Q. 55.0( 1 Vote )
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 ⇒
Thus, f(x) is invertible and
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