Given that f: R → R is defined as f(x) = x3 + 4
Injectivity of f:
Let x and y be two elements of domain (R),
Such that f (x) = f (y)
⇒ x3 + 4 = y3 + 4
⇒ x3 = y3
⇒ x = y
So, f is one-one.
Surjectivity of f:
Let y be in the co-domain (R),
Such that f(x) = y.
⇒ x3 + 4 = y
⇒ x3 = y – 4
⇒ x =\( 3\sqrt{ (x-4)}\) in R (domain)
⇒ f is onto.
So, f is a bijection and, hence, it is invertible.
Finding f-1:
Let f−1 (x) = y…… (1)
⇒ x = f (y)
⇒ x = y3 + 4
⇒ x − 4 = y3
⇒ y =\( 3\sqrt{ (x-4)}\)
So, f-1(x) =\(3\sqrt{ (x-4)}\) [from (1)]
f-1 (3) = \( 3\sqrt{ (x-4)}\)
=\( 3\sqrt{-1}\)
= -1
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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(ii) The Range of R.
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