F(x)= 1/x-3, g(x)= 3x+1/x prove whether or not the functions are inverse’s of each other and express the domain of the compositions using interval notation

so first lets say they are bijective so they can have inverse (you have to prove this in some exercices but here they ask you about inverse you can assume it s bijective )

1/x-3=y

1=(x-3)y

1/y=x-3

x=1-y/y

x=1/y-1 no because f (x) inverse=1-x/x it s not g(x)

let s see with g(x)

y=3x+1/x

xy=3x+1

3x-xy=-1

x(3-y)=-1

x=-1/3-y x=1/y-3 g(x) inverse=1/x-3 so f(x) is the inverse of g (x)

the domain for g:R\{0}->R because the denumitor can t be 0 x doesn t have to be 0

the domain of f(x) which is the same with g(x) inverse because we just prove that they are the same is f:R|{3}->R because x-3 is 0 for x=3.

step-by-step explanation: