the population of a town starts at 1,000 and doubles every decade. the graph below shows the towns…

the population of a town starts at 1,000 and doubles every decade. the graph below shows the towns population growth modeled by the function d = log₂ p, where p is the population in hundred - thousands and d is the time in decades. how do you compare the approximate rate of change of the function over the interval 400 < p < 800 with that of the interval 800 < p < 1600? what do these rates of change mean in the context of the problem? which of the statements below support the answers to the questions above?

the population of a town starts at 1,000 and doubles every decade. the graph below shows the towns population growth modeled by the function d = log₂ p, where p is the population in hundred - thousands and d is the time in decades. how do you compare the approximate rate of change of the function over the interval 400 < p < 800 with that of the interval 800 < p < 1600? what do these rates of change mean in the context of the problem? which of the statements below support the answers to the questions above?

Answer

Answer:

  1. Comparison of rates of change: The approximate rate of change of the function (d = \log_2P) over the interval (400<P<800) is greater than the rate of change over the interval (800 < P<1600).
  2. Interpretation in context: In the context of the problem, the rate of change represents the number of decades it takes for the population (in hundred - thousands) to increase by a certain amount. A higher rate of change over (400 < P<800) means that it takes fewer decades for the population to increase from 400,000 to 800,000 compared to the time it takes to increase from 800,000 to 1,600,000.
  3. Supporting statements (not provided in the problem - but in general): Since the function (y=\log_2x) is a concave - down function (its second derivative is negative), the slope of the secant line (which represents the average rate of change) between two points on the curve decreases as the (x) - values (in this case (P)) increase.

Explanation:

Step1: Recall the formula for average rate of change

The average rate of change of a function (y = f(x)) over the interval ([a,b]) is (\frac{f(b)-f(a)}{b - a}). For the function (d=\log_2P), the average rate of change over ([a,b]) is (\frac{\log_2b-\log_2a}{b - a}).

Step2: Analyze the function (y = \log_2x)

The function (y=\log_2x) has the property that its derivative (y'=\frac{1}{x\ln 2}). As (x) (or (P) in our case) increases, the value of the derivative decreases. This means the function is concave - down.

Step3: Calculate rate of change for intervals

For the interval (400 < P<800), let (a = 400) and (b = 800). The rate of change (r_1=\frac{\log_2800-\log_2400}{800 - 400}=\frac{\log_2\frac{800}{400}}{400}=\frac{\log_22}{400}=\frac{1}{400}). For the interval (800 < P<1600), let (a = 800) and (b = 1600). The rate of change (r_2=\frac{\log_21600-\log_2800}{1600 - 800}=\frac{\log_2\frac{1600}{800}}{800}=\frac{\log_22}{800}=\frac{1}{800}). Since (\frac{1}{400}>\frac{1}{800}), the rate of change over (400 < P<800) is greater.

Step4: Interpret in context

The rate of change (\frac{\Delta d}{\Delta P}) tells us how many decades ((\Delta d)) it takes for the population ((\Delta P)) to increase. A larger rate of change means a smaller increase in decades for a given increase in population in the first interval compared to the second interval.