00:01
For this problem, we are asked to find the critical numbers of the function f of x equals x squared minus 2x plus 1, all divided by x plus 1.
00:08
Then to find the open intervals on which the function is increasing or decreasing, and to locate all relative extrema.
00:15
We are to use a graphing utility to confirm our results.
00:19
To begin, we need to take the derivative of f of x.
00:23
We'll have to apply the quotient rule here.
00:27
So differentiating the numerator gives us 2x minus 2.
00:30
We're multiplying that by x plus 1.
00:33
Then when you have minus the numerator, x squared minus 2x plus 1, times just 1 from the derivative of the denominator, and that all is getting divided by x plus 1 squared.
00:45
So we can end up simplifying the numerator, simply to x squared plus 2x minus 3, divided by x plus 1, all squared.
00:55
Now we can then see that we'll have at least two different critical numbers, two different kinds of critical numbers at least.
01:04
We can see that we'll have the derivative will approach infinity when x equals negative 1, so we need to keep that in mind, as well as the fact that we'll have for the other one here.
01:19
Let's see.
01:20
We'll have that the function will be equal to zero, or not the function, the derivative will be equal to zero, so we'll have a turning point for when x squared plus 2 ,000, x minus 3 equals 0.
01:37
We can find this easily by first factoring the expression, writing it as x minus 1 times x plus 3.
01:46
So we can clearly see that that will be the case when we have x equals 1 and when x equals negative 3.
01:53
Excuse me.
01:54
So plugging in, or if we were to plug in a value that is in between negative 3 and negative 1.
02:01
So let's say f prime of negative 2.
02:04
Then we will find, one moment here, f prime of negative 2 would be negative 3.
02:14
So we can see that we are going to be decreasing in between negative 1 and negative 3...