Question

If $f\left(x\right)=3x-5$, then $f^{-1}\left(x\right)$

• A. is given by $\frac{1}{3x-5}$
• B. is given by $\frac{x+5}{3}$
• C. does not exist because $f$ is not one-one
• D. does not exist because $f$ is not onto

Answer 1

The correct option is B is given by $\frac{x+5}{3}$

Let $y=f\left(x\right)=3x-5$

Clearly $f\left(x\right)$ is one-one onto function so it is invertible.

$\Rightarrow y=3x-5\Rightarrow x=\frac{y+5}{3}$

$\therefore f^{-1}\left(x\right)=\frac{y+5}{3}$