Consider f : R → R given by f(x) = 4x + 3.

Injectivity of $f$ :

Let $x$ and $y$ be two elements of domain $(R)$, such that

$f(x)=f(y)$

$\Rightarrow 4 x+3=4 y+3$

$\Rightarrow 4 x=4 y$

$\Rightarrow x=y$

So, $f$ is one-one.

Surjectivity of $f$ :

Let $y$ be in the co-domain $(R)$, such that $f(x)=y$.

$\Rightarrow 4 x+3=y$

$\Rightarrow 4 x=y-3$

$\Rightarrow x=\frac{y-3}{4} \in R($ Domain $)$

$\Rightarrow f$ is onto.

So, $f$ is a bijection and, hence, is invertible.

Finding $f^{-1}$ :

Let $f^{-1}(x)=y$ $\ldots(1)$

$\Rightarrow x=f(y)$

$\Rightarrow x=4 y+3$

$\Rightarrow x-3=4 y$

$\Rightarrow y=\frac{x-3}{4}$

So, $f^{-1}(x)=\frac{x-3}{4} \quad[$ from (1) $]$