Question

estimate instantaneous rate of change at the point x=5 for f(x)=lnx

Answer 1

the derivative of lnx at x=5

Answer 2

1/5 maybe?

Answer 3

That is not what it is asking. Im supposed to use the average velocity formula to figure it out and its not working

Answer 4

the average velocity is for an average rate of change; your spose to be using the limit of that formula as h goes to 0 or some other denominator

Answer 5

I know that but we havent gotten that far. We are only using rates of change.

Answer 6

instantaneous velocity is the derivatie; which can be defined as:\[\lim_{h->0}\frac{ln(x+h)-ln(x)}{h }\]

Answer 7

then use the avg stuff with small changes

Answer 8

ln(5)-ln(5.0001)/.0001

Answer 9

or something akin to it

Answer 10

the way the book and cramster teaches it is way different.
im supposed to use numbers a little more and a little less than five

Answer 11

like 4.9 and 5.01

Answer 12

teh smaller the change, the more exact the estimate

Answer 13

(5)^3 +5-((4.9)^3 +4.9)/ 5-4.9

Answer 14

well it asks for estimates and all the ones I give are wrong

Answer 15

cubed stuff?

Answer 16

you said lnx

Answer 17

I end up with 68.75, 73.04,74.51,75.85

Answer 18

I dont know man. Ive been beating my head against the wall for the last two hours and the book is nigh on useless

Answer 19

ln(5.00001/5)/.00001 = .2000...... approx 1/5

http://www.wolframalpha.com/input/?i=ln%285.00001%2F5%29%2F.00001

Answer 20

it says that the instantaneous rate of change is estimated by computing the average rate of smaller intervals

Answer 21

thats what i did

Answer 22

well what am I doing then? because im lost. I dont understand how to set up the problem.

Answer 23

ln(5.00001) is a small change from ln(5) right?

Answer 24

in fact is .00001 small of a change

Answer 25

ok i follow that. but what then? how do I set up the problem to solve it? so its [5, 5.001]?

Answer 26

yes

Answer 27

or 4.99999 to 5 but its easier to see the change when it grows a little bigger

Answer 28

how do i plug it in? i see a formula for average rate of change, but it doesnt give one for instantaneous

Answer 29

im not sure what it is youre looking at ....

Answer 30

if its about inputing into a program, i got no idea what it wants then

Answer 31

the instantaneous rate of change at ln(5) = 1/5 = .2

Answer 32

so how you got those other vlaues I cant say

Answer 33

im sorry. the book does a very crappy job of explaining things