f: X $\rightarrow$Y, where X,$Y \subseteq R$. Let y $ \in $Y , arbitrarily.


By definition, y $ = $ 10x + 7 for x $ \in $ X


$\Rightarrow$ $x = \cfrac{{y – 7}}{{10}}$

We define, g : Y $\rightarrow$ X by g(y) $ = $$\cfrac{{y – 7}}{{10}}$


Now, (gof)(x) $ = g(f(x)) = $$\cfrac{{f(x) – 7}}{{10}} = \cfrac{{(10x + 7) – 7}}{{10}} = x$

and (fog) ( y ) $ = $ f (g(y)) $ = $ 10g (y) + 7 $ = $ 10 $\left( {\cfrac{{y – 7}}{{10}}} \right)$+ 7$ = $ y


Thus, gof $ = $ fog $ = $ IR.


Hence, f is invertible and g : Y $\rightarrow$ X such that g(y) $ = \cfrac{{y – 7}}{{10}}$

Purchase best math e books from : https://mathstudy.in/