Question

Find the inverse of the function $f\left(x\right)=ax+b$ if $a,b\in R$ and $f:R\to R$

Answer 1

Here the function

$f:R\to R$ is defined as $f\left(x\right)=ax+b=y\left(say\right)$. Then $x=\frac{y-b}{a}$

$x=\frac{y-b}{a}$

This leads to a function $g:R\to R$ defined as $g\left(y\right)=$ $\frac{y-b}{a}=x$

Therefore,$\left(gof\right)\left(x\right)=g\left(f\right(x)=g$($\frac{y-b}{a}$)=$\frac{ax+b-b}{a}$ $=x$ or

$g\left(of\right)=$ $I_R$

Similarly

$\left(fog\right)\left(y\right)=f\left(g\right(y\left)\right)=y$

$fog=$$I_R$

Hence $f$ is invertible and $f^{-1}$=$g$

which is given by $f^{-1}$$\left(x\right)=$ $\frac{y-b}{a}$