Question

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)

Answer 1

Check what I did for you a minute ago below.

Answer 2

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

Answer 3

Does that help. It is really, really important.

Answer 4

My columns are "change in y" and "change in change of y" for each interval of 2 in x

Answer 5

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

Answer 6

it is change in y / change in x

Your change in x is ALWAYS 2 in the table you gave me, not 8

Answer 7

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

Answer 8

so then for x = 8, u do the exact same steps by dividing the numbers by 2?

Answer 9

for x = 8 do u use between x =6 and x=8 and between x=8 and x=10?

Answer 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

Answer 11

calculus. please helppp!! - JJ, Wednesday, February 22, 2012 at 9:08pm

so then for x = 8, u do the exact same steps by dividing the numbers by 2?

calculus. please helppp!! - JJ, Wednesday, February 22, 2012 at 9:11pm

for x = 8 do u use between x =6 and x=8 and between x=8 and x=10?

Answer 12

i tried the for x = 2, -12.25
and for x =8, -8.5 and the website says it's wrong. it doesn't say which one is wrong. should the values be negative or positive?

Answer 13

I think you have it now. I want to make sure because if anything is important in this course, this is it.

Answer 14

its says the answers are wrong. doesn't specify which one. so I'm not sure if the actual answer is wrong or if its the negative sign.

Answer 15

They are both negative.
But your text may not be averaging as I did. They might be just taking one side (a poorer approximation)
so try
-12.5
and
-9.5

Answer 16

we could even get fancier by using the second derivative but I doubt if your text wants you to do that.

Answer 17

the values changed. now its table is: (0, 14) (2, 38)
(4, 60), (6, 77) (8, 93) (10, 104)
so i said the first derivative is positive, second is negative.
f^' (2) ---> between x=0 and x=2, it's 12
between x = 2 and x = 4, it's 11
average = 11.5
f^' (8) ----> between x = 6 and x = 8, it's 8
between x = 8 and x = 10, it's 5.5
average = 6.75

is that right?

Answer 18

agree + then -
agree dy/dx = 11.5 at x = 2

Answer 19

agree 6.75

Answer 20

urghh it says that's not right!

Answer 21

is there a way to use difference quotients for it?

Answer 22

probably can but forget method, hang on

Answer 23

K, using difference quotient you would just use the right side instead of averaging

Answer 24

so (2, 38) to (4,60)
would be
22/2 = 11 so use 11

Answer 25

and
(8,93) to (10,104)
would be
11/2 = 5.5

That is a much rougher approximation than we did but is
[f(x+h)-f(x)]/h

Answer 26

I will find you some online software for that

Answer 27

http://calculator.tutorvista.com/math/466/difference-quotient-calculator.html

Answer 28

that worked. thank u so much!!

Answer 29

Good, good luck !