Let f: R$-$R be defined as f(x)$=$ 10x + 7. Find the function g : R $\rightarrow$ R such that gof $=$ fog $=$ IR.

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) $= \cfrac{{y - 7}}{{10}}$