已知函数f(2x-1)=x^2,求f[f(x)]

函数f(2x-1)=x^2,
令2x-1 = k
x =( k+1 )/2
x² = ( k+1 )²/4

f(k)=( k+1 )²/4
所以 f(x)=( x+1 )²/4

f[f(x)]
=[ ( x+1 )²/4 +1 ]²/4
= ( x²+2x+1+ 4) / 16
= ( x²+2x+5) / 16

令2x-1=t，x=t+1/2
则f(t)=（t+1/2）^2
所以f(x)=(x+1/2)^2
f[f(x)]= (f(x)+1/2)^2
=[(x+1/2)^2+1 / 2 ]^2
=[(x+1)^2/4+ / 2]^2
=[(x+1)^2+4]^2 / 64

设t=2x-1,x=1/2(t+1)
f(t)=( t+1 )²/4
即f(x)=( x+1 )²/4
所以：f[f(x)]
=[ ( x+1 )²/4 +1 ]²/4
= ( x²+2x+1+ 4) / 16
= ( x²+2x+5) / 16

设t=2x-1,则x=(t+1)/2

y=f(t)=((t+1)/2))^2
f(f(t))=f(y)=((y+1)/2))^2=((((t+1)/2))^2+1)/2))^2=((t^2+2t+17)/16)^2
t再用x表示