Answer :

We have f : Q → Q and f(x) = 2x.

Recall that a function is a bijection only if it is both one-one and onto.

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

Let x_{1}, x_{2}ϵ Q (domain) such that f(x_{1}) = f(x_{2})

⇒ 2x_{1} = 2x_{2}

∴ x_{1} = x_{2}

So, we have f(x_{1}) = f(x_{2}) ⇒ x_{1} = x_{2}.

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

⇒ 2x = y

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 is a bijection and has an inverse.

We have f(x) = y ⇒ x = f^{-1}(y)

But, we found f(x) = y ⇒

Hence,

Thus,

Now, we have g : Q → Q and g(x) = x + 2.

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

Let x_{1}, x_{2}ϵ Q (domain) such that g(x_{1}) = g(x_{2})

⇒ x_{1} + 2 = x_{2} + 2

∴ x_{1} = x_{2}

So, we have g(x_{1}) = g(x_{2}) ⇒ x_{1} = x_{2}.

Thus, function g is one-one.

Now, we will prove that g is onto.

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

⇒ x + 2 = y

∴ x = y – 2

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

Thus, the function g is a bijection and has an inverse.

We have g(x) = y ⇒ x = g^{-1}(y)

But, we found g(x) = y ⇒ x = y – 2

Hence, g^{-1}(y) = y – 2

Thus, g^{-1}(x) = x – 2

We have (f^{-1}og^{-1})(x) = f^{-1}(g^{-1}(x))

We found and g^{-1}(x) = x – 2

⇒ (f^{-1}og^{-1})(x) = f^{-1}(x – 2)

We know (gof)(x) = g(f(x)) and gof : Q → Q

⇒ (gof)(x) = g(2x)

∴ (gof)(x) = 2x + 2

Clearly, gof is a bijection and has an inverse.

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

⇒ 2x + 2 = y

⇒ 2x = y – 2

We have (gof)(x) = y ⇒ x = (gof)^{-1}(y)

But, we found (gof)(x) = y ⇒

Hence,

Thus,

So, it is verified that (gof)^{-1} = f^{-1}og^{-1}.

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