LaddiusMaximus
  • LaddiusMaximus
estimate instantaneous rate of change at the point x=5 for f(x)=lnx
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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amistre64
  • amistre64
the derivative of lnx at x=5
amistre64
  • amistre64
1/5 maybe?
LaddiusMaximus
  • 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
  • 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
  • LaddiusMaximus
I know that but we havent gotten that far. We are only using rates of change.
amistre64
  • amistre64
instantaneous velocity is the derivatie; which can be defined as:\[\lim_{h->0}\frac{ln(x+h)-ln(x)}{h }\]
amistre64
  • amistre64
then use the avg stuff with small changes
amistre64
  • amistre64
ln(5)-ln(5.0001)/.0001
amistre64
  • amistre64
or something akin to it
LaddiusMaximus
  • 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
  • LaddiusMaximus
like 4.9 and 5.01
amistre64
  • amistre64
teh smaller the change, the more exact the estimate
LaddiusMaximus
  • LaddiusMaximus
(5)^3 +5-((4.9)^3 +4.9)/ 5-4.9
LaddiusMaximus
  • LaddiusMaximus
well it asks for estimates and all the ones I give are wrong
amistre64
  • amistre64
cubed stuff?
amistre64
  • amistre64
you said lnx
LaddiusMaximus
  • LaddiusMaximus
I end up with 68.75, 73.04,74.51,75.85
LaddiusMaximus
  • 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
  • amistre64
ln(5.00001/5)/.00001 = .2000...... approx 1/5 http://www.wolframalpha.com/input/?i=ln%285.00001%2F5%29%2F.00001
LaddiusMaximus
  • LaddiusMaximus
it says that the instantaneous rate of change is estimated by computing the average rate of smaller intervals
amistre64
  • amistre64
thats what i did
LaddiusMaximus
  • LaddiusMaximus
well what am I doing then? because im lost. I dont understand how to set up the problem.
amistre64
  • amistre64
ln(5.00001) is a small change from ln(5) right?
amistre64
  • amistre64
in fact is .00001 small of a change
LaddiusMaximus
  • LaddiusMaximus
ok i follow that. but what then? how do I set up the problem to solve it? so its [5, 5.001]?
amistre64
  • amistre64
yes
amistre64
  • amistre64
or 4.99999 to 5 but its easier to see the change when it grows a little bigger
LaddiusMaximus
  • LaddiusMaximus
how do i plug it in? i see a formula for average rate of change, but it doesnt give one for instantaneous
amistre64
  • amistre64
im not sure what it is youre looking at ....
amistre64
  • amistre64
if its about inputing into a program, i got no idea what it wants then
amistre64
  • amistre64
the instantaneous rate of change at ln(5) = 1/5 = .2
amistre64
  • amistre64
so how you got those other vlaues I cant say
LaddiusMaximus
  • LaddiusMaximus
im sorry. the book does a very crappy job of explaining things

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