Question

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

• A. $\frac{1}{3x-5}$
• B. $\frac{x+5}{3}$
• 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

$\frac{x+5}{3}$

Explanation for the correct option.

Find the $f^{-1}\left(x\right)$.

Let $y=f\left(x\right)$, so $x=f^{-1}\left(y\right)$. Now,

$y=3x-5$

Here, $f$ is one-one onto function. Now,

$\begin{array}{rcl}y& =& 3x-5\\ & \Rightarrow & x=\frac{y+5}{3}\end{array}$

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

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

Hence, option B is correct.