Question

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

Answer 1

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