设f(x)=lnx,g(x)=2(x+1)/x-1,则f[g-1(x)]=?则f[g-1(x)]=?中“-1”在g右上角

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设f(x)=lnx,g(x)=2(x+1)/x-1,则f[g-1(x)]=?则f[g-1(x)]=?中“-1”在g右上角

设f(x)=lnx,g(x)=2(x+1)/x-1,则f[g-1(x)]=?则f[g-1(x)]=?中“-1”在g右上角
设f(x)=lnx,g(x)=2(x+1)/x-1,则f[g-1(x)]=?
则f[g-1(x)]=?中“-1”在g右上角

设f(x)=lnx,g(x)=2(x+1)/x-1,则f[g-1(x)]=?则f[g-1(x)]=?中“-1”在g右上角
答:
y=g(x)=2(x+1)/(x-1)
=2(x-1+2)/(x-1)
=2+4/(x-1)
y-2=4/(x-1)
x-1=4/(y-2)
所以:
g-1(x)=1+4/(x-2)=(x+2)/(x-2)
因为:
f(x)=lnx
所以:f [g-1(x) ]=ln [(x+2)/(x-2)]