Question

函数f（x)是定义在实数集R上不恒为0的偶函数，且对任意实数x都有xf（x+1）=(1+x)f(x),求f[f(5/2)]的值

Answer 1

若,x不为0,则f(x+1)=(x+1)/xf(x),了取x=-1/2,
f(1/2)=f(-1/2+1)=[(1-1/2)/-1/2]f(-1/2)= -f(-1/2)= -f(1/2)
因f（x)是偶函数,有f(-1/2)= f(1/2),所以 f(1/2)=0
f(5/2)=f(3/2+1)=5/3f(3/2)=5/3f(1/2+1)=5/3{[(1+1/2)/1/2]}f(1/2)=5f(1/2)=0
f(5/2)=0
令x=-1则,-f(-1+1)=(-1+1)f(-x),-f(0)=0
f(0)=0
所以f[f(5/2)]=0