If a function f : R → R is defined by f(x)=x^2+1, then the pre images…

$f(x)=x^2+1 ~~~f: R \rightarrow R$

so  $y=x^2+1 ~~\Rightarrow ~~y-1=x^2$

so  $x= \pm \sqrt{y-1}$   →  image of y 

for y = -3

$x = \pm \sqrt{-3-1}$

$= \pm \sqrt{-4}$

X doesn't exist as x ∈ R

for y = 17

$x= \pm \sqrt{17-1}$

$x= \pm \sqrt{16}$

$x= \pm 4$

{4, -4}, $\phi$