Given f: R+ → R+ and g: R+ → R+
So, fog: R+ → R+ and gof: R+ → R+
Domains of fog and gof are the same.
Now we have to find fog and gof also we have to check whether they are equal or not,
Consider (fog) (x) = f (g (x))
= \( f (\sqrt{x})\)
= \( \sqrt{x^2}\)
= x
Now consider (gof) (x) = g (f (x))
= g (x2)
= \( \sqrt{x^2}\)
= x
So, (fog) (x) = (gof) (x), ∀x ∈ R+
Hence, fog = gof
Answered by Aaryan | 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?