Question

1. the table gives the population of a city from 1988 to 2016.

a) determine the average rate of change for p(t) between 1992 and 2000.
b) select two values of t that create an interval with a negative average rate of change. determine the average rate of change of this interval.

year, t	1988	1992	1996	2000	2004	2008	2012	2016
population, p(t)	35700	42700	33100	33700	45000	48400	40900	43000

Explanation:

Step1: Identify values for 1992 and 2000

From the table, for \( t = 1992 \), \( p(1992)=42700 \); for \( t = 2000 \), \( p(2000)=33700 \).

Step2: Apply average rate of change formula

The formula for average rate of change of a function \( p(t) \) over the interval \([t_1, t_2]\) is \( \frac{p(t_2)-p(t_1)}{t_2 - t_1} \). Here, \( t_1 = 1992 \), \( t_2 = 2000 \), so \( t_2 - t_1=2000 - 1992 = 8 \).

Substitute the values: \( \frac{33700 - 42700}{8}=\frac{-9000}{8}=-1125 \).

Step3: Select interval for negative rate (part b)

We can choose \( t = 2000 \) and \( t = 2004 \)? Wait, no, wait. Wait, to get negative rate, we need \( p(t_2)<p(t_1) \). Let's check values:

From table: 1992 (42700), 1996 (33100), 2000 (33700), 2004 (45000), 2008 (48400), 2012 (40900), 2016 (43000). Wait, 1992 to 1996: \( t_1=1992 \), \( t_2=1996 \), \( p(1992)=42700 \), \( p(1996)=33100 \). Then average rate: \( \frac{33100 - 42700}{1996 - 1992}=\frac{-9600}{4}=-2400 \). Let's confirm the interval selection. Let's pick \( t = 1992 \) and \( t = 1996 \) (or 1992 - 2000 as in part a, but let's choose another). Wait, the problem says "select two values of t" for negative rate. Let's take \( t_1 = 1992 \) and \( t_2 = 1996 \). Then \( t_2 - t_1 = 4 \), \( p(t_2)-p(t_1)=33100 - 42700=-9600 \). Average rate: \( \frac{-9600}{4}=-2400 \).

Step4: Calculate average rate for selected interval (part b)

Using \( t_1 = 1992 \), \( t_2 = 1996 \), the average rate is \( \frac{33100 - 42700}{1996 - 1992}=\frac{-9600}{4}=-2400 \). (We could also use 1992 - 2000 as in part a, but let's follow the selection. Alternatively, 2008 - 2012: \( p(2008)=48400 \), \( p(2012)=40900 \), \( t_2 - t_1=4 \), rate: \( \frac{40900 - 48400}{4}=\frac{-7500}{4}=-1875 \).)

Answer:

(part a): -1125 (part b example: interval 1992 - 1996, rate -2400)

(Note: For part a, the average rate of change between 1992 and 2000 is -1125. For part b, one possible interval is 1992 to 1996 with rate -2400, or 1992 to 2000 with rate -1125, etc.)