00:01
Here we want to use the first derivative test to find the local extremums.
00:05
So we have f of x is equal to x to the fifth over x to the fifth plus one.
00:13
So what we need to do is take the derivative so that we can find the critical points.
00:18
So f prime of x is going to be a quotient rule.
00:21
We have the derivative of the top, which is a power rule, 5x to the fourth, times the bottom, x to the fifth plus one, minus, now we're going to slip it, take the denominator's derivative, 5x to the fourth, and multiply it by just the numerator and divide that by this x to the fifth plus squared.
00:41
So now i just need to set that equal to zero.
00:44
Notice that since this denominator will never equal zero, let's just not worry about it.
00:49
Actually, it will equal zero and x equals negative one.
00:51
So we do have to consider that point.
00:53
However, then let's take the numerator here and notice that if x is zero, it's going to be zero as well.
01:00
So then after we do that, let's divide by five x to the fourth and we get x to the fifth plus one minus x to the fifth...