00:01
All right, so we want to show the absolute value of f of x is going to be less than some constant times g of x, and in this case, that is just x for this particular problem.
00:18
Okay, and our f of x is going to be equal to x squared plus 1 over x plus 1.
00:33
So first we're just going to do some long division, right? so take a look at this in other form.
00:45
So we consider what does it look like when we divide x squared plus 1 by x plus 1.
00:57
Okay, we get an x on top.
01:00
We minus x squared minus x here.
01:08
We get a minus 1.
01:10
Get negative x plus 1.
01:18
Let's take x plus 1 here, and that gives us a 2.
01:23
Okay, so the quotient win is then x minus 1 with remainder 2.
01:31
And so we can rewrite this fraction as being x minus 1 plus 2 over x plus 1.
01:48
Okay, so let's just get to a new page with that new, new form.
01:55
So that again was we have our x minus 1 plus the remainder, which is 2 over x plus 1.
02:07
We always have the property that x minus 1 is less than x...