In calculus, if you have an function [tex]y=f(t)[/tex] , to find the value of [tex]t[/tex] that minimizes or maximizes that function, you have to compute the first derivative of
[tex]y=f(t)\\=>y^\prime=\frac{df(t)}{dt}[/tex]
and then equate it to zero and solve for [tex]t[/tex].
The problem of finding when the position function will be maximum is best thought of as a problem of maximization in calculus. Consider the position function [tex]x(t)[/tex]. The derivative of this position function is,
[tex]v(t)=3+2t-t^2 .[/tex]
The position function will be maximized when the velocity function is zero, i.e
[tex]v(t)=3+2t-t^2 -0.[/tex]
The next step is to solve for the values of [tex]t[/tex] that satisfy this condition.
[tex]v(t)=3+2t-t^2&=0\\=>t^2-2t-3=0\\=>(t-3)(t+1)=0\\=>t=3,t=-1\\[/tex]
The value [tex]t=3[/tex] is the only value that makes sense in this physical situation.
The position will be maximized at 3 seconds.
