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Answer :

We have f : Q Q and f(x) = 3x + 5.


Recall that a function is invertible only when it is both one-one and onto.


First, we will prove that f is one-one.


Let x1, x2ϵ Q (domain) such that f(x1) = f(x2)


3x1 + 5 = 3x2 + 5


3x1 = 3x2


x1 = x2


So, we have f(x1) = f(x2) x1 = x2.


Thus, function f is one-one.


Now, we will prove that f is onto.


Let y ϵ Q (co-domain) such that f(x) = y


3x + 5 = y


3x = y – 5



Clearly, for every y ϵ Q, there exists x ϵ Q (domain) such that f(x) = y and hence, function f is onto.


Thus, the function f has an inverse.


We have f(x) = y x = f-1(y)


But, we found f(x) = y


Hence,


Thus, f(x) is invertible and


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