Question

problem 6
this table shows the population of a city from 1988
to 2016.
select two values of t that create an interval with a
negative rate of change.
1988
1992
1996
2000
2004
2008
2012
2016

Explanation:

Step1: Calculate the rate of change formula

The rate of change formula is $\frac{p(t_2)-p(t_1)}{t_2 - t_1}$.

Step2: Check each pair

• For \(t_1 = 1992\), \(p(1992)=42700\); \(t_2 = 1996\), \(p(1996)=33100\). Rate of change: \(\frac{33100 - 42700}{1996 - 1992}=\frac{-9600}{4}=-2400\)
• For \(t_1 = 2000\), \(p(2000)=33700\); \(t_2 = 2004\), \(p(2004)=45000\). Rate of change: \(\frac{45000 - 33700}{2004 - 2000}=\frac{11300}{4}=2825\)
• For \(t_1 = 2004\), \(p(2004)=45000\); \(t_2 = 2008\), \(p(2008)=48400\). Rate of change: \(\frac{48400 - 45000}{2008 - 2004}=\frac{3400}{4}=850\)
• For \(t_1 = 2008\), \(p(2008)=48400\); \(t_2 = 2012\), \(p(2012)=40900\). Rate of change: \(\frac{40900 - 48400}{2012 - 2008}=\frac{-7500}{4}=-1875\)
• For \(t_1 = 2012\), \(p(2012)=40900\); \(t_2 = 2016\), \(p(2016)=43000\). Rate of change: \(\frac{43000 - 40900}{2016 - 2012}=\frac{2100}{4}=525\)

Answer:

1992 and 1996, 2008 and 2012