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

⇒ 3x_{1} + 5 = 3x_{2} + 5

⇒ 3x_{1} = 3x_{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

⇒ 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

Rate this question :

| Let * be a binaMathematics - Board Papers

Find the idMathematics - Board Papers

Let f : A →Mathematics - Exemplar

Show that the binMathematics - Board Papers

Determine whetherRD Sharma - Volume 1

Fill in theMathematics - Exemplar