Application of differentiation

[huan.conchito]

Given a function f(t) = (2t-1) / (t^2-t+0.5)
max supported species = 120, and t is the time.
if 75 animals are introduced into te refuge the population changes at a rate of f(t)
What needs to be done to obtain
a ) Find when the population will be a minimum
and to find
b) When the rate of change of the population will be a maximum

 

[dextercioby]

Theoretically

a)f'(x)=0,f''(x)>0
b)f''(x)=0,f'''(x)<0...

Daniel.

 

[huan.conchito]

i still don't get it

 

[Data]

Are you sure you don't mean $f(t)$?

I'll assume that you do. In that case, what conditions are necessary for a differentiable function to have a minimum at a point? Look at your notes if you don't remember.

 

[dextercioby]

No need to look at the notes,i wrote them,before he's edited his post & spell out the function...

Daniel.

 

[huan.conchito]

yup is f(t)

 

[dextercioby]

Then solve it...Set those derivatives to zero and verify the nature of the critical points.

Daniel.

 

[huan.conchito]

i found the minimum at t= 1/2
how do i go about part b?
for part b i got
f''(t)= 2(2t-1)
2(2t-1)=120 -> t =30.5
is that the time of the max population?

 

[HallsofIvy]

Given a function f(x) = (2t-1) / (t^2-t+0.5)

Okay we have a function.

max supported species = 120, 75 are initially introduced.
and t is the time.

What needs to be done to obtain
a ) Find when the population will be a minimum
and to find
b) When the rate of change of the population will be a maximum

?? What happened to f? what does f have to do with the population?
Have you left something out- like f(t) is the population at time t? But I notice that f(0)= -2, not 75 and the maximum of f is 2, not 120. What does f have to do with the population?

 

[huan.conchito]

if 75 animals are introduced into te refuge the population changes at a rate of f(t)