Ap test prep: If dP/dt = 3.1(P-2)(P-10), P>0, for which P-values is pop. increasing?

[Snabe]

Ap test prep: If dP/dt = 3.1(P-2)(P-10), P>0, for which P-values is pop. increasing?

I need help with the problem;

If the rate of change of a population is given by

dP/dt = 3.1(P-2)(P-10), P>0,

for what values of P is the population increasing?

A) all P>o

B) only when 0<P<2

C) 2<P<10

D) 0<P<2 and 10<P

E) all P>0 expect P=2 and P=10

 

[Deleted member 4993]

I need help with the problem;

If the rate of change of a population is given by

dP/dt = 3.1(P-2)(P-10), P>0,

for what values of P is the population increasing?

A) all P>o

B) only when 0<P<2

C) 2<P<10

D) 0<P<2 and 10<P

E) all P>0 expect P=2 and P=10

You start with:

dP/dt = 3.1(P-2)(P-10)

dP/[P-2)(P-10)] = 3.1 dt

Now use partial fraction on the left-hand-side to convert it to an easily-integrable form.

continue...

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

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[HallsofIvy]

If the problem is just to determine "for what values of P is the population increasing?" then you don't need to solve the equation so don't need to integrate.

A differentiable function is increasing if and only if its derivative is positive.

You are given that dP/dt = 3.1(P-2)(P-10) so the function is increasing if and only if dP/dt = 3.1(P-2)(P-10)> 0. A product of numbers is positive if and only the number of negative factors is even. Since 3.1 is positive that means that this number is positive if and only if P- 2 and P-10 are both positive (so 0 negative factors) or if P- 2 and P- 10 are both negative (so 2 negative factors).

You need to solve the inequalities "P- 2< 0 and P- 10< 0" or "P- 2> 0 and P- 10> 0".