00:01
When computing the absolute maximum values on a given interval, we always want to check a certain number of points.
00:09
The points we always need to check are the end points at zero and four because we have a closed interval.
00:16
If we did not have a closed interval, we wouldn't check them, but we have a closed interval, so we have to check them.
00:21
And lastly, we also have to check all of the critical points, which means that all the points where the derivative is either zero or undefined.
00:30
So let's go ahead and try to find those.
00:33
If we take the derivative of this function with respect to x, we first off, lose the 9.
00:38
We're left with 81 minus 9x squared.
00:44
We now set this equal to 0.
00:47
If we do this, we would then get 81 is equal to 9x squared.
00:52
And then we would also get 9 is equal to x squared.
00:56
And lastly, we then take the square root of both sides.
00:59
We would get x is equal to plus or minus 3.
01:02
Now, if you notice we have two answers here.
01:05
One of them is actually not in our domain of interest.
01:07
So we really are only looking at x is equal to 3, not x equal to negative 3...