Question

Let $f:R\to R$ be defined as $f\left(s\right)=10x+7$. Find the function $g:R\to R$ such that $gof=fog=1_R$

Answer 1

Step:1

$f\left(x\right)=10x+7Letf\left(x\right)=yy=10x+7y-7=10x\therefore 10x=y-7whereg:Y\to Nx=\frac{y-7}{10}Letg\left(y\right)=\frac{y-7}{10}$

Step:2

$gof=g\left(f\left(x\right)\right)=g(10x+7)=\frac{(10x+7)-7}{10}=\frac{10x}{10}=x=I_R$

Step:3

$fog=f\left(g\left(y\right)\right)=f\left[\frac{y-7}{10}\right]=10\left(\frac{y-7}{10}\right)+y=y+0=y=I_R$

Since,$gof=fog=I_R$

$f$ is invertible.

and inverse of $f=g\left(y\right)=\frac{y-3}{4}$