Show that the function f : Q → Q, defined by f(x) = 3x + 5,

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) $]$