f(x) and g(x) are polynomials.
⇒ f: R → R and g: R → R.
So, fog: R → R and gof: R → R.
(i) (fog) (x) = f (g (x))
= f (x2 + 1)
= 2 (x2 + 1) + 5
=2x2 + 2 + 5
= 2x2 +7
(ii) (gof) (x) = g (f (x))
= g (2x +5)
= (2x + 5)2 + 1
= 4x2 + 20x + 26
(iii) (fof) (x) = f (f (x))
= f (2x +5)
= 2 (2x + 5) + 5
= 4x + 10 + 5
= 4x + 15
(iv) f2 (x) = f (x) x f (x)
= (2x + 5) (2x + 5)
= (2x + 5)2
= 4x2 + 20x +25
Hence, from (iii) and (iv) clearly fof ≠ f2
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?