If \(f: \mathbb{R} \rightarrow \mathbb{R}\) is defined by \(f(x)=2 x+3\), then \(\mathrm{f}^{-1}(\mathrm{x})\)

Let \(\mathrm{x}_{1}, \mathrm{x}_{2} \in \mathrm{R}\)
\[
\begin{array}{lrl}
\text { Let } & \mathrm{f}\left(\mathrm{x}_{1}\right) & =\mathrm{f}\left(\mathrm{x}_{2}\right) \\
\Rightarrow & 2 \mathrm{x}_{1}+3 & =2 \mathrm{x}_{2}+3 \\
\Rightarrow & \mathrm{x}_{1} & =\mathrm{x}_{2}
\end{array}
\]
Hence, \(\mathrm{f}\) is injective.
Let \(\mathrm{y} \in \mathrm{codomain} \mathrm{R}\)
\(\begin{array}{ll}\text { Let } & y=f(x) \\ \Rightarrow & y=2 x+3 \\ \Rightarrow & x=\frac{y-3}{2} \in \text { domain } R\end{array}\)
Hence, \(f\) is onto (surjective). \(\therefore \mathrm{f}^{-1}\) exists.
\[
\therefore \quad \mathrm{f}^{-1}(\mathrm{x})=\frac{\mathrm{x}-3}{2}, \mathrm{x} \in \mathrm{R}
\]