Question 1184854
{{{"f'(x)" = 4-2x}}} ===>  the turning point is at x = 2, since {{{"f'(2)" = 0}}}.


{{{"f''(x)" = -2 < 0}}}, and in particular {{{"f''(2)" = -2 < 0}}} ===> there is a maximum point at x = 2. 
(This is to be expected since the coefficient of {{{x^2}}} is negative, and so the parabola opens downward.)


For {{{x>=2}}}, f(x) has an inverse function because it is one-to-one over this set. The function is one-to-one because it is strictly decreasing over [2, {{{+infinity}}}).


The inverse function is obtained as follows:

{{{y = -x^2 + 4x + 1}}}  ===>  {{{0 = -x^2 + 4x + (1-y)}}} ===> {{{x = 2 +- sqrt(5-y)}}}  


after solving x in terms of y using the quadratic formula.  Choose {{{x = 2 + sqrt(5-y)}}} since {{{x>=2}}}.


Interchanging the places of x and y, the inverse function is therefore {{{f^(-1)(x) = 2 + sqrt(5-x)}}}.