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Consider $f: R \rightarrow R$ given by $f(x)=4 x+3$. Show that $f$ is invertible. Find the inverse of $f$.
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$\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ given by $\mathrm{f}(\mathrm{x})=4 \mathrm{x}+3$
$f\left(x_1\right)=4 x_1+3, f\left(x_2\right)=4 x_2+3$
If $\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)$, then $4 \mathrm{x}_1+3=4 \mathrm{x}_2+3$
or $4 x_1=4 x_2$ or $x_1=x_2 \therefore$ fis one-one
Also let $y=4 x+3$, or $4 x=y-3 \therefore x=\frac{y-3}{4}$
For each value of $\mathrm{y} \in \mathrm{R}$ and belonging to co-domain of $\mathrm{y}$ has a pre-image in its domain.
$\therefore f$ is onto i.e. $f$ is one-one and onto
$\therefore$ fis invertible and $\mathrm{f}^{-1}(\mathrm{y})=\mathrm{g}(\mathrm{y})=\frac{\mathrm{y}-3}{4}$.
$f\left(x_1\right)=4 x_1+3, f\left(x_2\right)=4 x_2+3$
If $\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)$, then $4 \mathrm{x}_1+3=4 \mathrm{x}_2+3$
or $4 x_1=4 x_2$ or $x_1=x_2 \therefore$ fis one-one
Also let $y=4 x+3$, or $4 x=y-3 \therefore x=\frac{y-3}{4}$
For each value of $\mathrm{y} \in \mathrm{R}$ and belonging to co-domain of $\mathrm{y}$ has a pre-image in its domain.
$\therefore f$ is onto i.e. $f$ is one-one and onto
$\therefore$ fis invertible and $\mathrm{f}^{-1}(\mathrm{y})=\mathrm{g}(\mathrm{y})=\frac{\mathrm{y}-3}{4}$.
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