Question

If $f:R\to R$ is defined by $f\left(x\right)=\frac{2x+1}{3}$ then $f^{-1}\left(x\right)=$

• A. $\frac{3x-1}{2}$
• B. $\frac{x-3}{2}$
• C. $\frac{2x-1}{3}$
• D. $\frac{x-4}{3}$

Answer 1

Answer: (A)

$\frac{3x-1}{2}$

$f\left(x\right)=\frac{2x+1}{3}$
Let $f\left(x\right)=y=\frac{2x+1}{3}$
$\Rightarrow 2x+1=3y$
$2x=3y-1$
$x=\frac{3y-1}{2}$
$\therefore f\left(x\right)=y\Rightarrow f^{-1}\left(y\right)=x$
$\Rightarrow f^{-1}\left(y\right)=\frac{3y-1}{2}$

$\Rightarrow f^{-1}\left(x\right)=\frac{3x-1}{2}$