Show that a function $\displaystyle \mathrm{f}: \mathbb{\mathrm{R}} \rightarrow \mathbb{\mathrm{R}}$ defined as $\displaystyle \mathrm{f}(\mathrm{x})=\frac{5 \mathrm{x}-3}{4}$ is both one-one and onto.

\multicolumn{6}{|c|}{ Let $\displaystyle \mathrm{x}_{1}, \mathrm{x}_{2} \in \mathrm{R}$ (Domain) such that $\displaystyle \mathrm{f}\left(\mathrm{x}_{1}\right)=\mathrm{f}\left(\mathrm{x}_{2}\right)$ $\displaystyle \Rightarrow \frac{5 \mathrm{x}_{1}-3}{4}=\frac{5 \mathrm{x}_{2}-3}{4}$\Rightarrow 5 \mathrm{x}_{1}=5 \mathrm{x}_{2} \Rightarrow \mathrm{x}_{1}=\mathrm{x}_{2}$\displaystyle \therefore \mathrm{f}$ is one-one.