Question

If $f\left(x\right)=(x-2{)}^3$ is a function defined in $R\to R$, then

• A. $f^{-1}\left(x\right)=\sqrt[3]{3x}-2$
• B. $f^{-1}\left(x\right)$ does not exist
• C. $f^{-1}\left(x\right)=(-x-2{)}^3$
• D. $f^{-1}\left(x\right)=\sqrt[3]{3x}+2$

Answer 1

The correct option is D $f^{-1}\left(x\right)=\sqrt[3]{3x}+2$


This function passes the horizontal line test which means it is a one to one function also it is an onto function, since it is cubic. So $f\left(x\right)$ is invertible.
$f\left(x\right)=(x-2{)}^3\Rightarrow y=(x-2{)}^3$
Switch $x$ and $y$
$\Rightarrow x=(y-2{)}^3\Rightarrow \sqrt[3]{3x}=y-2\Rightarrow y=\sqrt[3]{3x}+2$
Replace $y$ with $f^{-1}\left(x\right)$
$\Rightarrow f^{-1}\left(x\right)=\sqrt[3]{3x}+2$