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Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be defined as $f(x)=10 x+7$. Find the function $g:$ $\mathrm{R} \rightarrow \mathrm{R}$ such that $\mathrm{g}$ of $=\mathrm{f} 0 \mathrm{~g}=\mathrm{I}_{\mathrm{R}}$
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Let $\mathrm{f}: \mathrm{X} \rightarrow \mathrm{Y}$ where $\mathrm{X} \subseteq \mathrm{R}, \mathrm{Y} \subseteq \mathrm{R}$ consider an arbitrary element $y$ of $Y$.
By definition, $y=10 x+7$ for some $x \in X$ $\Rightarrow \mathrm{x}=\frac{\mathrm{y}-7}{10}$
Let us define $\mathrm{g}: \mathrm{Y} \rightarrow \mathrm{X}$ by $\mathrm{g}(\mathrm{y})=\frac{\mathrm{y}-7}{10}$
Nowgof $(x)=g[f(x)]=\frac{f(x)-7}{10}=\frac{(10 x+7)-7}{10}$
and $\operatorname{fog}(\mathrm{y})=\mathrm{f}[\mathrm{g}(\mathrm{y})]=10 \mathrm{~g}(\mathrm{y})+7$
$=10 \cdot\left(\frac{y-7}{10}\right)+7=y \Rightarrow \operatorname{gof}(x)=I_R$, fog $(y)=I_R$
$\Rightarrow$ f is invertible and $g: Y \rightarrow X$ such that $g(y)=\frac{y-7}{10}$.
By definition, $y=10 x+7$ for some $x \in X$ $\Rightarrow \mathrm{x}=\frac{\mathrm{y}-7}{10}$
Let us define $\mathrm{g}: \mathrm{Y} \rightarrow \mathrm{X}$ by $\mathrm{g}(\mathrm{y})=\frac{\mathrm{y}-7}{10}$
Nowgof $(x)=g[f(x)]=\frac{f(x)-7}{10}=\frac{(10 x+7)-7}{10}$
and $\operatorname{fog}(\mathrm{y})=\mathrm{f}[\mathrm{g}(\mathrm{y})]=10 \mathrm{~g}(\mathrm{y})+7$
$=10 \cdot\left(\frac{y-7}{10}\right)+7=y \Rightarrow \operatorname{gof}(x)=I_R$, fog $(y)=I_R$
$\Rightarrow$ f is invertible and $g: Y \rightarrow X$ such that $g(y)=\frac{y-7}{10}$.
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