已知g(x)=1+2x,f[g(x)]=x^2-2x

g(x)=1+2x,f[g(x)]=x^2-2x
求f(根号下2）的值
即g(x)=1+2x=根号2
=>x=(根号2-1)/2
所以f(根号2)=(根号2-1)^2/4-(根号2-1)=（2-根号2）/2

2)
令1+2x=t => x=(t-1)/2
f[g(x)]=f(1+2x)=f(t)=(t-1)^2/4-(t-1)=(t^2-6t+5)/4
所以f(x)=(x^2-6x+5)/4

f[g(x)]=x^2-2x=(1+2x)^2/4-x-1/4-2x=(1+2x)^2/4-3*(1+2x)/2+3/2-1/4
得到f(x)=x^2/4-3x/2+5/4
自己代入根号2算吧

g(x)=1+2x
x=1/2[g(x)-1]代入f[g(x)]=x^2-2x=(1/4)[g(x)]^2-(3/2)g(x)+5/4
f(x)=(1/4)x^2-(3/2)x+5/4
f(√2)=1/4*2-3/2*√2+5/4=7/4-3√2/2

f[g(x)]=1/4(2x+1)^2+1/2(2x+1)+1/4
f(x)=1/4x^2+1/2x+1/4
f(根号2）=3/4+根号2/2