Question

Find f ⁻¹ if it exists : f : A → B, where
(i) A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3 x.
(ii) A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x²

Answer 1

(i) A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3 x.
Given: f(x) = 3 x
So, f = {(0, 0), (-1, -3), (-3, -9), (2, 6)}
Clearly, this is one-one.
Range of f = Range of f =B
So, f is a bijection and, thus, f ^-1 exists.
Hence, f ^-1= {(0, 0), (-3, -1), (-9, -3), (6, 2)}

(ii) A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x²
Given: f(x) = x²
So, f = {(1, 1), (3, 9), (5, 25), (7,49), (9, 81)}
Clearly, f is one-one.
But this is not onto because the element 0 in the co-domain (B) has no pre-image in the domain (A) .
$\Rightarrow$f is not a bijection.
So, f ^-1does not exist.