设g（x）=1-2x，f[g（x）]=(1-x^2)\x ,x≠0,则f（1\2）=

f（1\2）即
g（x）]=1/2
即1-2x=1/2
则解得x=1/4
所以代入
即求
f[g（1/4）]=(1-(1/4)^2)/（1/4）
=15/4

f[g（x）]=(1-x^2)\x
=(1-x)(1+x)/x
=(1/2+1/2-x)(3/2-1/2+x)/(1/2-1/2+x)
=(1/2+1/2g(x))(3/2-1/2g(x))/(1/2-1/2g(x))
=(1+g(x))(3-g(x)/(2-2g(x))

f(x)=(1+x)(3-x)/(2-2x)
f(1/2)=3/2*5/2/1=15/4

g(x) = 1/2
=> x = 1/4

f(1/2) = [1-(1/4)^2]/(1/4) = 15/4