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§3.1 Introduction to the Family of Exponential Functions

In this section we will introduce growth/decay at a constant percentage rate. In these type of problems we are talking about growth/decay of money, population, drug, contaminants, etc. as a percentage of the existing amount during each period of time. If we look at this type of growth, we see a pattern form:

[{[Original(100% + r%)](100% + r%)}(100% + r%)]

Each time the original grows, the new growth is based on the same rate but “new original amount” is based upon previous results. What we see is (100% + r%) as a repeated factor. We know that repeated factors are represented by exponents, so we can represent the growth over t periods as:

Original (100% + r%)t

We have a name for (100% + r%), we call it the growth factor. The r% is the percentage rate of change. In forming a function to represent this type of scenario, we call the original amount and the growth factor parameters. Parameters are values that change from function to function, but are not the independent or dependent values of the function.

An Exponential Function

is any positive ( not equal to one raised to some ( valued exponent.

Q = f(t) = abt where b > 0, but b ( 0

a & b are parameters: a is initial amount & b = 1 + r

r is constant percent change

The domain of the function, the values for which x can equal, are all ( (-(,().

Why? Because an exponent can be any real number. And the exponent is the independent variable.

Note: For our applications we will only be interested in D: [0, ()

The range however, is all positive ( greater than zero (0, ().

Why? Because any number raised to some power will always be a positive number.

The quantity a represents the initial quantity (the vertical intercept)

Why? Because when t = 0 that makes “at” equal to one, which gives P0

For exponential growth b > 1, giving the increasing function you see below on the left

b = 1 + r where r = percent rate of change; growth is positive rate of change

For exponential decay 0 < b < 1, giving the decreasing function you see below on the rt.

b = 1 + r where r = percent rate of change; it is negative for decay

To recognize an exponential function from values, look for ratios of Q(t) values that are constantly spaced for constant intervals of t

Example: Consecutive Q(t) values are 40000, 42400, 44944, 47640.64

42400/40000=1.06

44944/42400=1.06

47640.64/44944=1.06

Which indicates that the growth factor is 1.06

Furthermore, the constant rate of growth is 6% (1.06 – 1)

To create an exponential equation, find the constant ratio, raise it to the power of t and multiply it by the initial value.

Example: Create an equation for the above scenario, knowing that the initial

value was 40000.

The shape of the curve will always be:

Just some notes:

1) An exponential graph will always approach the x-axis but will never touch the

axis.

2) The graph will grow larger and larger if the number (b) is greater than 1. This is

like population. Think of 2 people, they have children, then those have children

and so on, this is an exponential function, and you should be able to see that it

grows to an infinite number.

3) If the “b” is between one and zero (fractional; decimal less than one but greater than zero)

the shape is the same but starts at infinity and decreases to approach the x-axis.

Now, let’s see if we have the basics of the exponential function.

Example: The following functions give the amount of substance present at

time t. In each, give i) the amount present initially, ii) state

whether the function represents exponential growth or decay, iii)

give the growth factor and iv) give the constant percent growth or

decay rate.*(#2 p.43, Applied Calculus, Hughes-Hallet, et al, 4th ed.)

a) A = 100(1.07)t b) A = 3500(0.98)t

Example: A town has a population of 1000 people at t = 0. In each of the

following cases, write a formula for the population, P, of the town

as a function of year t.*(#4 p. 44, Applied Calculus, Hughes-Hallet, et al,

4th ed.)

a) The population increases by 50 people a year

b) The population increases by 5% per year

Note: One of these is linear.

Example: Find a possible formula for the function. *(#18 p. 45, Applied Calculus,

Hughes-Hallet, et al, 4th ed.)

Note: Write the equation from the general form and plug into it the ordered pair that you do have to solve for the growth rate.

Example: For the data in the table showing annual soybean production in

millions of tons, answer the following questions.

*(#26 p.45, Applied Calculus, Hughes-Hallet, et al, 4th ed.)

|Year |2000 |2001 |

|0 |30.8 |15,000 |

|1 |27.6 |9,000 |

|2 |24.4 |5,400 |

|3 |21.2 |3,240 |

3) Graphically

a) In a linear function, the graph will be a straight line and show

constant rate of change between any 2 points

b) In an exponential function, the graph will begin fairly flat and

increase more and more steeply or decrease sharply becoming

fairly flat

rate of change is increasing/decreasing

4) Pacing of Linear & Exponential

In the long run an exponential will always have a larger value than

the linear.

When the initial value of the linear is larger than the initial of the

exponential, the linear will produce larger values only to a point of

intersection at which time the exponential will increase at a rate that

will outpace the linear.

Creating Formulas – Modeling

1) Understand if the data conforms to linear or exponential characteristics (see above)

2) Form the model based on one of the following methodologies

a) Linear

i) Vertical Intercept known

Find the slope m = ∆y

∆x

b is initial quantity, quantity when x = 0

Put into y = mx + b

ii) Vertical Intercept Unknown

Find the slope as above

Must know 2 points

Use 1 point & slope and put into

y – y1 = m(x – x1)

b) Exponential

i) Vertical Intercept known

Find the growth factor : b = y2

y1

a is the initial quantity, quantity when x = 0

Put into y = abx

ii) Vertical Intercept Unknown

Find the growth factor as above

Find the initial quantity by solving for a

a = y2

y1

Solve the resulting exponential by taking the “x2 – x1” root

Let’s start with an example from your first exam, because I heard a lot about this problem.

A small business buys a computer for $4000. After 4 years the value of the

computer is expected to be $200. For accounting purposes the business uses linear

depreciation to assess the value of the computer at a given time. Find a model for

the linear depreciation of the computer for the small business.

Now the comparable exponential problem.

John invested $840 in an account paying interest compounded annually. If he had

$1200 in the account 3 years after investing, find a formula to model the amount

of money in his account t years after the original investment. Use a growth factor

rounded to the nearest thousandth.

This is an example from your book that do not have vertical intercepts given. (#24 p. 120, Functions Modeling Change, Connally, 3rd edition)

[pic]

Based on the above picture create a linear model for the linear function shown in the graph and an exponential model for the exponential function shown in the graph. Label them appropriately as f(t) or g(t).

One last example, using a table this time. (#6 & 9 p. 120, Functions Modeling Change, Connally, 3rd edition)

|x |F(x) |G(x) |

|1 |13.75 |14 |

|2 |15.125 |10 |

|3 |16.638 |6 |

|4 |18.301 |2 |

Based on the above table create a linear model for the linear function shown in the table and an exponential model for the exponential function shown in the table. Label them appropriately as f(x) or g(x).

§3.3 Graphs of Exponential Functions

We have already talked a great deal about the vertical intercept of the exponential function, and that is one topic of this section. Here is a reminder about information that you should know concerning the vertical intercept.

1) Different values of a in Q(t) = abt mean that a function will

cross the vertical axis at different locations when the value of b

remains the same.

2) Even when the vertical intercept remains the same the graph

will not look identical because the initial value effects the

growth curve.

Example: On your calculator graph the following

Q(t) = 10(1.2)t

R(t) = 20(1.2)t

S(t) = 40(1.2)t

Note: The graphs aren’t just a shift up as they would be if they were linear models because of the product of the vertical intercept by the base “b” to the power of “t”. When we see shifts on a linear function associated with a change in the vertical intercept we can simply add the value of the shift to the “general” function, but with exponential it is a multiplicative relationship. You should be able to see this either in your book’s graph or in yours (you might use a viewing window set -10, 10, 1, 0, 75, 10, 1).

When the growth rate changes it will effect the steepness of the curve. As discussed in §3.1, when the growth rate, b > 1 we see an increasing function and when 0 < b < 1 we see a decreasing function. See p. 2 of these notes for a review of the graphic representations.

Example: On your calculator graph the following

A(t) = 30(1.5)t

B(t) = 30(1.2)t

C(t) = 30(0.9)t

D(t) = 30(0.6)t

Note: This time we see that the greater the growth rate the steeper the faster the curve climbs when viewed left to right when b > 1 and when 0 < b < 1 we see that the smaller b becomes the more steeply the curve decreases when viewed left to right.

Our next discussion needs to be about another feature that we mentioned in our discussion in §3.1 – horizontal asymptotes. A horizontal asymptote is a horizontal line which the graph approaches but never crosses or touches. The basic exponential function has a horizontal asymptote that is the x-axis.

Increasing functions come from the horizontal asymptote and grow toward infinity. We say this in the following way:

As Q(t) ➜ 0, t ➜ -∞

This means that as the values input get smaller and smaller (more & more negative) the value of the function will get close to zero (but never quite to zero).

Decreasing functions come from infinitely large values and decline toward the horizontal asymptote. We say this in the following way:

As Q(t) ➜ 0, t ➜ ∞

This means that as the values input get larger and larger (more & more positive) the value of the function will get close to zero (but never quite to zero).

When we discuss any exponential function where the horizontal asymptote can be described by the line y = k, we can say the following.

Increasing functions

As Q(t) ➜ k, t ➜ -∞

Decreasing functions

As Q(t) ➜ k, t ➜ ∞

Furthermore in the future we will need to be able to read this information in a way that involves something called a limit. A limit is simply a value to which a function approaches as its input/independent value goes to a specified value. We use lim to represent the limit and a subscript indicating what values the input/independent are approaching and we say this is equal to the value which the function approaches (in this case our horizontal asymptote). Here’s how it looks:

Increasing lim Q(t) = k

Decreasing lim Q(t) = k

To practice the concepts in this section we will be doing a class work exercise.

§3.4 Continuous Growth and the Number e

The type of exponential function that we discussed in the last section can lead us to our discussion of the natural exponential base. If we investigate the growth of money over different compounding periods when the constant growth rate per year is 100% we will see that the amount of money (function value; y-value) approaches a constant value as the number of compounding periods are increased from yearly through by the second. (see p. 130&131 of your text) In other words, as the number of compounding periods approaches infinity the function approaches a constant. It is this constant that we see in many naturally occurring processes. This constant value is the natural base – e. Here are some facts about e and its relationship to other exponential functions:

1) Naturally occurring base in many processes including population

growth/decay, continuously compounding interest, decay of radioactive

substances.

2) Irrational number ( 2.718282

3) 2t < et < 3t

4) Along w/ e comes the parameter k, this is the continuous growth rate

5) Any base b (growth factor) can be converted to e

How? b = ek ∴ k = ln b

6) For e we can convert to b (growth factor)

How? bt = (ek)t

7) When k < 0, e behaves as 0 < b < 1 (decreasing f(n))

8) When k > 0, e behaves as b > 1 (increasing f(n))

Now, let’s formally state the continuous growth function

Q(t) = aekt a > 0

Let me show you the above information from 4-6 in a different format as well

P = P0at = P0(ek)t = P0ekt so, a = ek ∴ k = ln a = ln (ek) where

k is the continuous growth rate

a = ek where a = 1 + r and

r is growth rate per unit time

Example: Give the growth or decay rate and state if it is continuous or

annual.*(a is #18 & b is #20p. 50, Applied Calculus, Hughes-Hallet, et al,

4th ed.))

a) P = 7.7(0.92)t b) P = 3.2e0.03t

One thing that we should notice about annual growth rate vs. continuous growth rate is that a smaller continuous growth rate will produce a larger equivalent annual growth rate. A graphical representation of the 2 might help you to see this.

Example: Find the annual growth rate for P = 3.2e0.03t

How does it compare to the continuous growth rate? (use 6 decimals)

Write the function in terms of annual growth rate.

Graph both the original and the re-written and see that they

are the same (I suggest a viewing window -5, 105, 10, 0, 70, 10, 1 & make

sure to put the graph type on the second function to “0” so you can see it trace

out)

Example: For the function P = 100e0.06t, which represents a population size

after t years.

*(a-c are from #34p.50, Applied Calculus, Hughes-Hallet, et al, 4th ed.)

a) What is the continuous growth rate?

b) Write the function in terms of P = P0at

c) What is the annual growth rate per year?

d) What is the initial population?

e) To the nearest unit, give the population, P, after 10 years.

f) To the nearest 10th of a year, when will the population, P, reach

500?

g) The amount of time it takes for a population to double in size is

called the doubling time. This population will double when it

reaches what size?

h) What is the doubling time for this population to the nearest tenth of

a year?

Note: This means that the ratio between P & a is 2, so you can actually find doubling time no matter if you know final size or even initial size!

Example: In 2004, the gross world product, W, (total output in goods and services)

was $54.7 trillion and growing at a continuous rate of 3.8% per

year. Write a formula for W, in trillions of dollars, as a function of

years, t, since 2004. Estimate the value of t when the gross world

product is predicted to reach $80 trillion. (#22 p. 135, Functions

Modeling Change, Connally, 3rd edition)

One last topic before we move on – limits. We discussed the limits in the last section so we should re-iterate the limits in this section. For Q(t) = aekt + c

1) When k > 0 the function is increasing when a > 0 ∴

lim Q(t) = c

Note: We found out through investigation that when something is added to an

exponential function that value is where the asymptote lies. In other words, for

Q(t) = abt + c, the asymptote was y = c.

2) When k < 0 the function is decreasing when a > 0∴

lim Q(t) = c

3) When a < 0 & k > 0 this reflects the graph across the x-axis, but

the asymptotes don’t change.

lim Q(t) = c

4) When a < 0 & k < 0 this reflects the graph across the x-axis, but

the asymptotes don’t change.

lim Q(t) = c

Example: Find the limit

a) lim 2e0.1t + 6 b) lim 5 – e-0.07t

Note: When you have e0.1t + 6 this is the same as e0.1t • e6 and therefore the a = 2(e6), which means that the vertical intercept changes as a result of adding to the independent value (this is also known as a horizontal translation).

§3.5 Compound Interest

[{[Original(100% + r%)](100% + r%)}(100% + r%)]

Each time the original grows, the new growth is based on the same rate but “new original amount” is based upon previous results. What we see is (100% + r%) as a repeated factor. We know that repeated factors are represented by exponents, so we can represent the growth over t periods as:

Original • (100% + r%)t

Recall this example from the first section of the chapter. It is the same for compound interest with an annual growth factor (only being allowed to compound yearly). But, we want to be able to compound under different time periods than annually. This is how we create the compound interest formula based on what we already have:

Nominal Rate: Annual Percentage Rate

(rate per compounding period) x (# of compounding periods in a year)

Abbreviated r in the formula

Effective (Annual) Rate: Growth rate per compounding period to

exponent reflecting number of times

compounded annually

The growth rate money would have to earn if it were compounded annually

Shown as (1 + r/n)nt – 1 where (1 + r/n) is the base in the formula & nt

is the exponent

New Exponent: Number of Times Compound Annually

Daily: 365 x # of years (the original t)

Weekly: 52 x # of years (the original t)

Monthly: 12 x # of years (the original t)

Quarterly: 4 x # of years (the original t)

Shown as n • t in the formula where n is #of compounding periods

Example: For a deposit of $100 that earns 1/2% compounded monthly over a

1 year period:

Growth Rate is (1 + 0.005) = 1.005

Effective Rate is (1.005)12 • 1 – 1 = 1.061677 – 1

= 6.168% interest in a year

Note: Interest rates in your book are rounded to the nearest 1000th of a %.

What does the last example actually mean?

If we invest $100 at 1/2% compounded monthly or at 6.168% annual

percentage rate, the money earned will be approximately the same.

Check: 100(1.005)12 ≈ $106.17

100(1.06168)1 ≈ $106.17

We can also say that the nominal rate is the sum of percentages per compounding units in one year.

Example: In our last example this means:

1/2 x 12 = 6%

is the nominal rate

Now we are ready to revisit the formula from §3.1 where we only discussed a compounding unit of once per year. With the introduction of nominal vs. effective rate for different compounding periods within a year we can build a more complex model for compounded interest.

Now, we want to link compounded interest to the number e – our natural base. Your book shows you a table on p. 138 where the compounding units per year are increased from 1 through 365 (yearly to daily) ending with hourly. What you should notice is the effective annual rate creeps closer and closer to the same value. And that value is …

enominal rate

This is exactly how it will look for any nominal interest rate. This leads to a model for continuously compounded interest.

Remember, we can easily compare different nominal rates by converting to effect annual rates.

Example: Compare the following by finding the effective annual rates; list

them from smallest to largest

7% annual interest compounded quarterly

6.75% annual compounded monthly

6.75% annual compounded continuously

7% annually compounded

-----------------------

Where 0< b < 1

Decreasing Function

(0,1)

(-1,1/b)

Where b > 1

Increasing Function

(1,b)

(0,1)

(-1,1/b)

(1,b)

(0,30)

(25, 6)

10

20

30

y

t

10

20

30

y

t

10

20

30

x2 – x1

t ➜ -∞

t ➜ ∞

t ➜ -∞

t ➜ ∞

t ➜ -∞

t ➜ ∞

t ➜ ∞

t ➜ -∞

Compound Interest Formula

B = P • (1 + r/n) n • t

Where B = Balance at the end of time t (in years); the dependent variable

P = Amount of original deposit

r = nominal rate (sum of percentages per period)

n = # of compounding periods per year

t = time of investment in years; the independent variable

Note: When n = 1 this is the formula that we dealt with in §3.1 – 3.3

Continuously Compound Interest Formula

B = P • e r • t

Where B = Balance at the end of time t (in years); the dependent variable

P = Amount of original deposit/investment

e ≈ 2.718282

r = nominal rate

t = time of investment in years; the independent variable

Note: er = effective annual rate. Meaning that B = P(er)t is the same as the formula from §3.1 where b = er

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