Question:
Find $f^{-1}$ if it exists: $f: A \rightarrow B$, 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 $f(x)=x^{2}$
Solution:
(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 $f(x)=x^{2}$
Given: $f(x)=x^{2}$
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, $f^{-1}$ does not exist.