00:01
Here, our goal is to take the function, the cube root of x squared or x to the two thirds is another way to write that, over the interval 1 to 27, and apply the mvt, the mean value theorem, if it does apply.
00:12
This particular function, if you graph it, there's one way to start thinking about this.
00:19
It looks something like this.
00:21
And let's just say that 1 and 27 are two points here and here that we care about.
00:30
We need to show that this function is continuous over the interval, 1 to 27, and we need to show that it's differentiable over the open interval, 1 to 27.
00:47
The conditions are met.
00:48
I don't think you need to get too hot and heavy with this.
00:50
The function is clearly continuous.
00:52
If you look at the graph, you can see that you're defined at all x -virus from 1 to 27, so we check that box.
00:57
And you should be able to find slopes at all these points as well.
00:59
The only point on this whole curve where you aren't differential, couldn't find a derivative value, would be at zero, there's a cusp there.
01:06
But that's not true.
01:07
We don't really care of, at least, well, it's not an issue for the interval 1 to 27.
01:11
So we meet the credentials and the needs of the mean value theorem, and now we need to apply it.
01:18
And the mean value theorem says that f prime at some x value c is equal to f of b minus f of a, all over b minus a, which is the average rate of change.
01:30
So the average rate of change is that symbolic value is like connecting those two points and finding the slope there.
01:45
And then once you do, you want to find the point on the curve that also has that same slope as well.
01:52
And sometimes it's more than one point.
01:54
But in this case, i think it'll just be one.
01:57
Our a and or b are the x values that are the endpoints of the interval.
02:01
So we want to find f of 27 minus f of 1 over 27 minus 1...