Given f = {(1, −1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6), (4, 8)}
f: {1, 4, 9, 16} → {-1, -2, -3, 4} and g: {-1, -2, -3, 4} → {-2, -4, -6, 8}
Co-domain of f = domain of g
So, gof exists and gof: {1, 4, 9, 16} → {-2, -4, -6, 8}
(gof) (1) = g (f (1)) = g (−1) = −2
(gof) (4) = g (f (4)) = g (−2) = −4
(gof) (9) = g (f (9)) = g (−3) = −6
(gof) (16) = g (f (16)) = g (4) = 8
So, gof = {(1, −2), (4, −4), (9, −6), (16, 8)}
But the co-domain of g is not same as the domain of f.
So, fog does not exist.
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?