Question

If $f:Q\to Q$ is defined as $f\left(x\right)=x^2$, then $f^{-1}\left(9\right)$ is equal to

• A. 3
• B. - 3
• C. {-3, 3}
• D. $\varphi$

Answer 1

The correct option is C

{-3, 3}

If $f:A\to B$, such that $yϵB$, 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 $f^{-1}\left\{9\right\}=x$
Then, $f\left(x\right)=9$
$\Rightarrow x^2=9\Rightarrow x=\pm 3$
$\therefore f^{-1}\left\{9\right\}=\{-3,3\}$