Let $A = \{x ∈R: x≥1\}$. The inverse of the function $f: A→ A$ given by $f (x) = 2^{x(x-1)}$, is

$\frac{1}{2}\{1+\sqrt{1+4\log_2x}\}$

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

Let $f (x) = y$. Then,

$f(x) = y$

$⇒2^{x (x-1)}=y$

$⇒x(x-1)=\log_2 y$

$⇒ x^2-x-\log_2 y=0$

$x=\frac{1±\sqrt{1+4\log_2y}}{2}$

$⇒x=\frac{1}{2}\{1+\sqrt{1+4\log_2y}\}$  [∵ x > 1]

$⇒f^{-1}(y)=\frac{1}{2}\{1+\sqrt{1+4\log_2y}\}$

Hence, $f^{-1}:A→A$ is given by

$⇒f^{-1}(x)=\frac{1}{2}\{1+\sqrt{1+4\log_2x}\}$