SOLUTION: Y(t) is below
10e^(0.05t), 0<=t<=10
(600)/(20+10e^(-0.5(t-12))), t>12
f(t) is 10 is less than t which is less than or equal to 12
The amount of bacteria in a petri di
Algebra ->
Rational-functions
-> SOLUTION: Y(t) is below
10e^(0.05t), 0<=t<=10
(600)/(20+10e^(-0.5(t-12))), t>12
f(t) is 10 is less than t which is less than or equal to 12
The amount of bacteria in a petri di
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(600)/(20+10e^(-0.5(t-12))), t>12
f(t) is 10 is less than t which is less than or equal to 12
The amount of bacteria in a petri dish, measured in milligrams, at time t hours is modeled by the function G defined above where f is a continuous function such that f(12)=20
i) Find the limit of Y(t) as t approaches positive infinity. Explain the meaning of the the limit of Y(t) as t approaches positive infinity in the context of the problem.
ii) Is Y continuous at t=12? Please justify the answer.
iii) Y is continuous at t=10. Is there a t for 0<=t<=12, at which Y(t)=18? Please Justify answer
