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# Show that the function f : Q → Q defined by f(x) = 3x + 5 is invertible. Also, find f-1.

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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