高一数学 抽象函数

Question

已知f(x)为定义在实数集上不恒为零的偶函数，且对任意实数都有xf(x+1)=(1+x)f(x)，求f[f(5\2)]

Answer 1

先求 f(1/2) 和 f（0）。对于 x=-1/2 有 （-1/2）*f（【-1/2】+1）=（1+【-1/2】）*f（-1/2），于是 （-1/2）*f（1/2）=（1/2）*f（1/2）进一步有 f（1/2）=0 ；再求f（0），对于x=0有 0*f(1)=1*f(0),于是有 f（0）=0 。由 xf(x+1)=(1+x)f(x)可得 f（x）=xf（x-1）/（x-1），可推出f（5/2）= 0 ，于是 f[f(5\2)] =f（0）=0

Answer 2

f(x+1) = (x+1)f(x)/x
f(5/2) = f(2.5) = f(1.5 + 1) = (1.5+1)f(1.5)/1.5 = (5/2)f(3/2)/(3/2)
=(5/3)f(3/2)
f(3/2) = f(1.5) = f(0.5 + 1) = (0.5+1)f(0.5)/0.5 = (3/2)f(1/2)/(1/2)
= 3f(1/2)
f(5/2) = (5/3)*3f(1/2) = 5f(1/2)
f(1/2) = f(-1/2 + 1) = (-1/2 + 1)f(-1/2)/(-1/2) = -f(-1/2)
且f(x)是偶函数，所以f(-x)=f(x)，所以 f(1/2) = f(-1/2)
所以f(1/2) = -f(-1/2) = -f(1/2)
得到f(1/2) = 0
f(5/2) = 5f(1/2) = 0

f[f(5/2)] = f(0)

将x=0带入xf(x+1) = (1+x)f(x)
得到 0 * f(1) = 1 * f(0)，所以f(0) = 0

所以 f[f(5/2)] = 0

Answer 3

这很简单啊，f[f(5\2)] =5/3