The rate of change of a function is the first derivative of the function.

If p(t) is the population function, then `p'(t)=P(t)=10000e^(- t /4)`

To find the population function we integrate P(t) with respect to t:

`p(t)=int10000e^(-t/4)dt`

`=10000int e^(-t/4)dt` Let `u=-t/4,du=-1/4dt` then

`=-40000inte^u du`

`=-40000e^u+C_1`

`=-40000e^(-t/4)+C`

Now p(0)=80000 so `80000=-40000e^0+C==>C=120000`

...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

The rate of change of a function is the first derivative of the function.

If p(t) is the population function, then `p'(t)=P(t)=10000e^(- t /4)`

To find the population function we integrate P(t) with respect to t:

`p(t)=int10000e^(-t/4)dt`

`=10000int e^(-t/4)dt` Let `u=-t/4,du=-1/4dt` then

`=-40000inte^u du`

`=-40000e^u+C_1`

`=-40000e^(-t/4)+C`

Now p(0)=80000 so `80000=-40000e^0+C==>C=120000`

Then `p(t)=-40000e^(-t/4)+120000`

We are asked to find the limit of p(t) as t goes to infinity:

`lim_(t->oo)(-40000e^(-t/4)+120000)`

`=lim_(t->oo)-40000/e^(t/4)+120000`

`=120000`

**The population approaches 120000 as t goes to infinity.**