# Find f −1 if it exists : f : A → B, where

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.