SOLUTION: My question is 2 parts i solved the first part easy but the wecond part is got me stuck The growth of a population of bacteria can be bodelled using the function P=2xe^-x where

Question 5130: My question is 2 parts i solved the first part easy but the wecond part is got me stuck
The growth of a population of bacteria can be bodelled using the function P=2xe^-x where x is measured in hours and p is in millions
a) find the maximum population and when it occurs - i got 735,758 after 1 hour i think this is right
b) verify your value by finding the derivative and evaluating the maximum value - im not too sure how to do this and i cant figure out the derivative
if u could help me that would be sooo good thanks
Answer by khwang(438) (Show Source): You can put this solution on YOUR website!
[quote] P= 2xe^-x where x is measured in hours and p is in millions
a) find the maximum population and when it occurs - i got 735,758 after 1 hour [/quote]

dP/dx = P' = 2e^-x - 2xe^-x = 2e^-x(1-x)
Since,e^-x > 0, we have dP/dx = 0 -->x =1.
Note P" = -2e^-x(1-x) -2e^-x = 2e^-x(x-3), so P"(1) < 0.
By the 2nd derivative test,we see that P has relative max when x = 1.
And P(1) =0.735759377 *1 million (I really don't care about it)
Or you use the 1st derivative test by testing P'(1+) < 0 and P'(1-) > 0.

Kenny

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