00:01
Okay, we're going to go ahead and look at this equation right here.
00:05
X raised to the power of x equal to two.
00:09
And what we're going to be doing is using the intermediate value theorem to determine if this equation has a solution.
00:19
And i always like to rewrite the equation to set it equal to zero.
00:26
So that's what i'm going to go ahead and do.
00:28
For me it's just a little bit easy to work with you don't have to and one of the stipulations of the intermediate value theorem is that the function needs to be continuous over our region of interest and so i'm going to be looking for our domain restriction or appoint it where it's not continuous or discontinuous and so the function is continuous for all x values except x equal to zero and so let's talk about that and so i know that this function right here that is x raised to the power of x minus two is made up of two functions it's made up of x raised to the power of x and it's made up of the horizontal line two and then while we're doing is doing subtraction on it.
01:34
Well, of course, two, the horizontal line is continuous for all x values.
01:38
Where i'm getting my discontinuity point is for this x raised to the power of x.
01:44
And so if i have zero raised to the power of zero, that actually in calculus is an indeterminate form.
01:52
And so there is going to pose that discontinuity value.
01:55
So there's my exception.
01:57
And so now what i want to do is to evaluate my function at two x values on a closed region that cannot include x equal to zero.
02:14
And so what i'm looking for is that when i evaluate the function at those two x values, i get a y value that's positive and a y value that's negative.
02:24
So we're going to have to kind of be creative and logical about how we pick these...