Question

lesson practice
a1.4.06
name:
date:
period:
problems 5–7: this table shows the
population of a city from 1988 to 2016.

1. determine the average rate of change for

p(t) between 1992 and 2000.

1. state two values of t that create an

interval with a negative rate of change.

1. state two values of t that create an

interval with a positive rate of change.

year, t	population, p(t)
1992	42,700
1996	33,100
2000	33,700
2004	45,000
2008	48,400
2012	40,900
2016	43,000

Explanation:

Problem 5

Step1: Recall the formula for average rate of change

The average rate of change of a function \( p(t) \) over the interval \([t_1, t_2]\) is given by \(\frac{p(t_2)-p(t_1)}{t_2 - t_1}\).

Step2: Identify \( t_1, t_2, p(t_1), p(t_2) \)

For the interval between 1992 and 2000:

• \( t_1 = 1992 \), \( p(t_1)=42700 \)
• \( t_2 = 2000 \), \( p(t_2)=33700 \)

Step3: Calculate the average rate of change

Substitute the values into the formula:

$$ \frac{p(2000)-p(1992)}{2000 - 1992}=\frac{33700 - 42700}{8}=\frac{-9000}{8}=-1125 $$

A negative rate of change occurs when \( p(t_2)<p(t_1) \) for \( t_2 > t_1 \). Looking at the table:

• For \( t_1 = 1992 \) (\( p = 42700 \)) and \( t_2 = 1996 \) (\( p = 33100 \)), \( 33100<42700 \), so the rate of change is negative.
• Another example: \( t_1 = 1992 \), \( t_2 = 2000 \) (as in problem 5, we saw the rate is negative).

A positive rate of change occurs when \( p(t_2)>p(t_1) \) for \( t_2 > t_1 \). Looking at the table:

• For \( t_1 = 2000 \) (\( p = 33700 \)) and \( t_2 = 2004 \) (\( p = 45000 \)), \( 45000>33700 \), so the rate of change is positive.
• Another example: \( t_1 = 2004 \), \( t_2 = 2008 \) (since \( 48400>45000 \)).

Answer:

The average rate of change is \(-1125\) (people per year).

Problem 6