Question

Find $$f^{-1}$$, if it exists: $$f:A\to B,$$ where
(i) $$A=\{0,-1,-3,2\};B=\{-9,-3,0,6\}$$ and $$f\left(x\right)=3x$$
(ii) $$A=\{1,3,5,7,9\};B=\{0,1,9,25,49,81\}$$ 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 $$A=\{0,-1,-3,2\};B=\{-9,-3,0,6\}$$ 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 $$\{0,-1,-3,2\}$$ in $$f\left(x\right)$$ which is the domain $$\left(A\right)$$ of the function.
Therefore, $$f\left(0\right)=3\left(0\right)=0$$
$$f(-1)=3(-1)=-3$$
$$f(-3)=3(-3)=-9$$
$$f\left(2\right)=3\left(2\right)=6$$
Therefore, $$\{0,-3,-9,6\}$$ 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, $$f\left(x\right)=3x$$
$${\displaystyle \frac{f\left(x\right)}{3}}=x$$
By replacing $$x$$ by $$f^{-1}\left(x\right)$$ and $$f\left(x\right)$$ by $$x$$, we get
$$f^{-1}\left(x\right)={\displaystyle \frac{x}{3}}:B\to A$$ where $$B=\{-9,-3,0,6\}$$ and $$A=\{0,-1,-3,2\}$$
(ii) Here we have to find $$f^{-1}$$ for $$f:A\to B$$ where $$A=\{1,3,5,7,9\};B=\{0,1,9,25,49,81\}$$ 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 $$\{1,3,5,7,9\}$$ 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, $$\{1,9,25,49,81\}$$ 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 $$B:\{0,1,9,25,49,81\}$$ and $$A:\{0,1,3,5,7,9\}$$

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.