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

- LaddiusMaximus

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

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- amistre64

the derivative of lnx at x=5

- amistre64

1/5 maybe?

- LaddiusMaximus

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

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- amistre64

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

- LaddiusMaximus

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

- amistre64

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

- amistre64

then use the avg stuff with small changes

- amistre64

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

- amistre64

or something akin to it

- LaddiusMaximus

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

- LaddiusMaximus

like 4.9 and 5.01

- amistre64

teh smaller the change, the more exact the estimate

- LaddiusMaximus

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

- LaddiusMaximus

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

- amistre64

cubed stuff?

- amistre64

you said lnx

- LaddiusMaximus

I end up with 68.75, 73.04,74.51,75.85

- LaddiusMaximus

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

- amistre64

ln(5.00001/5)/.00001 = .2000...... approx 1/5
http://www.wolframalpha.com/input/?i=ln%285.00001%2F5%29%2F.00001

- LaddiusMaximus

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

- amistre64

thats what i did

- LaddiusMaximus

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

- amistre64

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

- amistre64

in fact is .00001 small of a change

- LaddiusMaximus

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

- amistre64

yes

- amistre64

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

- LaddiusMaximus

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

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

so how you got those other vlaues I cant say

- LaddiusMaximus

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

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