The population of a colony of bacteria is modeled by the function p(x)=50(e^-x - e^-x^2)+10 ,for 0 ≤ x, where population P is in thousands, x is in hours, and x = 0 corresponds to the moment of introduction of a certain chemical into the colony's environment. At which time(s) is the bacteria population growing at an instantaneous rate of 0, going from positive growth just prior, to negative growth right after?
3 answers
arg
well, that would be when p' = 0. So,
p' = 50(-e^-x + 2xe^-x^2)
You can use your favorite graphing utility or numerical method to find where p'(x) = 0.
The graph of p(x) is at
http://www.wolframalpha.com/input/?i=50%28e^-x+-+e^-x^2%29%2B10+for+0%3C%3Dx%3C%3D2
If you erase the domain from the input box, it will show the approximate values of the local min and max.
p' = 50(-e^-x + 2xe^-x^2)
You can use your favorite graphing utility or numerical method to find where p'(x) = 0.
The graph of p(x) is at
http://www.wolframalpha.com/input/?i=50%28e^-x+-+e^-x^2%29%2B10+for+0%3C%3Dx%3C%3D2
If you erase the domain from the input box, it will show the approximate values of the local min and max.
1.7
2 answers