If f : R

If $f: A \rightarrow B$ is such that $y \in B$, then $f^{-1}\{y\}=\{x \in A: f(x)=y\}$.

In other words, $f^{-1}\{y\}$ is the set of pre - images of $y$.

Let $f^{-1}\{17\}=x$.

Then, f (x) =17 .

$\Rightarrow x^{2}+1=17$

$\Rightarrow x^{2}=17-1=16$

$\Rightarrow x=\pm 4$

$\therefore f^{-1}\{17\}=\{-4,4\}$

Again,

let $f^{-1}\{-3\}=x$.

Then, $f(x)=-3$

$\Rightarrow x^{2}+1=-3$

$\Rightarrow x^{2}=-3-1=-4$

$\Rightarrow x=\sqrt{-4}$

Clearly, no solution is available in R.

So $f^{-1}\{-3\}=\Phi$.