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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

User English
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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.



User Ross Moody
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