Let `f: R ->R`be defined as `f(x) = 10 x + 7`. Find the function `g: R ->R`such that`gof=fog=1_R`

Explanation

In `F: R rarr R`, f(x) = 10 x +7
Let x,y `in`R
and f(x)=f(y)
`rArr 10 x +7 =10 y +7`
`rArr 10x = 10 y `
`rArr x=y`
`therefore` f is one - one
Again let f(x) =y where `y in R`
`rArr 10x +7 =y rArr x(y-7)/(10)i R`
`therefore` f is onto
Now `(fog )(x) = f((y-7)/(7))`
`=10 ((y-7)/(10))+7=y`
`therefore go f =I_R "and" fog =I_R`
`therefore g: R rarr R ` is defined as g(y) `=(y-7)/(10)`