00:01
Okay, we want to find all about this function, f of x equals 1 plus 5 over x minus 8 over x squared.
00:09
Okay, so first, vertical asthmatote, set the bottoms equal to 0, and you get x equals 0.
00:17
Is the vertical asthmot? horizontal asymptote.
00:21
Well, i think before i do this, i'm going to get a common denominator so i can add these together.
00:26
And i get 8x or x squared plus 5x minus 8 over x squared.
00:34
Okay, because this one i multiply by x squared over x squared and this one x over x.
00:41
Okay, so to find the horizontal last one, you take the limit as x goes to infinity, x squared plus 5x minus 8 over x squared.
00:54
So i'm going to divide everything by the highest power of x, which is x squared.
01:01
1 plus 5 over x minus 8 over x squared over 1.
01:08
So if x goes to infinity, that goes to 0, and that goes to 0, and we get the limit is equal to 1.
01:14
So y equals 1 is the horizontal asymptote.
01:18
All right, now we're going to find this first derivative so that we can find out where it's increasing and decreasing in max and mint.
01:25
So we need to decide would it be easier to take the derivative the way it was to start with, or would it be easier to take it? it in its new form.
01:35
Okay, well in its new form, i'm going to have to use the quotient rule.
01:37
So i'm going to go back to the original one.
01:40
So now i'm going to write it like this.
01:41
One plus five x to the minus one minus eight x to the minus two.
01:48
Okay, so f prime would be minus five x to the minus two plus 16 x to the minus three or minus five over x squared plus 16 over x cubed.
02:02
And now multiply top and bottom of the first one by x.
02:06
So you get 16 minus 5x over x cubed.
02:12
All right, critical values, where it's equal to zero, which would be 16 equals 5x or x is 16 fifths.
02:27
And where it's undefined.
02:29
Well, it's undefined at x equals zero, but that's a vertical asymptote.
02:33
So that can't be a critical value.
02:35
All right, now we want to find where it's increasing and decreasing, so we have to put the critical value on there and the vertical asymptote.
02:45
So here will be zero, and then here will be 16 fifths, and then we're checking the derivative, which is 16 minus 5x over x cubed.
02:57
Pick a number bigger than 16 fifths, 100.
03:01
16 minus 500, that's negative, over 100 cubed, that's positive.
03:08
So a negative divided by a positive, that's negative, so that is decreasing.
03:14
Pick a number between 0 and 16 .5s.
03:16
Well, 16 5ths is 3 something, so i'm going to pick 1.
03:21
16 minus 5, that's positive.
03:23
1 cube, that's positive.
03:25
Positive times positive is positive, so that's increasing.
03:30
Pick a number less than 0, like negative 10.
03:34
So 16 plus 50, that's positive.
03:37
Negative 10 cube, that's negative.
03:39
So that's decreasing.
03:42
So it's increasing from 0 to 16 fifths.
03:47
It's decreasing minus infinity to 0 and 16 fifths to infinity.
03:55
Okay, it has a maximum here at x equals 16 fifths, okay, because it goes up and then goes down.
04:06
So let's see what the y value is there...