Given f: R → R+ → [4, ∞) given by f(x) = x2 + 4.
Now we have to show that f is invertible,
Consider injection of f:
Let x and y be two elements of the domain (Q),
Such that f(x) = f(y)
⇒ x2 + 4 = y2 + 4
⇒ x2 = y2
⇒ x = y (as co-domain as R+)
So, f is one-one
Now surjection of f:
Let y be in the co-domain (Q),
Such that f(x) = y
⇒ x2 + 4 = y
⇒ x2 = y – 4
⇒ x = \( \sqrt{ (y-4)}\) in R
⇒ 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 = y2 + 4
⇒ x − 4 = y2
⇒ y =\( \sqrt{ (x-4)}\)
So, f-1(x) = \( \sqrt{ (x-4)}\)
Now substituting this value in (1) we get,
So, f-1(x) = \( \sqrt{ (x-4)}\)
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.
A function f: R → R is defined as f(x) = x3 + 4. Is it a bijection or not? In case it is a bijection, find f−1 (3).
If f: R → R be defined by f(x) = x3 −3, then prove that f−1 exists and find a formula for f−1. Hence, find f−1 (24) and f−1 (5).
Consider f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with f-1(x) = \( \dfrac{\sqrt{(x+6)-1}}{3}\)
If f(x) = \(\dfrac{ (4x + 3)}{(6x – 4)}\), x ≠ (\( \dfrac{2}{3}\)) show that fof(x) = x, for all x ≠ (\( \dfrac{2}{3}\)). What is the inverse of f?