Given f: R → R given by f(x) = 4x + 3
Now we have to show that the given function is invertible.
Consider injection of f:
Let x and y be two elements of domain (R),
Such that f(x) = f(y)
⇒ 4x + 3 = 4y + 3
⇒ 4x = 4y
⇒ x = y
So, f is one-one.
Now surjection of f:
Let y be in the co-domain (R),
Such that f(x) = y.
⇒ 4x + 3 = y
⇒ 4x = y -3
⇒ \( \dfrac{y-3}{4}\) in R (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 = 4y + 3
⇒ x − 3 = 4y
⇒ y = \( \dfrac{x-3}{4}\)
Now substituting this value in (1) we get
So, f-1(x) = \( \dfrac{x-3}{4}\)
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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