Question

Find ${{f}^{-1}}$, if it exists: $f:A\to B,$ where
(i) A=\left\\{ 0,-1,-3,2 \right\\};B=\left\\{ -9,-3,0,6 \right\\} and $f\left( x \right)=3x$
(ii) A=\left\\{ 1,3,5,7,9 \right\\};B=\left\\{ 0,1,9,25,49,81 \right\\} and $f\left( x \right)={{x}^{2}}$

Answer 1

Hint: Check if the function is one-one and onto. If yes, then write $x$ in terms of $f\left( x \right)$, then replace $x$ by ${{f}^{-1}}\left( x \right)$ and $f\left( x \right)$ by $x$.

(i) Here we have to find ${{f}^{-1}}$ for $f:A\to B$, where A=\left\\{ 0,-1,-3,2 \right\\};B=\left\\{ -9,-3,0,6 \right\\} and $f\left( x \right)=3x$
We know that $f\left( x \right)$ is invertible only when $f\left( x \right)$ is one-one and onto.
Now, we will check $f\left( x \right)$ for one-one and onto.
Now, we will put \left\\{ 0,-1,-3,2 \right\\} in $f\left( x \right)$ which is the domain $\left( A \right)$ of the function.
Therefore, $f\left( 0 \right)=3\left( 0 \right)=0$
$f\left( -1 \right)=3\left( -1 \right)=-3$
$f\left( -3 \right)=3\left( -3 \right)=-9$
$f\left( 2 \right)=3\left( 2 \right)=6$
Therefore, \left\\{ 0,-3,-9,6 \right\\} is in the range of the function.
As $A$ have different $f$ images in $B$, therefore the function is one – one
Also, range $=$ co-domain, therefore function is onto.
Therefore, to find ${{f}^{-1}}\left( x \right)$, we will write $x$ in terms of $f\left( x \right)$ and then replace $x$ by ${{f}^{-1}}\left( x \right)$ and $f\left( x \right)$by $x$.
So, $f\left( x \right)=3x$
$\dfrac{f\left( x \right)}{3}=x$
By replacing $x$ by ${{f}^{-1}}\left( x \right)$ and $f\left( x \right)$ by $x$, we get
${{f}^{-1}}\left( x \right)=\dfrac{x}{3}:B\to A$ where B=\left\\{ -9,-3,0,6 \right\\} and A=\left\\{ 0,-1,-3,2 \right\\}
(ii) Here we have to find ${{f}^{-1}}$ for $f:A\to B$ where A=\left\\{ 1,3,5,7,9 \right\\};B=\left\\{ 0,1,9,25,49,81 \right\\} and $f\left( x \right)={{x}^{2}}$.
First of all, we will check $f\left( x \right)$ for one-one and onto.
Now, we will put \left\\{ 1,3,5,7,9 \right\\} in $f\left( x \right)$ which is the domain of the function.
Therefore, $f\left( 1 \right)={{\left( 1 \right)}^{2}}=1$
$f\left( 3 \right)={{\left( 3 \right)}^{2}}=9$
$f\left( 5 \right)={{\left( 5 \right)}^{2}}=25$
$f\left( 7 \right)={{\left( 7 \right)}^{2}}=49$
$f\left( 9 \right)={{\left( 9 \right)}^{2}}=81$
Therefore, \left\\{ 1,9,25,49,81 \right\\} is the range of the function.
As$A$ have different $f$ images in $B$, therefore the function is one – one.
Also range $=$ co-domain. Therefore, the function is onto.
Therefore, to find ${{f}^{-1}}\left( x \right)$, we will write $x$ in terms of $f\left( x \right)$ and then replace $f\left( x \right)$ by $x$ and $x$ by ${{f}^{-1}}\left( x \right)$.
So, $f\left( x \right)={{x}^{2}}$
$\sqrt{f\left( x \right)}=x$
By replacing $x$ by ${{f}^{-1}}\left( x \right)$ and $y$ by $x$, we get
${{f}^{-1}}\left( x \right)=\sqrt{x}:B\to A$ where B:\left\\{ 0,1,9,25,49,81 \right\\} and A:\left\\{ 0,1,3,5,7,9 \right\\}

Note: Students must check if the function is one-one and onto before finding the inverse of the function. One – one function means different elements of the domain have different $f$ images in the codomain. Onto function means the range of the function should be equal to its codomain.