Posted by Sarah on Sunday, April 11, 2010 at 2:56pm.
If P'(x)=xe^(-x^2) , then
P(x) = (-1/2) e^(-x^2) + c
when t= 3 , P(3) = 10000
10000= (-1/2)e^(-9) + c
10000 = -.00006 + c
c = 10000.0000617
1. P(x) = (-1/2) e^(-x^2) + 10000.0000617
2. "in the long run" implies x --->∞
which means (-1/2) e^(-x^2) ---> 0
so P(∞) = 10000.0000617
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