Transcript Math 1314 College Algebra

Module 4
Section 4.1
Exponential Functions
and Models
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
n – 1 Terms

Define function f(n) by: f(n) = an = a1 + (n – 1)2
Let k = n – 1
k=n–1
so that g(k) = a1 + 2k = 2k + a1 = 3 + k(2) = 2k + 3
A Linear Function
Question:
How do we know this is linear ?

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Section 4.3 v5.0.1
2
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



n – 1 Factors

Define function f(n) by: f(n) = an = a1 rn–1 = 3  2n–1
k=n–1

Letting k = n – 1 this becomes g(k) = a1rk = 3  2k
An Exponential Function
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3
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 : x a
Base x
Exponent a
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 :
f(x) = Cax
where C is the constant coefficient , C > 0 , and base is a
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4
Function Comparisons
f(x) = 2x
f(x) = 2x
f(x) = x2
y
y
y
x
Linear Function
x
x
Exponential Function
Power Function
x –3 –2 –1 0 1 2 3
2x –6 –4 –2 0 2 4 6
x –3 –2 –1 0 1 2 3
2x ⅛ ¼ ½ 1 2 4 8
x –3 –2 –1 0 1 2 3
x2 9 4 1 0 1 4 9
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
y
f(x) = C2x
g(x) = f(–x) = Ca–x , 1 < a
... a reflection of f(x) ... OR
1
h(x) = Cbx , 0 < b < 1 , b =
(0, C)
●
x
a
Domain = R
Range = { x  x > 0 }
As ordered pairs (C = 1) :
f = { (x, 2x)  x  R }
g = { (x, 2–x )  x  R }
Questions:
Intercepts ?
Asymptotes ?
In tabular form (C = 1) :
x –3 –2 –1 0 1 2 3
2x ⅛ ¼ ½ 1 2 4 8
2–x 8 4 2 1 ½ ¼ ⅛
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g(x) = f(–x) = C2–x
Section 4.3 v5.0.1
Effects of larger/smaller a ?
a>1
Growth factor a ?
Decay factor a ? 0 < a < 1
6
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 } = ( 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
WHY ?
Graphs of ax and bx intersect at (0, 1)

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If ax = ay then x = y
WHY ?
Section 4.3 v5.0.1
7
Exponential Equations

Solve


1. 25x = 125
(52)x = 53
52x = 53
2x = 3 WHY ?
x = 3/2 Solution set is {
3
2
}
2. 9x – 2 = 27x
(32)x – 2 = (33)x
32x – 4 = 33x
2x – 4 = 3x WHY ?
x = –4 Solution set is { –4 }
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Exponential Decay

Radioactive Decay

Radioactive isotopes of some elements such as
14C , 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


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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
… or just A0(½)
A(2k) = (½)A(k) = (A0(½))(½) = A0(½)2
A(3k) = (½)A(2k) = (A0(½)2)(½) = A0(½)3
A(4k) = A0(½)4 , … , A(nk) = A0(½)n , …

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
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A(x) = A0(½)x/k
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10
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
a = (½)1/k
Hence
f(x) = Cax = C((½)1/k)x = C(½)x/k
Question: What does this look like graphically ?
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Exponential Decay

Radioactive Decay Graph
A(t)
Let :
A(t) = amount at time t
A0

A0 = initial amount
A0 = A(0)
k = half-life
A(t) = A0(½)t/k
Amount is reduced by
half in each half-life
1
2
A0
1
4
A0

After n half-lives t = nk
A(t) = A0(½)n
Since n = t/k
A(t) = A0
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(½)t/k
1
32
1
64
A0
A0


1
A0
8
1
A
16 0
0
Section 4.3 v5.0.1
k
2k
3k

4k


5k
6k
t
12
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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Compound Interest in General
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:
A=P
A = P + Pr = P(1 + r)1
A = P(1 + r) + P(1 + r)r
= P(1 + r)(1 + r) = P(1 + r)2
A = P(1 + r)2 + P(1 + r)2r
= P(1 + r)2(1 + r) = P(1 + r)3
A = P(1 + r)t
... well ...
Now
Is this
isobvious?
this obvious?
Question: What if compounding is quarterly ?
What if compounding is n times per year ?
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Compound Interest in General
Compounding n times per year we annualize interest to r/n
A=P
0 period:
A = P + P(r/n) = P(1 + r/n)1
1 period:
2 periods: A = P(1 + r/n) + P(1 + r/n)(r/n) = P(1 + r/n)2
3 periods: A = P(1 + r/n)2 + P(1 + r/n)2(r/n) = P(1 + r/n)3
k periods: A = P(1 + r/n)k
In t years k = nt
A = P(1 + r/n)nt IsAh,
... consider
...
now
it’s obvious
!
this
obvious?
t years:
Example: $1000 for 20 years at 5% compounded quarterly
Here P = 1000, r = .05, n = 4 and t = 20
A = 1000(1 + .05/4)4(20) = 1000(1.0125)80 = 1000(2.701484941)
≈ $2,701.48
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Natural Exponential Function
Compute the first few terms of the sequence an =
n
an
1
2
3
4
5
6
10
100
200
400
2000
10000
100000
1000000
2.000000000
2.250000000
2.370370370
2.441406250
2.488320000
2.521626372
2.593742460
2.704813829
2.711517123
2.714891744
2.717602569
2.717942121
2.718268237
2.718280469
1
1+ n
(
)
n
Question:
Does an approach a value as n
∞?
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 exponential
functions ... and natural logarithms
Natural exponential functions are of form
f(x) = ex
?
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Natural Exponential Function
Graph of f(x) =
x
ex

ex
5
1
2.7182 81828
2
7.3890 56098
3
20.0855 36923
4
54.5981 50033
5
148.4131 59105
6
403.4287 93492
7
1096.6331 58428
8
2980.9579 87041
9
8103.0839 27575
10
22026.4657 94806
11
59874.1417 15197
12 162754.7914 19003
13 442413.3920 08920
14 1202604.2841 64776
?
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ex
ex

1200
f(x) = ex
4

3

1000
2
800

600
–1

1


x
0
1
2
3

400
200





1
2
3
Section 4.3 v5.0.1

4
5
6
7
x
17
Continuous Compounding
1
We have shown that 1 + n
1 nx
1
Thus 1 +
= 1+
n
n
(
)
(
((
n
)
))
n
e as n
x
e x as n
∞
1
and n
∞
0
1
and n
0
Recall that amount P compounded n times per year at
annual interest rate r for t years is given by
Then
A = P(1 + r/n)nt
A = P(1 +
= P((1 + r/n)(n/r))rt
r/n)(n/r)rt
Pert
What does this mean ?
As the number of compounding periods per year (n) increases
periodic compounding approaches continuous compounding
Thus an amount P compounded continuously for t years at
annualized interest rate r yields amount A given by
A = Pert
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Continuous Compounding
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
With simple
annual interest
The accumulated amount is
For 20 years this would be:
For 30 years this would be:
For 40 years this would be:
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$1,648.72
$2,718.28
$4,481.69
$7,389.06
Section 4.3 v5.0.1
$1,628.89
$2,653.30
$4,321.94
$7,039.99
19
Regular Saving
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? 20 years? 30 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)
Geometric Series
( Ak = Sk+1 )
1 – (e.05/12)k+1
= A0
1 – e.05/12
At 10 years, k = 120, A120 = 3,925.44
At 20 years, k = 240, A240 = 10,356.18
At 30 years, k = 360, A360 = 20,958.68 At 40 years, k = 360, A480 = 38,439.26
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Section 4.3 v5.0.1
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Population Growth
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
How do we know this ?
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
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Population Growth
P(t) = P02kt
50k = 1
k = 1/50
P(t) = 150(2t/50)
Thus
In 1975, t = 25 so
P(25) = P02kt = (150)(225/50)) = 150  2 ≈ 212.1 M
When the population is 600 M we have
P(t) = 600 = 150(2t/50)
2t/50 = 600/150 = 4 = 22
Thus
t/50 = 2 and t = 100 years after 1950
Hence the population will be 600 M in the year 2050
Sound familiar ?
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Think about it !
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