Question

Let $f:R\to R$ be defined by $f\left(x\right)=2x+6$ which is a bijective mapping then $f^{-1}\left(x\right)$ is given by

• A. $\frac{x}{2}-3$
• B. $2x+6$
• C. $x-3$
• D. $6x+2$

Answer 1

The correct option is A $\frac{x}{2}-3$

Let $y=f\left(x\right)=2x+6$

$\Rightarrow 2x=y-6$

$\Rightarrow x=\frac{y}{2}-\frac{6}{3}$

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

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