00:01
So in this problem, what we're going to be doing is we're going to be taking a number of steps to use the first derivative test to determine any local maximum or minimum that exist within this function, f of x here.
00:17
So what the first derivative test does is by looking at the derivative of f of x, f prime of x, if we look at the open intervals, which we find from the critical numbers, by finding the zeros of f prime of x.
00:35
What we can do is we can look at on either side of the interval whether the sign of f prime of x changes when it goes across these critical numbers.
00:45
So if we see that we have a positive to negative change in f prime of x across a critical number across those intervals, then that tells us that we have a relative maximum.
01:01
And conversely, if we see, that f prime of x undergoes a change from negative to positive cross that interval.
01:13
Then we see that we have a relative minimum.
01:18
And if f of x doesn't change, that means we have neither a relative maximum or minimum.
01:25
So let's take a look at our problem here, and we can go through the steps that we need to take.
01:30
So first what we want to do is we want to find the critical numbers.
01:34
And the way that we're going to do that is we're going to take the derivative of f of x and then find where f prime of x is equal to zero.
01:43
So we have an absolute value here and just to refresh the derivative of an absolute value, a function with an absolute value, that will be as such.
01:55
So we have, we take the derivative of you, the absolute value of you with respect to x.
02:04
What we're going to have is u over the absolute value of u times the derivative of u with the respect to x.
02:20
So now if we're to go ahead and apply this to our function, we'll find that f prime of x is going to be equal to x minus five over the absolute value of x minus five.
02:42
And we can simplify this actually.
02:48
So we'll have a piecewise function where we have f prime of x equal to 1 at when x is less than 5...