00:02
Here we have a f of x is given to us at 3 times ln of x.
00:15
To find the derivative of f of x, so f prime of x, that's negative that's 3 over x.
00:30
So the second derivative f double prime, that's negative 3 over x squared.
00:45
The third derivative, f triple prime of x, that's 6 over x cubed and the fourth derivative that will be negative 18 over x to the fourth and the intervals are one is less not equal to x and x is less not equal to two for part b we have the derivative of the fourth derivative with this interval in mind then we can calculate for a you know the absolute eight the absolute value of negative 18 over x to the fourth has to be less than or equal to a.
02:12
Now you can solve for a, which is substitute 1 for our x into our this equation, we have the absolute value of negative 18.
02:30
It's less than not equal to a.
02:34
So that means a is equal to 18.
02:44
For part c, given our boundaries, using the midpoint rule, n is equal to two.
02:58
If you calculate the midpoint, we have a times b minus a squared.
03:05
I mean a to the fifth, divided by 2880 times two to the fourth...