题目

g(x)=x2-1且当x≠0时,f[g(x)]=(2-x2)/x2,求（1/3）

提问时间：2021-03-02

答案

f(1/3)=1/2
f[g(x)]=(2-x2)/x2
=-(x^2-2)/x^2
=-((x^2-1)-1)/((x^2-1)+1)
=-((gx)-1)/(g(x)+1)
=(1-g(x))/(g(x)+1)
所以f(x)=(1-x)/(x+1)
f(1/3)=(1-1/3)/(1/3+1)
=(2/3)/(4/3)
=1/2