Consider $f:R \to R$ given by $f(x) = 4x + 3.$ Show that f is invertible. Find the inverse of f.

First, we show that f is invertible
We know that, $f:R \to R,\;f(x) = 4x + 3$

Injectivity

Now, let ${x_1},\;{x_2} \in R.\;\;\;\;\;\;\therefore \;\;f({x_1}) = 4{x_1} + 3\;\;and\;\;f({x_2}) = 4{x_2} + 3$
For one-one, $f({x_1}) = f({x_2}),\;\;4{x_1} + 3 = 4{x_2} + 3 \Rightarrow {x_1} = {x_2}$
Therefore, $f(x)$ is one-one.

Surjectivity
$f:R \to R,$ given for every $y \in R$ (co-domain of f) there exist

$x \in R$ (domain of f) such that f( x) $=$ y
Therefore, $y = 4x + 3 \Rightarrow x = \cfrac{{y - 3}}{4}$
Now, $f\left( {\cfrac{{y - 3}}{4}} \right) = 4\left( {\cfrac{{y - 3}}{4}} \right) + 3 = y\;\;\;\;\;\therefore \;\;f(x) = y$
Thus, f(x) is onto function.
Thus, f is bijective and hence invertible.

Now, we find the inverse of f. We have f(x) $=$ y

$\Rightarrow$ $x = {f^{ - 1}}(y) = \cfrac{{y - 3}}{4}\;\;or\;\;{f^{ - 1}}(x) = \cfrac{{x - 3}}{4}$