Question:
Show that the function $f: Q \rightarrow Q$, defined by $f(x)=3 x+5$, is invertible. Also, find $f^{-1}$
Solution:
Injectivity of $f$ :
Let $x$ and $y$ be two elements of the domain $(Q)$, such that
$f(x)=f(y)$
$\Rightarrow 3 x+5=3 y+5$
$\Rightarrow 3 x=3 y$
$\Rightarrow x=y$
So, $f$ is one-one.
Surjectivity of $f$.
Let $y$ be in the co-domain $(Q)$, such that $f(x)=y$
$\Rightarrow 3 x+5=y$
$\Rightarrow 3 x=y-5$
$\Rightarrow x=\frac{y-5}{3} \in Q($ domain $)$
$\Rightarrow f$ is onto.
So, $f$ is a bijection and, hence, it is invertible.
Finding $f^{-1}$ :
Let $f^{-1}(x)=y$
$\Rightarrow x=f(y)$ ...(1)
$\Rightarrow x=3 y+5$
$\Rightarrow x-5=3 y$
$\Rightarrow y=\frac{x-5}{3}$
So, $f^{-1}(x)=\frac{x-5}{3} \quad[$ from (1) $]$