00:02
Okay, so in this problem, we're given a function, and we're asked to find the t that gives the minimum value of this function, as well as finding out what this value is.
00:16
So the first step i would do is take the derivative of this function with respect to t.
00:22
So the derivative with respect to t is going to be 3t squared minus 12t.
00:37
And the 40 just goes away.
00:41
And so now we're going to set this of a derivative equal to zero and then solve for the critical points of this function.
00:51
And in this case, since we have a t to the power of two, we're going to have two critical points.
00:58
So let's set this equal to zero, 3t squared minus 12t.
01:08
So the first thing i would do is actually pull out the t, from both of these.
01:14
So we'll have a 3t minus 12.
01:18
And so we know that in order for the right side of this function to be true, t has t can be zero, which would make the entire function on the right side zero.
01:31
So if this t was zero, then this whole thing would be true.
01:37
Zero equals zero.
01:39
So that's one of our critical points.
01:41
So now we can find the other.
01:44
The other one is going to be 3t minus 12.
01:51
And if we move 12 over equals 3t, t is going to be 4.
01:59
So we have t equals 4 and t equals 0.
02:03
Those are going to be two critical points.
02:06
So now what i would do is actually graph the original equation as well as the derivative of the equation to just see where those critical points are going to be.
02:17
And if that corresponds to a maximum or a minimum...