The population of a suburb grows at a rate proportional to the population. suppose the population doubles in size from 3000 to 6000 in a 6- month period and continues at the current rate of growth. find the particular solution to the differential equation with the initial condition p(0)=3000.

Step-by-step explanation:

The growth of the population can be modeled by the following differential equation:

Where r is the growth rate, P is the population, and t is the time measures in months.

I am going to solve the above differential equation with the separation of variables method.

Integrating both sides:

Where P(0) is the initial condition

We need to isolate P in this equation, so we do this

So

The problem states that P(0)=3000, so:

The problem wants us to find the value of r:

It states that the population doubles in size from 3000 to 6000 in a 6- month period, meaning that P(6) = 6000. So

To isolate 6r, we apply ln to both sides.

r = 0.1155

The particular solution to the differential equation with the initial condition P(0)=3000 is:

it is probably the graph on the bottom right.

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).