Given $f:R \to R$ defined by $\; f(x)=2x+3$
To check if a function is invertible or not ,we see if the function is both one-one and onto.
$\textbf {Step 1: Checking one-one}$
A function $f: X \rightarrow Y$ where for every $x1, x2 \in X, f(x1) = f(x2) \Rightarrow x1 = x2$ is called a one-one or injective function.
Let $f(x)=f(y) \rightarrow 2x+3=2y+3 \rightarrow x = y$.
Therefore f is one-one or injective.
$\textbf {Step 2: Checking onto}$
A function$ f : X \rightarrow Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
Let $y= f(x) = 2x+3 \rightarrow x=\large \frac{y-3}{2}$
$\Rightarrow$ $f(x)=f( \large\frac{y-3}{2})$$=2(\large \frac{y-3}{2})$$+3=y-3+3=y$
Therefore f is onto. Therefore, is invertible, since it both one-one and onto.
$\textbf {Step 3: To calculate}\; f^{-1}, \textbf {we must first define g(y):}$
We know that a function $g$ is called inverse of $f:x \to y$, then exists $g:y \to x$ such that $ gof=I_x\;and\; fog=I_y$, where $I_x, I_y$ are identify functions.
Let us define a function $g:R \to R$ such that $g(y)=\large \frac{y-3}{s}$
$\textbf {Step 4: Calculate gof}$
$\Rightarrow (gof)(x)=g(f(x)) =g(2x+3) =\large \frac{(2x+3)-3}{2}=\frac{2x}{2}$$=x$
$\textbf {Step 5: Calculate fog}$
$\Rightarrow (fog)(y)=f\large (\frac{y-3}{2})$$=2\large (\frac{y-3}{2})$$+3$ $= y$
$\textbf {Step 6: Calculating} \; f^{-} \textbf{ from} \; gof = fog$:
$gof=I_R\;and\;fog=I_R$
$\Rightarrow$ The required inverse function $f^{-1} = g:R \to R$ $g(x)=\large \frac{x-3}{2}$