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Module 4
Section 4.1
Exponential Functions and Models
10/26/2012 Section 4.3 v5.0.1 2
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 ?
10/26/2012 Section 4.3 v5.0.1 3
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
10/26/2012 Section 4.3 v5.0.1 4
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
10/26/2012 Section 4.3 v5.0.1 5
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 ?
10/26/2012 Section 4.3 v5.0.1 6
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
10/26/2012 Section 4.3 v5.0.1 7
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 ?
10/26/2012 Section 4.3 v5.0.1 8
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 ?
10/26/2012 Section 4.3 v5.0.1 9
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
10/26/2012 Section 4.3 v5.0.1 10
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
10/26/2012 Section 4.3 v5.0.1 11
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
10/26/2012 Section 4.3 v5.0.1 12
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
10/26/2012 Section 4.3 v5.0.1 13
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
10/26/2012 Section 4.3 v5.0.1 14
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?
10/26/2012 Section 4.3 v5.0.1 15
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 !
10/26/2012 Section 4.3 v5.0.1 16
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
10/26/2012 Section 4.3 v5.0.1 17
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
10/26/2012 Section 4.3 v5.0.1 18
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
10/26/2012 Section 4.3 v5.0.1 19
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
10/26/2012 Section 4.3 v5.0.1 20
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 )
10/26/2012 Section 4.3 v5.0.1 21
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 ?
10/26/2012 Section 4.3 v5.0.1 22
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 ?
10/26/2012 Section 4.3 v5.0.1 23
Think about it !
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