Question:
If $t: \mathrm{Q} \rightarrow \mathrm{Q}$ is defined as $f(x)=x^{2}$, then $f^{-1}(9)$ is equal to
(a) 3
(b) −3
(c) {−3, 3}
(d) ϕ
Solution:
(c) {−3, 3}
If $f: A \rightarrow B$, 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 $f^{-1}\{9\}=x$
Then, $f(x)=9$
$\Rightarrow x^{2}=9$
$\Rightarrow x=\pm 3$
$\therefore f^{-1}\{9\}=\{-3,3\}$
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