Quiz 3-1aQuiz 3-1a1.1. Write the equation that Write the equation that models the data under the models the data under the column labeled g(x).column labeled g(x).

xabxg )(

2. Write the equation that models the data under the column2. Write the equation that models the data under the column labeled f(x) above.labeled f(x) above. xxf )25.0(8)(

3.3. Without using your calculator, determine if the following Without using your calculator, determine if the following function growth or decay?function growth or decay?

xexf 5.03)(

xxg )3(4)(

xabxf )(

xxf 2)(4. Without using your calculator, determine if the following 4. Without using your calculator, determine if the following function growth or decay?function growth or decay?

3.1B3.1BApplications of Exponential Applications of Exponential Functions Functions

Exponential FunctionExponential Functionxabxf )(

Initial valueInitial valueGrowth factor:Growth factor:

What does ‘b’ equal What does ‘b’ equal In order for it to be “growth”?In order for it to be “growth”?

Input variableInput variable

What does ‘b’ equal What does ‘b’ equal In order for it to be “decay”?In order for it to be “decay”?

What is the value of ‘y’ whereWhat is the value of ‘y’ where the graph crosses the y-axis?the graph crosses the y-axis?

Your turn:Your turn:

xxf )4(3)(

Graph the functions:Graph the functions:

1. 1. Where does it cross the y-axis?Where does it cross the y-axis?

2. 2. What is the “intial value of f(t) ? What is the “intial value of f(t) ?

time)is t'' (where )5(2)( ttf

Population GrowthPopulation Growthpopulation initial0 P

If population grows at a constant percentage rate over a If population grows at a constant percentage rate over a year time frame, (the final population is the initial population year time frame, (the final population is the initial population plusplus a percentage of the orginial population) then the a percentage of the orginial population) then the population at the end of the first year would be:population at the end of the first year would be:

001 rPPP

At the end of the second year the population would be:At the end of the second year the population would be:

Percent rate of changePercent rate of change (in decimal form)(in decimal form)

112 rPPP

)1(01 rPP

)1(12 rPP

Population GrowthPopulation Growthpopulation initial0 P

)1(01 rPP

112 rPPP

year 1after population 1 P

years 2after population 2 P

rrPrPP )1()1( 002

20002 2 rPrPPP Quadratic equation!Quadratic equation!

202 )1( rPP

Population GrowthPopulation Growthpopulation initial0 P

)1(01 rPP

)1(12 rPP

year 1after population 1 P

years 2after population 2 P

rrPPrPPP )()( 00002

20002 2 rPrPPP Quadratic equation!Quadratic equation!

202 )1( rPP

Population GrowthPopulation Growthpopulation initial0 P

)1(01 rPP 2

02 )1( rPP

year 1after population 1 Pyears 2after population 2 P

rPPP 223

rrrPrPP ))1(()1( 20

203

Special cubic!Special cubic!

)221( 32203 rrrrrPP

)133( 2303 rrrPP

303 )1( rPP ?4 P ?nP

Population GrowthPopulation GrowthtrPtP )1()( 0

PopulationPopulation (as a (as a function of time)function of time)

InitialInitial populationpopulation

GrowthGrowth raterate

time time

Percent rate of changePercent rate of change (in decimal form)(in decimal form)

xabxf )(

Initial valueInitial valueGrowth factor:Growth factor:

Word problemsWord problems trPtP )1()( 0 There are 4 quantities in the equation:There are 4 quantities in the equation:

2. 2. InitialInitial population population

3. 3. GrowthGrowth rate rate

1. Population “t” years/min/sec in the future1. Population “t” years/min/sec in the future

4. 4. timetime

The words in the problem will give you The words in the problem will give you threethree of the four of the four quantities. You just have to “plug them in” to the equationquantities. You just have to “plug them in” to the equation and solve for the unknown quantity.and solve for the unknown quantity.

Population GrowthPopulation GrowthtrPtP )1()( 0

PopulationPopulation (at time (at time “ “t”) in the futuret”) in the future

InitialInitial populationpopulation

GrowthGrowth raterate

time time

The initial population of a colony of bacteria is 1000. The The initial population of a colony of bacteria is 1000. The population increases by 50% every hour. What is the population increases by 50% every hour. What is the population after 5 hours?population after 5 hours?

5)50.01(1000)5( P

7593)5( P

Percent rate of changePercent rate of change (in decimal form)(in decimal form)

5)5.1(1000)5( P

Unknown valueUnknown value

Simple Interest (savings Simple Interest (savings account)account)

trPtA )1()( AmountAmount (as a (as a function of time)function of time)

Initial amountInitial amount (“principle”)(“principle”)

Interest Interest raterate

time time

A bank account pays 3.5% interest per year. If you initially A bank account pays 3.5% interest per year. If you initially invest $200, how much money will you have after 5 years? invest $200, how much money will you have after 5 years?

5)035.1(200$)5( A 54.237$)5( AUnknown valueUnknown value

Your turn:Your turn: trPtA )1()( A bank account pays 14% interest per year. If you initially A bank account pays 14% interest per year. If you initially invest $2500, how much money will you have after 7 years? invest $2500, how much money will you have after 7 years?

3. 3.

4. 4. The population of a small town was 1500 in 1990. The The population of a small town was 1500 in 1990. The population increases by 3% every year. What is the population increases by 3% every year. What is the population in 2009?population in 2009?

Solve bySolve by graphing graphing

Year PopulationYear Population1990 782,2481990 782,2482000 895,193 2000 895,193

San Jose, CASan Jose, CA

Assuming exponential growth, when willAssuming exponential growth, when will the population the population equalequal 1 million? 1 million?

Let ‘t’ = years since 1990Let ‘t’ = years since 1990tbPtP 0)(

193,895248,782)10( 10 bPWe must find the growth factor ‘b’We must find the growth factor ‘b’

‘‘b’ = 1.0136b’ = 1.0136 ttP )0136.1(248,782)( 248,782

193,89510 b 1010 10

248,782

193,895b

t)0136.1(248,782000,000,1 Unknown valueUnknown value

ExampleExamplettP )0136.1(248,782)(

1,000,0001,000,000

‘‘t’ = approximately 18 t’ = approximately 18

18 years 18 years AFTERAFTER 1990 1990 2008 2008

000,000,1)(? yearsP

Later in the chapter we will learn how to solve for theLater in the chapter we will learn how to solve for the unknown unknown exponentexponent algebraically. algebraically.

Your Turn:Your Turn:

5. 5. When did the population reach 50,000 ?When did the population reach 50,000 ?

The population of “Smallville” in the year 1890 was 6250. The population of “Smallville” in the year 1890 was 6250. Assume the population increased at a rate of 2.75% per year.Assume the population increased at a rate of 2.75% per year.

Your turn:Your turn:Year PopulationYear Population1990 248,709,8731990 248,709,8732009 307,006,550 2009 307,006,550

USAUSA

6. 6. Assuming exponential growth, when willAssuming exponential growth, when will the population exceed 400 million?the population exceed 400 million?

tbPtP 0)(

0307,006,553248,709,87)19( 19 bP

We must find the growth factor ‘b’We must find the growth factor ‘b’

‘‘b’ = 1.0111b’ = 1.0111

ttP )0111.1(3248,709,87)(

3248,709,87

0307,006,5519 b

1919 19

3248,709,87

0307,006,55b

t)0111.1(3248,709,87000,000,400 43 yrs after43 yrs after t = 0 (1990)t = 0 (1990)

20332033

Your turn:Your turn:Year PopulationYear Population1900 76.21 million1900 76.21 million2000 248.71 million 2000 248.71 million

USAUSA

7. 7. Assuming exponential growth, when willAssuming exponential growth, when will the population exceed 400 million?the population exceed 400 million?

tbPtP 0)(

71.24821.76)100( 100 bP

We must find the growth factor ‘b’We must find the growth factor ‘b’

‘‘b’ = 1.0119b’ = 1.0119

ttP )0119.1(21.76)(

21.76

71.248100 b

100100 100

21.76

71.248b

t)0119.1(21.76000,000,400 140.2 yrs after140.2 yrs after t = 0 (1900)t = 0 (1900)

2040.2 2040.2

Finding an Exponential Finding an Exponential FunctionFunction

$500 was deposited into an account that pays “simple $500 was deposited into an account that pays “simple interest” (interest paid at the end of the year).interest” (interest paid at the end of the year).

trPtA )1()( 0 5000 P25)1(5001250 r

25 years later, the account contained $1250. What was 25 years later, the account contained $1250. What was the percentage rate of change?the percentage rate of change?

1250)25( A

25)1(500

1250r

)1(500

125025 r

)1(037.1 r

037.0r

% 7.3r

Unknown valueUnknown value

Your Turn:Your Turn:8. 8. The population of “Smallville” in the year 1890 was 6250. The population of “Smallville” in the year 1890 was 6250.

Assume the population increased at a rate of 2.75% per year. Assume the population increased at a rate of 2.75% per year. What is the population in 1915 ? What is the population in 1915 ?

9. 9. The population of “Bigville” in the year 1900 was 25,200. In The population of “Bigville” in the year 1900 was 25,200. In 1955 the population was 37,200. 1955 the population was 37,200. What was the percentage What was the percentage rate of change?rate of change?

10. 10. The population of “Ghost-town” in the year 1900 was 3500. The population of “Ghost-town” in the year 1900 was 3500. In 1935 the population was 200. In 1935 the population was 200. What was the percentage What was the percentage rate of change?rate of change?

Finding Growth and Decay Finding Growth and Decay RatesRates

Is the following population model an Is the following population model an exponential growth or decayexponential growth or decay function? function?

ttP )0136.1(248,782)(

trPtP )1()( 0 ttP )0136.01(248,782)(

‘‘r’ > 0, therefore this is r’ > 0, therefore this is exponential growthexponential growth..

‘‘r’ = 0.0136 or 1.36%r’ = 0.0136 or 1.36%

Find the constant percentage growth (decay) rate.Find the constant percentage growth (decay) rate.

Your turn:Your turn:

11. 11. Is it growth or decay?Is it growth or decay?

ttP )5.1(50)(

trPtP )1()( 0 ttP )5.01(50)(

‘‘r’ > 0, therefore this is r’ > 0, therefore this is exponential growthexponential growth..

‘‘r’ = 0.5 or 50% r’ = 0.5 or 50% % rate of % rate of growthgrowth is 50% is 50%

12. 12. Find the constant percentage growth (decay) rate.Find the constant percentage growth (decay) rate.

xabxf )( b = 1.5b = 1.5 b > 0b > 0 Growth!Growth!

Finding an Exponential Finding an Exponential FunctionFunction

Determine the exponential function with initial value = 10, Determine the exponential function with initial value = 10, increasing at a rate of 5% per year.increasing at a rate of 5% per year.

trPtP )1()( 0 100 P ‘‘r’ = 0.05r’ = 0.05

ttP )05.1(10)( xxf )05.1(10)( or or

Modeling Bacteria GrowthModeling Bacteria Growth

Suppose a culture of 100 bacteria cells are put into a petri dish Suppose a culture of 100 bacteria cells are put into a petri dish and the culture and the culture doublesdoubles every hour. every hour.

Predict whenPredict when the number of bacteria will be 350,000. the number of bacteria will be 350,000.

trPtP )1()( 0

P(0) P(0) = 100 = 100 P(t) P(t) = 350000 = 350000 t)2(100350000

What is the What is the growth factorgrowth factor??

3500105.3101

105.32 3

2

5

t 35002 t

tPtP )2()( 0

trPtP )1()( 0

Modeling Bacteria GrowthModeling Bacteria GrowthSuppose a culture of 100 bacteria cells are put into a petri dish Suppose a culture of 100 bacteria cells are put into a petri dish and the culture and the culture doublesdoubles every hour. every hour.

Predict whenPredict when the number of bacteria will be 350,000. the number of bacteria will be 350,000.

35002 t

3500

2

2

1

y

y x

Where do the two Where do the two graphs cross?graphs cross?

t = 11 hours + 0.77hrst = 11 hours + 0.77hrs

77.11x

t = 11 hours + 0.77hrst = 11 hours + 0.77hrshr

min 60*

t = 11 hours + 46 mint = 11 hours + 46 min

Your turn:Your turn:

13. 13. A family of 10 rabbits doubles every 2 years. WhenA family of 10 rabbits doubles every 2 years. When will the family have 225 members?will the family have 225 members?

trPtP )1()( 0

t = 7 years 6 months t = 7 years 6 months

t = 7.8 yearst = 7.8 years

t)2(10225 b = 2b = 2

xabxf )(

Modeling U.S. Population Modeling U.S. Population Using Exponential Using Exponential RegressionRegression

Use the 1900-2000 data and Use the 1900-2000 data and exponential regression to predict exponential regression to predict the U.S. population for 2003. the U.S. population for 2003. ((Don’t enter the 2003 valueDon’t enter the 2003 value).).

Let P(t) = population, Let P(t) = population, “ “t” years after 1900.t” years after 1900.

Enter the data into yourEnter the data into your calculator and use calculator and use exponential regressionexponential regression to determine the model (equation).to determine the model (equation).

Exponential RegressionExponential RegressionStat p/b Stat p/b gives lists gives lists

Enter the data: Enter the data: Let L1 be years since initial valueLet L1 be years since initial value

Let L2 be populationLet L2 be population

Stat p/b Stat p/b calc p/b calc p/bscroll down to exponential regressionscroll down to exponential regression

““ExpReg” displayed:ExpReg” displayed:enter the lists: “L1,L2”enter the lists: “L1,L2”

xabxf )( The calculator will display the The calculator will display the values for ‘a’ and ‘b’.values for ‘a’ and ‘b’.

Your turn:Your turn:14. 14. What is your equation?What is your equation?

15. 15. What is your predicted population in 2003 ?What is your predicted population in 2003 ?

16. 16. Why isn’t your predicted value the same as the Why isn’t your predicted value the same as the actual value of 290.8 million?actual value of 290.8 million?

Find the amout of material after ‘20’ Find the amout of material after ‘20’ days if the days if the initial mass is 5 grams and it doubles initial mass is 5 grams and it doubles every 4 days:every 4 days:tabtf )(

The issue is units !!!The issue is units !!!

Initial value ‘a’ Initial value ‘a’ units of grams units of grams

Can the exponent have any units?Can the exponent have any units?

This doubles every 4 days. How many times does itThis doubles every 4 days. How many times does it double in 20 days?double in 20 days? 5(2) grams 5)( xf

The mass (# of grams) at some time “t” in the future is The mass (# of grams) at some time “t” in the future is the initial mass (# of grams) times some number .the initial mass (# of grams) times some number .tb

NO !!!NO !!!

Units of the exponentUnits of the exponent

The input value is The input value is timetime (with units of seconds, minutes, hrs, etc.). (with units of seconds, minutes, hrs, etc.).

How can the input value have units and the exponent not haveHow can the input value have units and the exponent not have any units (since that is where the input value is inserted into theany units (since that is where the input value is inserted into the equation)?equation)?

IF the input value has the units of IF the input value has the units of time in secondstime in seconds , then the exponent , then the exponent really has the units of really has the units of “# of times the base is used as a factor / day” “# of times the base is used as a factor / day” to make the units work out.to make the units work out.

tA )3(8)t(

7)3(8)days 7( A

Since the base is a 3, then this could be shortened toSince the base is a 3, then this could be shortened to “ “# of triples/ day” # of triples/ day” This could be shortened to This could be shortened to “per day” “per day” which in math is which in math is “1/day” “1/day”

Find the amout of material after 20 Find the amout of material after 20 days if the days if the initial mass is 5 grams and it doubles initial mass is 5 grams and it doubles every 4 days:every 4 days: Initial mass = 5 gramsInitial mass = 5 grams

mass doubles every 4 daysmass doubles every 4 days

tabtA )(

)days 4 1days*( t""

grams(2) 5)days "(" tA

(amount (grams) as (amount (grams) as a function of time)a function of time)

)days 4 1days*( 20

grams(2) 5)days 16( A 5grams(2) 5)days 16( A

grams 160)days 16( A

No units remain in the No units remain in the exponent.exponent.

Find the amout of material after 20 Find the amout of material after 20 days if the days if the initial mass is 5 grams and it doubles initial mass is 5 grams and it doubles every 4 days:every 4 days: Initial mass = 5 gramsInitial mass = 5 grams

mass doubles every 4 daysmass doubles every 4 days

tabtA )(

)days 4 1days*( t""

grams(2) 5)days "(" tA

(amount (grams) as (amount (grams) as a function of time)a function of time)

)days 4 1days*( 20

grams(2) 5)days 16( A 5grams(2) 5)days 16( A

grams 160)days 16( A

So you could just write it as:So you could just write it as:

Your turn:Your turn:17. 17. The crowd in front of the Tunisian parlament The crowd in front of the Tunisian parlament

building increased by a factor of 4 every 3 hours.building increased by a factor of 4 every 3 hours. If the initial crowd had 500 people in it, how manyIf the initial crowd had 500 people in it, how many people would there be after 12 hours? people would there be after 12 hours?

18. 18. The amount of radioactive Rubidium 88 The amount of radioactive Rubidium 88 decreases decreases by a factor of 2 every 8 minutes. If there was 5 by a factor of 2 every 8 minutes. If there was 5 grams of the material at the start, how much would grams of the material at the start, how much would there be after 30 minutes?there be after 30 minutes?

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