00:03
In this video, we are going to look at how to locate the absolute min and absolute max of the function ff x equals 2 over 2 minus x on the interval that includes the n .0 up in 2 but not including 2, if they exist.
00:17
Now recall, if we have a continuous function on a closed interval, then we are guaranteed that the absolute men and absolute max of that function both exist.
00:28
But here i have a function on a half open interval.
00:33
And there are values of x.
00:35
And specifically in this one, the function is undefined at 2.
00:40
But again, my interval doesn't include 2.
00:44
So i'm not guaranteed that the absolute min and absolute max exist, but they could.
00:49
So i still want to approach this problem looking for them in the manner that we did when they were guaranteed.
00:56
So first we want to find the critical values of f of x equals 2 over 2 minus x.
01:02
Now i can rewrite this function f of x as 2 times a quantity 2 minus x to the negative 1 power to take its derivative.
01:12
Certainly i could use a quotient rule but i could also bring up the denominator in our parentheses to the negative 1 power and use the chain rule.
01:23
So now my derivative f prime of x will the derivative of a constant times of function is the constant times the derivative of the function and the derivative of a letter base expression to a number power is the power rule so i'm going to bring down the negative one keep the base of 2 minus x subtract 1 from the exponent so negative 1 minus 1 is negative 2 and multiply by the derivative of the variable expression base so the derivative of 2 minus x is negative 1.
01:57
Rewriting that, i get f prime of x is equal to, we'll think of this as over 1.
02:03
This factor of 2 minus x to the negative 2 power goes in the denominator as 2 minus x now to the second power.
02:12
When you move it, that factor to the other part of the fraction, the sign changes on the exponent.
02:18
And then in the numerator, i have 2 times negative 1 times negative 1, so i get a positive 2.
02:24
Now, this derivative is only undefined when x is 2, but the function itself was undefined there.
02:32
So it's not a critical value plus the fact that's not in the interval.
02:36
Remember, critical values are values in the domain of the function for which the first derivative is zero or undefined.
02:44
And specifically, we're talking about the domain from zero up and two, but not including two.
02:50
So i don't have any values that are critical values from the derivative being undefined with those restrictions.
02:59
And also, my derivative cannot be zero.
03:02
If i set zero equal to two over two minus x quantity squared, the fraction is only zero when the numerator is zero and the denominator is not.
03:12
The numerator is two.
03:13
There's no values of x that would make two equal to zero.
03:16
So that doesn't give me any values.
03:19
So i don't have any critical values for this function in that interval, so there's no values to evaluate my function at.
03:29
Now remember our function is 2 over 2 minus x...