What is the inverse of f(x)=(2x−4)2 for x≥2 where function g is the inverse of function f?

Then, if function 'f' is one-to-one and onto function (a needed condition for inverses to exist), then, the inverse of the considered function is

[tex]f^{-1}: Y \rightarrow X[/tex]

such that:

[tex]\forall \: x \in X : f(x) \in Y, \exists \: y \in Y : f^{-1}(y) \in X[/tex]

(and vice versa).

It simply means, inverse of 'f' is undo operator, that takes back the effect of 'f'

For this case, the given function is:

[tex]f(x)=(2x-4)^2 \: \: \: (\text{for }x \geq 2)[/tex]

Take y = f(x).

Expressing x (which is input to f(x) ) in terms of y (which is output f(x) ) as:

[tex]y = (2x-4)^2\\\text{Taking positive sq. root}\\\\\sqrt{y} = (2x-4)[/tex]
(we could legally take positive root since 2x - 4 ≥ 0 because of x ≥ 2 (if 2x - 4 wouldn't be ≥ 0, then as √y ≥0, we would have to write [tex]\pm \sqrt{y} = (2x-4)[/tex] )

Proceeding further:

[tex]\sqrt{y} = (2x-4)\\\\x = \dfrac{\sqrt{y} + 4}{2}[/tex] = g(y)

(this 'g' is the inverse function of y = f(x) )

The g(y) has its own existance, so better not relate it depending on y = f(x).

Since we usually symbolize inputs by variable x, so we get:

[tex]g(x) = \dfrac{\sqrt{x} + 4}{2} = \dfrac{\sqrt{x}}{2} + 2[/tex]

This is defined only if x ≥ 0 (because sq. root isn't defined for negative values, (if working with real numbers only) ).

Thus, the inverse of the function [tex]f(x)=(2x-4)^2[/tex] for x≥2 is [tex]g(x) = \dfrac{\sqrt{x}}{2} + 2, \text{(for x } \geq 0)[/tex]

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