令t=x-1,则1-x=-t,1+x=t+2
因为f(x)是偶函数,所以
f(1-x)=f(-t)=f(t)
f(1+x)=f(t+2)
f(1+x)=f(1-x)
=》f(t+2)=f(t)
因此f(x)是周期为2的函数
f(0.6)=f(-0.6)=-(-0.6)=0.6
所以f(8.6)=f(0.6+2*4)=f(0.6)=0.6
-1
设f(x)是定义在R上的偶函数,且f(1+x)= f(1－x),当－1≤x≤0时,f (x) = － x,则f (8.6 ) = _________ （
设f(x)是定义在R上的偶函数,且f(1+x)= f(1－x),当－1≤x≤0时,f (x) = － x,则f (8.6 ) = _________ （
为什么周期是2
∵f(x)是定义在R上的偶函数∴x = 0是y = f(x)对称轴；
又∵f(1+x)= f(1－x) ∴x = 1也是y = f (x) 对称轴。故y = f(x)是以2为周期的周期函数，∴f (8.6 ) = f (8+0.6 ) = f (0.6 ) = f (－0.6 ) = 0.3
为什么周期是2
∵f(x)是定义在R上的偶函数∴x = 0是y = f(x)对称轴；
又∵f(1+x)= f(1－x) ∴x = 1也是y = f (x) 对称轴。故y = f(x)是以2为周期的周期函数，∴f (8.6 ) = f (8+0.6 ) = f (0.6 ) = f (－0.6 ) = 0.3
数学人气：976 ℃时间：2019-09-29 07:51:08
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