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
Estimate (f prime) f^' (2) and f^' (8)
Please show work so I can understand. I'm really stuck.
if the table comes out messed up, the values are:
(0, 137) (2, 112) (4, 88) (6, 68) (8, 49) (10, 34)
Check what I did for you a minute ago below.
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
Does that help. It is really, really important.
My columns are "change in y" and "change in change of y" for each interval of 2 in x
yeah for 2 but when I tried that for x = 8, I have a huge gap between the two numbers for 8. I got between x=0 and x =8, it is -88. then dividing that by 8, i got -11. then when i did between x= 8 and x =10, i got -55. then dividing that by 8, i got -1.875 and now I'm stuck
it is change in y / change in x
Your change in x is ALWAYS 2 in the table you gave me, not 8
from 6 to 8 dy = -19
so dy/dx = -19/2 = -9.5
from 8 to 10 dy = -15
so dy/dx = -7.5
average at x = 8 is -8.5
so then for x = 8, u do the exact same steps by dividing the numbers by 2?
for x = 8 do u use between x =6 and x=8 and between x=8 and x=10?
Remember dy/dx is the slope at that point, like at x = 8 here
find the slope between x = 6 and x = 8
find the slope between x = 8 and x = 10
average them
sketch what you are doing on a rough graph