已知函数f(x)的定义域为R,满足f(x+2)=-f(x),当0≤x≤1时,f(x)=x,则f(8.5)等于( )
A.-0.5
B.0.5
C.-1.5
D.1.5
∵函数f(x)的定义域为R,满足f(x+2)=-f(x),∴f(x+4)=f(x),
故函数的周期等于4.
又当0≤x≤1时,f(x)=x,故f(8.5)=f(4×2+0.5)=f(0.5)=0.5,
故答案为 B.
已知函数f(x)的定义域为R,满足f(x+2)=-f(x),当0≤x≤1时,f(x)=x,则f(8.5)等于( )
A.-0.5
B.0.5
C.-1.5
D.1.5
∵函数f(x)的定义域为R,满足f(x+2)=-f(x),∴f(x+4)=f(x),
故函数的周期等于4.
又当0≤x≤1时,f(x)=x,故f(8.5)=f(4×2+0.5)=f(0.5)=0.5,
故答案为 B.