00:01
10 of y equals x to the a minus x to the b on 0 1 and notice that a is smaller than b is but they're both positive and after you find that then use it to find the maximum of x to the 5th minus x to the 10th okay so we need to find the critical value so we're going to take the derivative so that's a x to the a minus 1 minus b x to the b minus 1 and we're going to set that equal to zero.
00:35
Okay, so a minus one is less than b minus one.
00:38
So let's factor out x to the a minus one.
00:43
And we get a minus b.
00:46
Okay, so let's think about if we had x to the seventh and we factored out an x squared, then we would have x to the fifth left.
00:56
So if we have x to the b minus one and we factor out an x to the a minus one, we just have to support.
01:04
It from there.
01:06
So we get x to the b minus a because when you add them you get a minus 1 plus b minus a which is b minus 1.
01:20
Oh yeah yeah that's right.
01:26
Okay so we get minus b x to the b minus a.
01:38
Okay so either x is zero which we're going to check anyway because it's an endpoint or x to the b minus a equals a over b so x is a over b to the one over b minus a power like i took the b minus a root that's not a h that's a one there okay so then we have to check and see which is the max and min 0, 1, or a over b to the 1 over b minus a power...