Module 4

Section 4.1

Exponential Functions and Models

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Exponential Functions Growth Rates

Linear Growth Consider the arithmetic sequence: 3, 5, 7, 9, 11, 13, ...

a1 = 3

a2 = a1 + 2 = 3 + 2

a3 = a2 + 2 = 3 + 2 + 2

a4 = a3 + 2 = 3 + 2 + 2 + 2

an = a1 + (n – 1)2 = 3 + 2 + 2 + … + 2

Define function f(n) by:

Let k = n – 1

n – 1 Terms

= 3 + k(2) = 2k + 3k = n – 1

f(n) = an = a1 + (n – 1)2

g(k) = a1 + 2k = 2k + a1 so that

A Linear FunctionQuestion:How do we know this is linear ?

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Exponential Functions Growth Rates

Exponential Growth Consider the geometric sequence: 3, 6, 12, 24, 48, 96, ... a1 = 3

a2 = a1 2 = 3 21

a3 = a2 2 = a1 2 2 = 3 22

an = an–1 2 … 2 = 3 2n–1

Define function f(n) by: f(n) = an = a1 rn–1 = 3 2n–1

Letting k = n – 1

n – 1 Factors

k = n – 1

this becomes g(k) = a1rk = 3 2k

An Exponential Function

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Exponential and Power Functions

What’s the difference ? For any real number x , and rational number a

we write the ath power of x as :

Function f(x) = xa is called a power function

For any real numbers x and a , with a ≠ 1 and a > 0 ,

the function f(x) = ax is called an exponential function

The general form is :

where C is the constant coefficient , C > 0 , and base is a

x a

Base x Exponent a

f(x) = Cax

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Function Comparisons

x

y

x

y

x

y

f(x) = 2x f(x) = 2x f(x) = x2

Linear Function Exponential Function Power Function

–2

–4

–1

–2

0 0

1 2

2 4

3 6

–3

–6

x 2x

–2

¼

–1

½

0 1

1 2

2 4

3 8

–3 ⅛

x 2x

–2 4

–1 1

0 0

1 1

2 4

3 9

–3

9

x x2

f = { (x, 2x) x R } f = { (x, 2x) x R } f = { (x, x2) x R }

Question: What is f(5) ? ... and f(10) ? ... and f(20) ?

Which function grows fastest as x ?

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Exponential Functions Increasing/Decreasing Exponential Functions

Exponential growth function :

f(x) = Cax , a > 1 Exponential decay function

g(x) = f(–x) = Ca–x , 1 < a

... a reflection of f(x)

h(x) = Cbx , 0 < b < 1 , x

y

f(x) = C2x

–2

¼4

–1

½2

0 11

1 2½

2 4¼

3 8⅛

–3 ⅛8

f = { (x, 2x) x R }

g(x) = f(–x) = C2–x

x 2x 2–x

●(0, C)

g = { (x, 2–x ) x R }

As ordered pairs (C = 1) :

In tabular form (C = 1) :

Questions:Intercepts ?

Asymptotes ?

Domain = R

Range = { x x > 0 }

... OR1ab =

Effects of larger/smaller a ?

Growth factor a ?

Decay factor a ?

a > 1

0 < a < 1

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Exponential Function Basics Let f(x) = ax with a > 0 , a ≠ 1

f(0) = 1

Domain-of-f = R

Range-of-f = { y x > 0 }

Graph is increasing for a > 1 and decreasing for a < 1

f is 1–1

For a > b > 1 : ax > bx for x > 0 and ax < bx for x < 0

Graphs of ax and bx intersect at (0, 1)

If ax = ay then x = y

= ( – , ) ∞ ∞= ( 0 , )∞

WHY ?

WHY ?

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Exponential Equations Solve

1. 25x = 125

(52)x = 53

52x = 53

2x = 3

x = 3/2 Solution set is

2. 9x – 2 = 27x

(32)x – 2 = (33)x

32x – 4 = 33x 2x – 4 = 3x

x = –4 Solution set is { –4 }

{ }32

WHY ?

WHY ?

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Exponential Decay Radioactive Decay

Radioactive isotopes of some elements such as14C , 16N , 238U , etc decay spontaneously into more stable forms (12C , 14N , 236U , 232U , etc)

Decay times range from a few microseconds to thousands of years

Decay measurement Often can’t measure whole decay time Can measure limited decay, then calculate half-life

Half-life = time for decay to half original measured amount Model with exponential functions Facts: Decay rate proportional to amount present Same proportion decays in equal time

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Exponential Decay Radioactive Decay (continued)

Let initial amount of radioactive of substance Q be A0 and A(x) the amount after x years of decay

After half-life of k years, A(k) = ½A0

A(2k) = (½)A(k)

After n half-lives, x = nk so the amount left is

A(x) = A(nk) = A0(½)n

Since x = nk, then n = x/k

and A(x) = A0(½)x/k

A(3k) = (½)A(2k)

, … , A(nk) = A0(½)n , …

= (A0(½))(½)= A0(½)2

… or just

A0(½)

= (A0(½)2)(½) = A0(½)3

A(4k) = A0(½)4

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Exponential Decay Radioactive Decay (continued)

Alternative view: just recognize that half-life can be modeled by an exponential function

f(x) = Cax Then initial amount is f(0) = C and, for half-life k,

f(k) = (½)C = Cak

Dividing out the constant C gives: ak = ½

and solving for a , we get Hence

f(x) = Cax = C((½)1/k)x = C(½)x/k

Question: What does this look like graphically ?

a = (½)1/k

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A(t)

t

Exponential Decay Radioactive Decay Graph

Let :

A0

A012

A014

A018

A0116

A0164 k 2k 3k 4k 5k 6k0

A0132

A(t) = amount at time t

A0 = initial amount

k = half-lifeA(t) = A0(½)t/k

Since n = t/k

A(t) =

A0(½)n

A0 = A(0)

Amount is reduced by half in each half-life

After n half-lives t = nk

A(t) = A0(½)t/k

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Compounding $200 is deposited and earns 5% interest compounded

annually. How much is in the account after three years ?

After 1 year: Balance = 200 + (.05)(200) = 200(1 + .05)

After 2 years: Balance = 200(1 + .05) + 200(1 + .05)(.05)

= 200(1 + .05)(1 + .05)

= 200(1 + .05)2

After 3 years: Balance = 200(1 + .05)2 + 200(1 + .05)2(.05)

= 200(1 + .05)2(1 + .05)

= 200(1 + .05)3 = 231.525 ≈ $231.53

Question: What if interest is simple interest?

Balance = $230.00

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Suppose amount P draws r per cent interest (expressed as a decimal fraction) compounded annually for t years

What is the amount A accumulated after t years?

0 year:

1 year:

2 years:

3 years:

t years:... well ... Is this obvious?

Compound Interest in General

A = P A = P + Pr = P(1 + r)

A = P(1 + r) + P(1 + r)r

= P(1 + r)(1 + r) = P(1 + r)2

A = P(1 + r)t

A = P(1 + r)2 + P(1 + r)2r

= P(1 + r)2(1 + r) = P(1 + r)3

= P(1 + r)1

Question: What if compounding is quarterly ?What if compounding is n times per year ?

Now is this obvious?

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Compound Interest in General Compounding n times per year we annualize interest to r/n

0 period:

1 period:

2 periods:

3 periods:

k periods:

In t years k = nt

t years:

A = P A = P + P(r/n)

A = P(1 + r/n) + P(1 + r/n)(r/n) = P(1 + r/n)2

A = 1000(1 + .05/4)4(20)

A = P(1 + r/n)2 + P(1 + r/n)2(r/n) = P(1 + r/n)3

= P(1 + r/n)1

Example: $1000 for 20 years at 5% compounded quarterly

Here P = 1000, r = .05, n = 4 and t = 20

A = P(1 + r/n)k

= 1000(1.0125)80 = 1000(2.701484941)

A = P(1 + r/n)nt

≈ $2,701.48

Is this obvious? ... consider ...Ah, now it’s obvious !

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Natural Exponential Function

Compute the first few terms of the sequence

an n

= 1 + n1( )

n an

1 2.0000000002 2.2500000003 2.3703703704 2.4414062505 2.4883200006 2.52162637210 2.593742460100 2.704813829200 2.711517123400 2.7148917442000 2.717602569

10000 2.717942121

100000 2.718268237

?

Does an approach a value as n ∞ ?

Question:

1000000 2.718280469

In fact,

an 2.7182 81828 45904 52353 60287 ....

We call this number e

e is irrational (in fact transcendental)and is the base for natural exponentialfunctions Natural exponential functions are of form

f(x) = ex

... and natural logarithms

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Natural Exponential Function

Graph of

x ex

1 2.7182 818282 7.3890 560983 20.0855 369234 54.5981 500335 148.4131 591056 403.4287 934927 1096.6331 584288 2980.9579 870419 8103.0839 2757510 22026.4657 9480611 59874.1417 15197

12 162754.7914 19003

13 442413.3920 08920

?14 1202604.2841 64776

f(x) = ex

x

ex

1000

800

600

400

200

1200

1 2 3 4 5 6 7

x–1 0 1 2 3

ex

1

2

3

4

5

f(x) = ex

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We have shown that

Thus

Recall that amount P compounded n times per year at annual interest rate r for t years is given by

Then

As the number of compounding periods per year (n) increasesperiodic compounding approaches continuous compounding

Thus an amount P compounded continuously for t years atannualized interest rate r yields amount A given by

Continuous Compoundingn

1 + n1( ) e

nx)1 + n1( e x

A = P(1 + r/n)nt

x=

n1 +( )( )n

1

as n ∞ n1

and 0

as n ∞ n1

and 0

A = P(1 + r/n)(n/r)rt

= P((1 + r/n)(n/r))rt

What does this mean ?

A = Pert

Pert

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Example:$1000 compounds continuously at 5% interest for 10 years

What is the accumulated amount ?

A = Pert

= 1000e(.05)10

= 1000(1.648721271)

≈ 1648.72

The accumulated amount is $1,648.72

For 20 years this would be: $2,718.28

For 30 years this would be: $4,481.69

For 40 years this would be: $7,389.06

Continuous Compounding

With simple annual interest

$1,628.89

$2,653.30

$4,321.94

$7,039.99

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Example: $25 is deposited at the end of each month in an account paying 5% annualized interest compounded continuously

How much is in the account after 10 years?

Let An be the amount in the account at the end of month n and A0 be the initial deposit

A1 = A0 + A0(e.05/12) = A0(1 + (e.05/12))

A2 = A0 + A1(e.05/12) = A0 + A0(1 + (e.05/12))e.05/12

= A0(1 + e.05/12 + (e.05/12)2)

Ak = A0(1 + e.05/12 + (e.05/12)2 + ... + (e.05/12)k)

Regular Saving

20 years? 30 years?

= A0

1 – (e.05/12)k+1

1 – e.05/12

At 20 years, k = 240, A240 = 10,356.18 At 10 years, k = 120, A120 = 3,925.44

At 30 years, k = 360, A360 = 20,958.68 At 40 years, k = 360, A480 = 38,439.26

Geometric Series

( Ak = Sk+1 )

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Example:The population of a certain country doubles every 50 years

In 1950 the population was 150 M (million)

What was the population in 1975 ?

When will the population reach 600 M ?

Solution:Let P(t) be the population at time t years

Let t = 0 represent 1950 and P(0) = P0 = 150 M

P(t) = P02kt

where k is a growth control

In 2000, t = 50 the population is doubled:

P(50) = 2P0 = 300 = 150(250k)

250k = 300/150 = 21

50k = 1

Population Growth

How do we know this ?

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Thus

In 1975, t = 25 so

P(25) = P02kt = (150)(225/50))

When the population is 600 M we have

P(t) = 600 = 150(2t/50)

2t/50 = 600/150 = 4 = 22

Thus t/50 = 2

Hence the population will be 600 M in the year 2050

Population Growth

P(t) = P02kt

50k = 1

P(t) = 150(2t/50)

= 150 2 ≈ 212.1 M

k = 1/50

and t = 100 years after 1950

Sound familiar ?

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Think about it !