Question

what is the practical significance of the number 200 in the equation:
(200(1-.956e^(-.18t)))

Answer 1

the y values only make it to 200 but why?

Answer 2

\(\lim_{t\to\infty}200\left(1-.956e^{-.18x}\right)\) = 200

So the function can never go past 200

Answer 3

im confused still

Answer 4

how do you know that?

Answer 5

Sure let me explain


\(\lim_{t\to\infty}200\left(1-.956e^{-.18x}\right)\)
=\(lim_{t\to\infty}200\left(1-\frac{.956}{e^{.18t}}\right)\)
As t approaches infinity, the denominator with e^.18t gets larger, so the term is 1/∞, which = 0
\(lim_{t\to\infty}200(1-0)=200\)

Answer 6

why did you do step 2, I mean why did you take (1−.956e^−.18t) and do (1−(.956/e^.18t))?

Answer 7

Exponent rules...a negative power on a variable means that it's going in the denominator

\( x^{-n} = \frac{1}{x^n}\)

Answer 8

oh, didn't see that

Answer 9

one more thing

Answer 10

whats this last thing
As t approaches infinity, the denominator with e^.18t gets larger, so the term is 1/∞, which = 0
limt→∞200(1−0)=200

Answer 11

I get this

As t approaches infinity, the denominator with e^.18t gets larger,

Answer 12

So think of it this way.

1/∞ means you have fractions getting smaller and smaller..closer to 0. You have 1/2, then 1/4, then 1/200, then 1/10000000. It's approaching 0, right? So the LIMIT of 1/∞ is 0

Answer 13

ok

Answer 14

oh, cuz it gets infinitely smaller

Answer 15

The e^.18t is getting larger, meaning that the denominator is getting larger...towards infinity. It looks like 1/∞ now. So when we take the LIMIT of that, we're going to get 0.

Answer 16

okay, I understand now

Answer 17

you are a genius

Answer 18

Tyty I try :)