Is f(x)=1/(x-3) the inverse of f(g)=1/(x-3)

Question

explain if it is an inverse or not:))

Answer 1

(1) Since g has not been defined I'm assuming that g is the inverse of f.

Let y=f(x)=1/(x-3), so x-3=1/y and x=3+1/y=(3y+1)/y.

If x=g(y)=(3y+1)/y, then g(x)=f^-1(x)=(3x+1)/x, that is, g is the inverse of f.

f(g(x))=1/(g(x)-3)=1/((3x+1)/x-3)=x/(3x+1-3x)=x, so f(g)=f(f^-1(x))=x. 

Let z=f(g(x))=x, then x=z. Let x=h(z)=z, so h(x)=(f(g))^-1(x)=x, that is, h is the inverse of f(g).

Therefore 1/(x-3) is not the inverse of f(g). The inverse of f(g) is x.

(2) If f(x)=f(g(x))=1/(x-3)=1/(g(x)-3), then g(x)=x. The inverse of f(g(x)) is the same as the inverse of f(x) which is (3x+1)/x.

Whichever of solutions (1) or (2) applies, the inverse of f(g) is not 1/(x-3).