Let $A=\{x\in R:x\ge 1\}$.The inverse of the function $f:A\to A$ is given by $f\left(x\right)=2^{x(x-1)}$, is

It can be easily verified that $f:A\to A$ is a bijection.

Let $f\left(x\right)=y.$ Then,

$\begin{array}{l} f\left(x\right)=y\\ \Rightarrow 2^{x(x-1)}=y\\ \Rightarrow x(x-1)={\log }_{2}y\\ \Rightarrow x^2-x-{\log }_{2}y=0\end{array}$

$x = \frac{1\pm \sqrt{1+4{\log }_{2}y}}{2}$

$\Rightarrow x=\frac{1}{2}\left\{1+\sqrt{1+4{\log }_{2}y}\right\} [\because x>1]$

$\Rightarrow f^{-1}\left(y\right)=\frac{1}{2}\left\{1+\sqrt{1+4{\log }_{2}y}\right\}$

Hence, $f^{-1}:A\to A$ is given by

   $f^{-1}\left(x\right)=\frac{1}{2}\left\{1+\sqrt{1+4{\log }_{2}x}\right\}$