Find out the absolute extrema for the given function and interval.

 g (t ) = 2t ³ + 3t ² -12t + 4 on [-4, 2]

Solution : All we actually need to do here is follow the process given above.  Thus, first notice that it is a polynomial and hence in continuous everywhere and in specific is then continuous on the given interval.

Now, we have to get the derivative so that we can determine the critical points of the function.

                      g ′ (t ) = 6t ² + 6t -12 = 6 (t + 2) (t -1)

It looks like we'll contain two critical points, t = -2 and t = 1.  Note as well that we in fact desire something more than just the critical points. We just desire the critical points of the function which lie in the interval in question.  Both of these do fall within the interval as so we will employ both of them. That might seem like a silly thing to indicate at this point, however it is frequently forgotten, usually while it becomes important, and thus we will indicate it at every opportunity to ensure it's not forgotten.

Now we evaluate the function at the critical points & the end points of the interval.

g ( -2) = 24                                         g (1) = -3

g ( -4) = -28                                        g ( 2) = 8

Absolute extrema are the largest & smallest the function will ever be & these four points show the only places within the interval where the absolute extrema can take s place.  Thus, from this list we illustrates that the absolute maximum of g(t) is 24 & it takes place at t = -2 (a critical point) and the absolute minimum of g(t) is -28 that occurs at t = -4 (an endpoint).

In this instance we saw that absolute extrema can and will take place at both endpoints & critical points.