Question

Let $f:R\to R$ be a real valued function, defined by $f\left(x\right)=2x+3$. Then find $f^{-1}\left(7\right)$.

Answer 1

Given, $f:1R\to iR$ be defined by

$f\left(x\right)=(2x+3)$

Let, $f^{-1}\left(7\right)=y$

or, $f\left(y\right)=7$

or, $2y+3=7$

or, $2y=(7-3)$

or, $y=2$

So, $f^{-1}\left(7\right)=2$.