Let f: R$ – $R be defined as f(x)$ = $ 10x + 7. Find the function g : R $\rightarrow$ R such that gof $ = $ fog $ = $ IR. ~~~~[NCERT Misc.,Q.1, Page 29]By admin / April 29, 2021 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$ = $ yThus, 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/