Question

If $f:R\to R$ be defined by $f\left(x\right)=x^2+1$, then find $f^{-1}\left\{17\right\}$ and $f^{-1}\{-3\}$.

Answer 1

We know that,
if $f:A\to 13$
such that $yϵ3$. Then,
$f^{-1}\left(y\right)=\{xϵA:f(x)=y\}$. In other words, $f^{-1}\left(y\right)$ is the set of pre-images of y.
Let $f^{-1}\left(17\right)=x$. Then, f(x) = 17
$\Rightarrow x^2+1=17$
$\Rightarrow x^2=17-1=16$
$\Rightarrow x=\pm 4$
Let $f^{-1}(-3)=x$. Then, f(x) = -3
$\Rightarrow x^2+1=-3$
$\Rightarrow x^2=-3-1=-4$
$\Rightarrow x=\sqrt{-4}$
$\therefore f^{-1}(-3)=\theta$