Question

Given that $f\left(x\right)=\frac{2x}{5}+\frac{7}{3}$, find $f^{-1}\left(x\right)$.

• A. $\frac{15}{6x+35}$
• B. $\frac{6}{15x+35}$
• C. $\frac{15x-6}{35}$
• D. $\frac{15x-35}{6}$
• E. $\frac{15-6x}{35}$

Answer 1

The correct option is D $\frac{15x-35}{6}$

Given, $f\left(x\right)=\frac{2x}{5}+\frac{7}{3}$
Let $f\left(x\right)$ be $y$ , then $f^{-1}\left(y\right)$ is $x$
$\therefore f\left(x\right)=\frac{2x}{5}+\frac{7}{3}=y$

$\therefore \frac{2x}{5}=y-\frac{7}{3}=\frac{(3y-7)}{3}$
$\therefore x=\frac{(15y-35)}{6}=f^{-1}\left(y\right)$
$\therefore f^{-1}\left(x\right)=\frac{(15x-35)}{6}$