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

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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the derivative of lnx at x=5

LaddiusMaximus
 4 years ago
Best ResponseYou've already chosen the best response.0That is not what it is asking. Im supposed to use the average velocity formula to figure it out and its not working

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the 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
 4 years ago
Best ResponseYou've already chosen the best response.0I know that but we havent gotten that far. We are only using rates of change.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1instantaneous velocity is the derivatie; which can be defined as:\[\lim_{h>0}\frac{ln(x+h)ln(x)}{h }\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1then use the avg stuff with small changes

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1ln(5)ln(5.0001)/.0001

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1or something akin to it

LaddiusMaximus
 4 years ago
Best ResponseYou've already chosen the best response.0the 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
 4 years ago
Best ResponseYou've already chosen the best response.0like 4.9 and 5.01

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1teh smaller the change, the more exact the estimate

LaddiusMaximus
 4 years ago
Best ResponseYou've already chosen the best response.0(5)^3 +5((4.9)^3 +4.9)/ 54.9

LaddiusMaximus
 4 years ago
Best ResponseYou've already chosen the best response.0well it asks for estimates and all the ones I give are wrong

LaddiusMaximus
 4 years ago
Best ResponseYou've already chosen the best response.0I end up with 68.75, 73.04,74.51,75.85

LaddiusMaximus
 4 years ago
Best ResponseYou've already chosen the best response.0I dont know man. Ive been beating my head against the wall for the last two hours and the book is nigh on useless

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1ln(5.00001/5)/.00001 = .2000...... approx 1/5 http://www.wolframalpha.com/input/?i=ln%285.00001%2F5%29%2F.00001

LaddiusMaximus
 4 years ago
Best ResponseYou've already chosen the best response.0it says that the instantaneous rate of change is estimated by computing the average rate of smaller intervals

LaddiusMaximus
 4 years ago
Best ResponseYou've already chosen the best response.0well what am I doing then? because im lost. I dont understand how to set up the problem.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1ln(5.00001) is a small change from ln(5) right?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1in fact is .00001 small of a change

LaddiusMaximus
 4 years ago
Best ResponseYou've already chosen the best response.0ok i follow that. but what then? how do I set up the problem to solve it? so its [5, 5.001]?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1or 4.99999 to 5 but its easier to see the change when it grows a little bigger

LaddiusMaximus
 4 years ago
Best ResponseYou've already chosen the best response.0how do i plug it in? i see a formula for average rate of change, but it doesnt give one for instantaneous

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1im not sure what it is youre looking at ....

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1if its about inputing into a program, i got no idea what it wants then

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the instantaneous rate of change at ln(5) = 1/5 = .2

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1so how you got those other vlaues I cant say

LaddiusMaximus
 4 years ago
Best ResponseYou've already chosen the best response.0im sorry. the book does a very crappy job of explaining things
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