Question

Let $f:R\to R$ be defined by $f\left(x\right)=2x+3$, find $f^{-1}\left(4\right)$.

Answer 1

Given that:

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

To find:

$f^{-1}\left(4\right)=?$

Solution:

A function is one-to=one for which every element of the range of the function corresponds to exactly one element of the domain.

Since, $f$ is a linear function, we can say $f$ is one-to-one and hence $f^{-1}$ exits.

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

or, $y-3=2x$

or, $\frac{y-3}{2}=x$

or, $y=\frac{x-3}{2}$ (Interchange $x$ and $y$)

or, $\therefore f^{-1}\left(x\right)=y=\frac{x-3}{2}$

or, $f^{-1}\left(4\right)=\frac{4-3}{2}$

or, $f^{-1}\left(4\right)=\frac{1}{2}$