anonymous
  • anonymous
compute the difference quotient of the given function; A(t)=2t/ 3-t
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
/\y difference quotient = ------- right? /\x
anonymous
  • anonymous
what is that????
amistre64
  • amistre64
the change in "y" that is the difference quotient; ------------- the change in "x"

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
2t ---- is your equation right? 3-t
amistre64
  • amistre64
when we divide everything by"t" we end up with -2....not that I know what thats sposed to help :)
anonymous
  • anonymous
gr8
anonymous
  • anonymous
but you can't divide evrything by t.....that's false maths
amistre64
  • amistre64
yes you can... multiply by 1/t//1/t it equals 1
amistre64
  • amistre64
2t*(1/t) = 2 --------- ------- (3-t)(1/t) = 3/t - t/t
amistre64
  • amistre64
they are equal equations
amistre64
  • amistre64
as t-> infinity; we get 2/-1 = -2
amistre64
  • amistre64
but how that helps out, I dont know :)
amistre64
  • amistre64
perhaps you problem here is that you want to find: A(t+h) - A(t) ---------- is that right? h
anonymous
  • anonymous
amistres.....exactly....u are a genius....pliz continue
amistre64
  • amistre64
lol.... im a genius when compared to fish :) ok; we just fill in our equation with this new jargon...
amistre64
  • amistre64
[2(t+h)/ 3-(t+h)] - [2t/ 3-t] //h ; the "//h" just means all on top of h
amistre64
  • amistre64
2t+2h 2t ------ - ---- //h 3-t+h 3-t lets get like demoninators and add away
amistre64
  • amistre64
(3-t)(2t+2h) - (3-t-h) 2t ---------------------- //h (3-t+h)(3-t)
amistre64
  • amistre64
gotta adjust for my own stupidity here... (3-t-h) is what it should have been... 3t +6h -2t^2 -2th - [6t -2t^2 -2th] ------------------------------ //h (3-t-h)(3-t)
amistre64
  • amistre64
3t +6h -6t ---------- //h (3-t-h)(3-t)
amistre64
  • amistre64
-3t +6h ---------- //h multiply top and bottom by 1/h (3-t-h)(3-t) -3t +6h ------------- (3-t-h)(3-t)(h) we good so far?
anonymous
  • anonymous
i kind of gat lost how does (3-t)(1/t) =1
amistre64
  • amistre64
when we get a variable stuck on the bottom, then the limit goes to 0 as that variable gets really big: for example; 1 --------------------- = .000000000...000000001 10000000...00000000 which is very very tiny
amistre64
  • amistre64
3/t goes to zero; and -t/t = -1
anonymous
  • anonymous
oooh ok didn know that
amistre64
  • amistre64
I gotta recheck something, I think I forgot how to add :)
amistre64
  • amistre64
right here....... 6t +6h -2t^2 -2th - [6t -2t^2 -2th] ------------------------------ //h (3-t-h)(3-t) 6h ---------- //h ; thats better, told ya I forgot how to add :) (3-t-h)(3-t)
amistre64
  • amistre64
6h ---------- //h ; now we multiply (1/h) top and bottom (3-t-h)(3-t) 6h ------------- //(h/h) (3-t-h)(3-t)(h) ---------------------- 6 ------------- thats our answer (3-t-h)(3-t)
amistre64
  • amistre64
when h=0; we get 6/(3-t)^2
amistre64
  • amistre64
A = 2t/ 3-t A' = (3-t)(2) - (2t)(-1)//(3-t)^2 A' = 6-2t +2t // (3-t)^2 A' = 6/(3-t)^2 ....see much quicker :)
anonymous
  • anonymous
wow thanx alot

Looking for something else?

Not the answer you are looking for? Search for more explanations.