Question

$f:(-\infty ,0]\to [0,\infty )$ is defined as $f\left(x\right)=x^2$. The domain and range of its inverse is

• A. Domain of $\left(f^{-1}\right)=[0,\infty )$, Range of $\left(f^{-1}\right)=(-\infty ,0]$
• B. Domain of $\left(f^{-1}\right)=[0,\infty )$, Range of $\left(f^{-1}\right)=(-\infty ,\infty )$
• C. Domain of $\left(f^{-1}\right)=[0,\infty )$, Range of \left( { f }^{ -1 } \right) =[0,\infty )$
• D. $f^{-1}$ does not exist

Answer 1

The correct option is A Domain of $\left(f^{-1}\right)=[0,\infty )$, Range of $\left(f^{-1}\right)=(-\infty ,0]$

Given, $f\left(x\right)=x^2$

$\Rightarrow y=x^2$

$\Rightarrow x=\pm \sqrt{y}$

$\Rightarrow x=-y$
$f^{-1}\left(x\right)=-\sqrt{-x}$
Domain $f^{-1}=[0,\infty )$
Range of $f^{-1}=(-\infty ,0]$