00:01
Consider the function f of x equal to 3 plus 2 over x plus 1 over x squared.
00:12
We want to discuss the domain of the function, find its horizontal and vertical asymptotes, intervals of increase or decrease, extreme points, concavity, and inflection points.
00:23
And all this information can be determined by either analyzing f, f prime of x, and f prime prime of x.
00:33
So let's go ahead and write away and calculate these first and second derivatives.
00:37
F prime of x will give us minus 2 over x squared minus 2 over x cubed.
00:51
Where it's easier to evaluate these derivatives by writing 2 over x as 2 times x power minus 1 and 1 over x as x power minus 2.
01:05
Let's bring this to common denominator and we find minus 2 times x plus 1 over x cubed.
01:15
Let's take this expression for f prime and differentiate again.
01:20
And we find 4 over x cubed plus 6 over x power 4, which simplifies to 4x plus 6 over x to the power of 4.
01:46
And the reason why i'm writing it like this in common denominator is because to find the critical points and inflection points, we want to solve for when f prime of x and f prime prime of x is equal to 0.
02:04
So now that we have f prime of x and f prime prime of x, let's analyze our two functions here.
02:10
Let's start with f of x to discuss the domain and horizontal vertical asymptotes.
02:16
We simply need to look where f of x is defined.
02:19
So we see that f of x is defined for all values of x.
02:32
Again, it might be easier to write in common denominator 3x squared plus 2x plus 1 over x squared.
02:42
And we see that this function will be defined everywhere for all values of x as long as x is not equal to 0.
02:52
So here i'm removing the 0 point.
02:54
Why? because when x is equal to 0, we have a division by 0, which is a vertical asymptote.
03:19
And there's no indications for a...
03:31
Excuse me, that's not true.
03:33
We do have a horizontal asymptote.
03:35
And to find this, we need to evaluate the limit for when x approaches to plus or minus infinity.
03:42
So when x approaches to plus infinity, f of x is equal to 3.
03:49
And similarly, when x approaches to minus infinity, f of x is equal to 3.
03:58
And this corresponds to a horizontal asymptote.
04:07
So we have a vertical asymptote at x equal to 3.
04:11
We have two horizontal asymptotes at y equal to 3 when x approaches to plus and minus infinity.
04:21
Let's keep this in mind when sketching the graph.
04:23
Now let's look for critical points.
04:31
So critical points correspond to values of x for which f prime of x is equal to 0...