Question

If $f:(0,\infty )\to (0,\infty )$ is defined by $f\left(x\right)=x^2$, then $f^{-1}\left(x\right)=$

• A. $\sqrt{x}$
• B. $\frac{1}{\sqrt{x}}$
• C. Not invertible
• D. $\frac{2}{\sqrt{x}}$

Answer 1

The correct option is A $\sqrt{x}$

As $f$ is defined from $(0,\infty )\to (0,\infty )$
$\therefore f$ is onto and one - one.
Let $f\left(x\right)=x^2=y$
$\Rightarrow x=f^{-1}\left(y\right)$
Also, $x^2=y\Rightarrow x=\sqrt{y}$ as $x>0$
$\therefore f^{-1}\left(y\right)=\sqrt{y}$

Hence, $f^{-1}\left(x\right)=\sqrt{x}$