若函数f(x)对任意的x∈R都有f(x+3)=-f(x+1),且f(2)=2014,则f[f(2014)+2]+3=?
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/07 20:48:26
若函数f(x)对任意的x∈R都有f(x+3)=-f(x+1),且f(2)=2014,则f[f(2014)+2]+3=?
若函数f(x)对任意的x∈R都有f(x+3)=-f(x+1),且f(2)=2014,则f[f(2014)+2]+3=?
若函数f(x)对任意的x∈R都有f(x+3)=-f(x+1),且f(2)=2014,则f[f(2014)+2]+3=?
函数f(x)对任意的x∈R都有f(x+3)=-f(x+1),且f(2)=2014,
当x=-1时,f(2)=-f(0)=2014,∴f(0)=-2014
∵f(x+4)=-f(x+2)=f(x)
∴f(x)是周期为4的周期函数,
∴f(2014)=f(503×4+2)=f(2)=2014
∴f[f(2014)+2]+3=f(2014+2)+3=f(2016)+3=f(504×4)+3=f(0)+3=-2014+3=-2011
由已知f(x+3)=-f(x+1)得:
1°当x=-1时,f(2)=-f(0)=2014;
2°f(x+5)=-f(x+3)=f(x+1),即有f(x+4)=f(x),即f(x)周期为4.
综上:f[f(2014)+2]+3=f[f(2)+2]+3=f(2014+2)+3=f(2016)+3=f(0)+3=2017.f(2)=-f(0)=2014,所以f(0)=-2014。...
全部展开
由已知f(x+3)=-f(x+1)得:
1°当x=-1时,f(2)=-f(0)=2014;
2°f(x+5)=-f(x+3)=f(x+1),即有f(x+4)=f(x),即f(x)周期为4.
综上:f[f(2014)+2]+3=f[f(2)+2]+3=f(2014+2)+3=f(2016)+3=f(0)+3=2017.
收起