函数f(x)=cx/2x+3(x≠-3/2),满足f[f(x)]=x,则常数c=已知g(x)=1-2x,f[g(x)]=(1-x^2)/x^2(x≠0),那么f(1/2)=?

函数f(x)=cx/2x+3(x≠-3/2),满足f[f(x)]=x,则常数c=已知g(x)=1-2x,f[g(x)]=(1-x^2)/x^2(x≠0),那么f(1/2)=?
函数f(x)=cx/2x+3(x≠-3/2),满足f[f(x)]=x,则常数c=
已知g(x)=1-2x,f[g(x)]=(1-x^2)/x^2(x≠0),那么f(1/2)=?

函数f(x)=cx/2x+3(x≠-3/2),满足f[f(x)]=x,则常数c=已知g(x)=1-2x,f[g(x)]=(1-x^2)/x^2(x≠0),那么f(1/2)=?
3

3

答：设y=f(x)
则y=f(x)=cx/(2x+3)
y=cx/(2x+3)
x=f[f(x)]=f(y)=cy/(2y+3)
所以
cx=2xy+3y
cy=2xy+3x
两式相减得：
c(x-y)=3(y-x)
所以c=-3