if f:[1,infinity)-[1,infinity) is defined by f(x)-2^(x(x-1)) then find inverse of f(x)
Certainly y = f(x) is strictly increasing in [1,+infty) and
lim_ {x to +infty} f(x) = +infty
This means that
f : [1,+infty) -> [1,+infty)
is bijective, as a consequence there exists f^{ -1 } .
Take log in both sides of y = f(x) , solve the quadratic equation on x and choose the positive branch.
There is some problem of 'interpretation', so I suppose that the function is...
y= 2^[x (x-1)] (1)
... so that its inverse function is the solution(s) of the equation...
x^2 - x - log2 y=0 (2)
... that are...
x= 1/2 [1 +/- (1+ 4 log2 y)^(1/2)] (3)
It has to be noted that (3) has two distinct brantches and is defined for y> 2^(- 1/4)...
Kind regards
chi sigma