Given f: R → R be defined by f(x) = x3 −3
Now we have to prove that f−1 exists
Injectivity of f:
Let x and y be two elements in domain (R),
Such that, x3 − 3 = y3 − 3
⇒ 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 – 3 = y
⇒ x3 = y + 3
⇒ x = \(3\sqrt{(y+3)}\) in R
⇒ 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 − 3
⇒ x + 3 = y3
⇒ y =\( 3\sqrt{(x + 3)}\) = f-1(x) [from (1)]
So, f-1(x) = \(3\sqrt{(x + 3)}\)
Now, f-1(24) = \( 3\sqrt{(24+3)}\)
=\( 3\sqrt{27}\)
= \( 3\sqrt{3^3}\)
= 3
And f-1(5) =\( 3\sqrt{(5 + 3)}\)
= \( 3\sqrt{8}\)
= \( 3\sqrt{2^3}\)
= 2
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
Express R and R-1 as sets of ordered pairs. Determine also
(i) the domain of R‑1
(ii) The Range of R.
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