00:01
So we need to find the inverse's derivative at this x value of 3.
00:09
So there's a lot of things that are important in this problem.
00:12
First of all, you want to make sure that this function is one to one.
00:16
And i appreciate that they restrict our bounds to just be x is greater than 1.
00:26
Because when i find the derivative of this, you would take the derivative of the top, which is one, leave x minus 1 alone minus the derivative of the bottom, which is one, leave the top alone all over the denominator squared.
00:42
So as i distribute into this problem, you have an x minus an x, so those things are canceling out, but then you have negative 1 minus 3, which is negative 4.
00:55
I guess what's important to this is the denominator is always positive.
01:02
So what you have is a negative.
01:03
Let me just put arrows here, divided by a positive.
01:07
Because once you square a number, you'll get a positive, and they exclude one because it's not even a value.
01:15
So you always get a number less than zero...