1.) values of f(t) are given in the following table:
t 0 2 4 6 8 10
f(t) 137 112 88 68 49 34
a. Does this function appear to have a positive or negative first derivative? Second derivative?
* I think both derivatives are negative
b. Estimate (f prime) f^' (2) and f^' (8)
Please show work for part b so I can understand. I'm really stuck.
i just saw the table is messed up so the values are:
(0, 137) (2, 112) (4, 88) (6, 68) (8, 49) (10, 34)
it is gettin smaller each increment, first derivative is negative.
Now, the rate of decrease is 25,24,20,19,15 getting smaller, so the second derivative is positive.
oh ok. but wut about f^'(2). how do i find that from the table?
0 137
====== -25
2 112 ****** +1
====== -24
4 88 ******* +4
======= -20
6 68 ******* +1
======= -19
8 49 ********* +4
======== -15
10 34
so
dy/dx is negative
but the change of dy/dx with x is positive right down the table
Between x = 0 and x = 2, y changes -25
dy/dx = -25/2 = -12.5
between x = 2 and x = 4, y changes -24
dy/dx = -24/2 = -12
so averaging to get an instantaneous dy/dx at x = 2
I would say dy/dx = -12.25 at x = 2
go through the same routine at x = 8
I'm getting confused on x= 8
would you do between x=0 and x= 8, the change is
-88 and then divide that by 8 = -11
between x= 8 and x= 10, y changes -15 and then divide that by 8 to get -1.875 but then what?
always divide by the change in x which is always 2 in your table
To determine whether a function has a positive or negative first derivative, we need to examine the trend of the function values. Likewise, to determine the sign of the second derivative, we need to look at the trend of the first derivative.
Let's start with part a:
To determine whether the first derivative is positive or negative, we compare the function values at different points. In this case, we can see that as t increases, the corresponding f(t) values decrease. This indicates a negative slope, meaning the function has a negative first derivative.
To determine whether the second derivative is positive or negative, we need to observe the trend in the first derivative values. Since we do not have the exact values of the first derivative, we can estimate the trend by comparing the slopes of the tangent lines connecting each pair of points.
Using the given table, we can calculate the average rate of change between adjacent points to approximate the first derivative (slope) at those points:
For t = 0 to t = 2:
(f(2) - f(0)) / (2 - 0) = (112 - 137) / (2 - 0) = -25 / 2 = -12.5
For t = 2 to t = 4:
(f(4) - f(2)) / (4 - 2) = (88 - 112) / (4 - 2) = -24 / 2 = -12
For t = 4 to t = 6:
(f(6) - f(4)) / (6 - 4) = (68 - 88) / (6 - 4) = -20 / 2 = -10
For t = 6 to t = 8:
(f(8) - f(6)) / (8 - 6) = (49 - 68) / (8 - 6) = -19 / 2 = -9.5
For t = 8 to t = 10:
(f(10) - f(8)) / (10 - 8) = (34 - 49) / (10 - 8) = -15 / 2 = -7.5
From these approximations, we can see that the slopes between successive points are all negative. This suggests that the first derivative is decreasing, implying a negative second derivative.
Moving on to part b:
To estimate f'(2), we can use the average rate of change between the two closest points, t = 0 and t = 2:
(f(2) - f(0)) / (2 - 0) = (112 - 137) / (2 - 0) = -25 / 2 = -12.5
Therefore, the estimated value of f'(2) is -12.5.
To estimate f'(8), we can use the average rate of change between the two closest points, t = 6 and t = 8:
(f(8) - f(6)) / (8 - 6) = (49 - 68) / (8 - 6) = -19 / 2 = -9.5
Therefore, the estimated value of f'(8) is -9.5.
Note that these are only estimates based on the given data points, and the actual values may differ.