Question

Let $f:R\to R$ be defined by $f\left(x\right)=10x-7$. If $g=f^{-1},$then $g\left(x\right)=?$

• A. $\frac{1}{\left[10x-7\right]}$
• B. $\frac{1}{\left[10x+7\right]}$
• C. $\frac{\left[X+7\right]}{10}$
• D. $\frac{\left[X-7\right]}{10}$

Answer 1

The correct option is C

$\frac{\left[X+7\right]}{10}$

Determine the value of $g\left(x\right)$

We have, $f\left(x\right)=10x-7$ and $g=f^{-1}$

$f^{-1}\left(x\right)$ is calculated by making $x$ as the subject of the equation.

$$\begin{gathered} y=10x-7 \\ \Rightarrow y+7=10x \\ \Rightarrow x=\frac{y+7}{10} \\ \therefore f^{-1}\left(x\right)=\frac{x+7}{10} \\ g\left(x\right)=f^{-1}\left(x\right) \\ \therefore g\left(x\right)=\frac{x+7}{10} \end{gathered}$$

Hence, the correct answer is Option (C).