Given function f: Q → Q, defined by f(x) = 3x + 5
Now we have to show that the given function is invertible.
Injection of f:
Let x and y be two elements of the domain (Q),
Such that f(x) = f(y)
⇒ 3x + 5 = 3y + 5
⇒ 3x = 3y
⇒ x = y
so, f is one-one.
Surjection of f:
Let y be in the co-domain (Q),
Such that f(x) = y
⇒ 3x +5 = y
⇒ 3x = y – 5
⇒ x = \( \dfrac{ (y -5)}{3}\) belongs to Q domain
⇒ f is onto.
So, f is a bijection and, hence, it is invertible.
Now we have to find f-1:
Let f-1(x) = y…… (1)
⇒ x = f(y)
⇒ x = 3y + 5
⇒ x −5 = 3y
⇒ y =\( \dfrac{ (x -5)}{3}\)
Now substituting this value in (1) we get
So, f-1(x) = \( \dfrac{ (x -5)}{3}\)
Answered by Sakshi | 1 year agoLet A be the set of first five natural and let R be a relation on A defined as follows: (x, y) R x ≤ y
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