)=2f(0)f(y),所以f(y)=f(-y),偶函数得证;
(2)令y=1/2,等式化为:f(x+1/2)+f(x-1/2)=0;
再x-1/2=z,则f(z)=-f(z+1)=f(z+2),周期函数,最小正周期T=2;
(3)令x=1/3...
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(1)令y=0,等式变为:f(x)+f(x)=2f(x)f(0),所以f(0)=1;
再x=0,等式化为:f(y)+f(-y)=2f(0)f(y),所以f(y)=f(-y),偶函数得证;
(2)令y=1/2,等式化为:f(x+1/2)+f(x-1/2)=0;
再x-1/2=z,则f(z)=-f(z+1)=f(z+2),周期函数,最小正周期T=2;
(3)令x=1/3;y=1/6,等式为:f(1/3+1/6)+f(1/3-1/6)=2f(1/3)f(1/6),
即f(1/6)[2f(1/3)-1]=0.因为单调,且f(0)=1>f(1/2)=0,即单调减函数,所以f(1/6)>f(1/3)>f(1/2)=0;
f(1/3)=1/2;
令x=y=1/3,等式为:f(1/6+1/6)+f(1/6-1/6)=2f(1/6)f(1/6),
f(1/6)=√3/2.
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这个我还真的忘了,都大学毕业工作几年了
(1)将x=y=0代入 f(x+y)+f(x-y)=2f(x)f(y)得,f(0)+f(0)=2f(0)f(0),即
2f(0)=2f(0)f(0),由f(0)≠0得,f(0)=1,
再将x=0代入f(0+y)+f(0-y)=2f(0)f(y)得,
f(y)+f(-y)=2f(y), f(-y)=f(y),故f(x)是偶函数。
(2)将y=1/2代入f(x+y)...
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(1)将x=y=0代入 f(x+y)+f(x-y)=2f(x)f(y)得,f(0)+f(0)=2f(0)f(0),即
2f(0)=2f(0)f(0),由f(0)≠0得,f(0)=1,
再将x=0代入f(0+y)+f(0-y)=2f(0)f(y)得,
f(y)+f(-y)=2f(y), f(-y)=f(y),故f(x)是偶函数。
(2)将y=1/2代入f(x+y)+f(x-y)=2f(x)f(y)得
f(x+1/2)+f(x-1/2)=2f(x)f(1/2)=0.由f(1/2)=0 得 f(x+1/2)+f(x-1/2) =0.
f(x+1/2)=-f(x-1/2) , 同理f(x-1/2)= f(x-1+1/2)=- f(x-1-1/2)= - f(x-3/2),故f(x+1/2)= -f(x-1/2) =f(x-3/2),
设y= x-3/2,则x=y+3/2,x+1/2=y+2,代入上式得f(y+2)=f(y),故f(x)是周期为2的周期函数。
(3)设x=y代入f(x+y)+f(x-y)=2f(x)f(y)得f(2x)+f(0)=2f(x)f(x),
f(2x) =2f(x)f(x)-1, 将x=1/2代入得f(1) =2f(1/2)f(1/2)-1=-1,即
f(1) =-1,
设y=2x代入f(x+y)+f(x-y)=2f(x)f(y)得,f(3x)+f(-x)=2f(x)f(2x),故
f(3x)=2f(x)f(2x)-f(x)=f(x)( 2f(2x)-1)
=f(x)( 4f(x)f(x)-3)=4f(x)f(x)f(x)-3f(x)
将x=1/3代入上式得
f(1)=4 f(1/3) f(1/3) f(1/3)-3 f(1/3),设a=f(1/3)得
4a^3-3a+1=0,解得a=1/2,即f(1/3)=1/2,
再x=1/6代入f(2x) =2f(x)f(x)-1得f(1/3) =2f(1/6)f(1/6)-1,设b=f(1/3)得2bb-1= f(1/3)=1/2, 4bb-3=0,b=√3/2. f(x)在[0,1]内单调t得f(1/6)=√3/2。
所以f(1/3)=1/2 f(1/6)=3/2
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