Find the rate of change, in bacteria per day, of N with respect to t when the following values are true. (Round your answers to the nearest tenth.)

Question

This is due tonight so please help!

N=600[1-3/(t^2+2^2]

when t=0, what does the bacteria per day equal?

when t=1 what does the bacteria per day equal?

when t =2 what does the bacteria per day equal?

when t=3 what does the bacteria per day equal?

when t=4 what does the bacteria per day equal?

Answer 1

assuming N(t)=600(1-3/(t²+2)²)

rate of change is the derivative dN/dt

here dN/dt=N'=7200t/(t²+2)³

t=0 N'=0

t=1 N'=7200/27

t=2 N'=14400/6³

...

it is kind of interesting

N starts at 150 and ends by approaching 600 after enough days

N' (the rate of change) is always positive, high on day1, and

near zero after enough days