P' = 50(2x e^-x^2 - e^-x)
its rate of change is a maximum where its derivative is zero
P" = 50((-4x^2+2) e^-x^2 + e^-x) = 0
x = 0.84843
The population of a colony of bacteria is modeled by the function P(x) = 50(e^-x -e^-x^2)+10, for x is greater than or equal to 0, 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 enviorment. At which time below is the rate of population growth the greatest?
A) x=1.2
B) x=2.2
c) x=3
D) x=.2
E) x=0.8
2 answers
dP/dx = 50 [ -e^-x - (e^-x^2) * -2x) ] = 50 [ -e^-x + 2 x e^-x^2 ]
zero when
e^-x = 2 x e^-x^2
1/2 = x e^(x-x^2)
try x = 3
3 e^(-6) way too small , needs to be 0.5
try x = 0.8
.8 e^(0.16 )= 9.4 , too big
try 1.2
etc
zero when
e^-x = 2 x e^-x^2
1/2 = x e^(x-x^2)
try x = 3
3 e^(-6) way too small , needs to be 0.5
try x = 0.8
.8 e^(0.16 )= 9.4 , too big
try 1.2
etc
2 answers