Question

高一数学，急！！！！

已知函数f(x)为定义在R上不恒为零的偶函数，且对于定义域上任意的x，恒有xf(x+1)=(1+x)f(x)成立，求f(f(2.5）的值

Answer 1

令x=-0.5
x+1=0.5
则-0.5f(0.5)=0.5f(-0.5)
偶函数则f(0.5)=f(-0.5)
所以f(0.5)=0

令x=0.5
则0.5f(1.5)=1.5f(0.5)
所以f(1.5)=0
同理令x=1.5则f(2.5)=0

令x=-1
-f(0)=0*f(-1)=0
f(0)=0

所以f[f(2.5)]
=f(0)
=0

Answer 2

x=-0.5,-0.5*f(0.5)=(1-0.5)*f(-0.5),偶函数,f(-0.5)=f(0.5),f(0.5)=0
1.5f(1+1.5)=2.5f(1.5),
0.5f(1.5)=1.5*f(0.5)
f(2.5)=0,
0f(0+1)=(0+1)f(0)
f(0)=0
f(f(2.5))=f(0)=0

Answer 3

答案就是0