已知函数f(x)定义域为R,对任意x,y属于R有f(x+y)+f(x-y)=2f(x)(y),且f(0)不等于0.若存在常数C,使f(c/2)=0.求证:对任意x属于R,有f(x+c)=-f(x).
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![已知函数f(x)定义域为R,对任意x,y属于R有f(x+y)+f(x-y)=2f(x)(y),且f(0)不等于0.若存在常数C,使f(c/2)=0.求证:对任意x属于R,有f(x+c)=-f(x).](/uploads/image/z/3629426-50-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%E5%AE%9A%E4%B9%89%E5%9F%9F%E4%B8%BAR%2C%E5%AF%B9%E4%BB%BB%E6%84%8Fx%2Cy%E5%B1%9E%E4%BA%8ER%E6%9C%89f%28x%2By%29%2Bf%28x-y%29%3D2f%28x%29%28y%29%2C%E4%B8%94f%280%29%E4%B8%8D%E7%AD%89%E4%BA%8E0.%E8%8B%A5%E5%AD%98%E5%9C%A8%E5%B8%B8%E6%95%B0C%2C%E4%BD%BFf%28c%2F2%29%3D0.%E6%B1%82%E8%AF%81%EF%BC%9A%E5%AF%B9%E4%BB%BB%E6%84%8Fx%E5%B1%9E%E4%BA%8ER%2C%E6%9C%89f%EF%BC%88x%2Bc%EF%BC%89%3D-f%EF%BC%88x%EF%BC%89.)
已知函数f(x)定义域为R,对任意x,y属于R有f(x+y)+f(x-y)=2f(x)(y),且f(0)不等于0.若存在常数C,使f(c/2)=0.求证:对任意x属于R,有f(x+c)=-f(x).
已知函数f(x)定义域为R,对任意x,y属于R有f(x+y)+f(x-y)=2f(x)(y),且f(0)不等于0.
若存在常数C,使f(c/2)=0.求证:对任意x属于R,有f(x+c)=-f(x).
已知函数f(x)定义域为R,对任意x,y属于R有f(x+y)+f(x-y)=2f(x)(y),且f(0)不等于0.若存在常数C,使f(c/2)=0.求证:对任意x属于R,有f(x+c)=-f(x).
令y=0代入f(x+y)+f(x-y)=2f(x)(y)
则2f(x)=2f(x)^2,所以f(x)=f(x)^2,所以f(x)=0或1
令y=c/2代入f(x+y)+f(x-y)=2f(x)(y)
则f(x+c/2)+f(x-c/2)=2f(x)f(c/2)=0
所以f(x+c/2)^2+f(x-c/2)^2=0
所以f(x+c/2)=f(x-c/2)=0{x不等于正负c/2}
令x=x+c/2所以f(x+c)=f(x)=0=-f(x){x不等于0或-c}
下面只需证f(c)=-f(0)=-1和-f(-c)=f(0)=1
即证f(c)=f(-c)=-1
令x=y=c/2,代入f(x+y)+f(x-y)=2f(x)(y)
得f(c)+f(0)=2f(c/2)^2=0,因为f(0)=1,所以f(c)=-1
令x=c,y=c,代入f(x+y)+f(x-y)=2f(x)(y)
得f(2c)+f(0)=2f(c)^2=2
令x=c,y=-c,代入f(x+y)+f(x-y)=2f(x)(y)
得f(0)+f(2c)=2f(c)f(-c)=2,因为f(c)=-1,所以f(-c)=-1
所以对任意x属于R,有f(x+c)=-f(x)