Lesson 14: Exponential Functions

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Section3.1ExponentialFunctions

V63.0121.034, CalculusI

October19, 2009

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Outline

Definitionofexponentialfunctions

PropertiesofexponentialFunctions

Thenumber e andthenaturalexponentialfunctionCompoundInterestThenumber eA limit

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DefinitionIf a isarealnumberand n isapositivewholenumber, then

an = a · a · · · · · a︸︷︷︸n factors

Examples

I 23 = 2 · 2 · 2 = 8I 34 = 3 · 3 · 3 · 3 = 81I (−1)5 = (−1)(−1)(−1)(−1)(−1) = −1

. . . . . .

FactIf a isarealnumber, then

I ax+y = axay

I ax−y =ax

ayI (ax)y = axy

I (ab)x = axbx

wheneverallexponentsarepositivewholenumbers.

Proof.Checkforyourself:

ax+y = a · a · · · · · a︸︷︷︸x + y factors

= a · a · · · · · a︸︷︷︸x factors

· a · a · · · · · a︸︷︷︸y factors

= axay

. . . . . .

Let’sbeconventional

I Thedesirethatthesepropertiesremaintruegivesusconventionsfor ax when x isnotapositivewholenumber.

I Forexample:

an = an+0 != ana0

DefinitionIf a ̸= 0, wedefine a0 = 1.

I Notice 00 remainsundefined(asalimitform, it’sindeterminate).

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Conventionsfornegativeexponents

If n ≥ 0, wewant

an · a−n != an+(−n) = a0 = 1

DefinitionIf n isapositiveinteger, wedefine a−n =

1an.

Fact

I Theconventionthat a−n =1an

“works”fornegative n aswell.

I If m and n areanyintegers, then am−n =am

an.

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Conventionsforfractionalexponents

If q isapositiveinteger, wewant

(a1/q)q!= a1 = a

DefinitionIf q isapositiveinteger, wedefine a1/q = q

√a. Wemusthave

a ≥ 0 if q iseven.

Fact

I Nowwecansay ap/q = (a1/q)p withoutambiguity

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Conventionsforirrationalpowers

I So ax iswell-definedif x isrational.I Whataboutirrationalpowers?

DefinitionLet a > 0. Then

ax = limr→x

r rational

ar

Inotherwords, toapproximate ax forirrational x, take r closeto xbutrationalandcompute ar.

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Graphsofvariousexponentialfunctions

. .x

.y

.y = 1x

.y = 2x.y = 3x.y = 10x .y = 1.5x.y = (1/2)x.y = (1/3)x .y = (1/10)x.y = (2/3)x

. . . . . .


. .x

.y

.y = 1x

.y = 2x

.y = 3x.y = 10x .y = 1.5x.y = (1/2)x.y = (1/3)x .y = (1/10)x.y = (2/3)x

. . . . . .


. .x

.y

.y = 1x

.y = 2x.y = 3x

.y = 10x .y = 1.5x.y = (1/2)x.y = (1/3)x .y = (1/10)x.y = (2/3)x

. . . . . .


. .x

.y

.y = 1x

.y = 2x.y = 3x.y = 10x

.y = 1.5x.y = (1/2)x.y = (1/3)x .y = (1/10)x.y = (2/3)x

. . . . . .


. .x

.y

.y = 1x

.y = 2x.y = 3x.y = 10x .y = 1.5x

.y = (1/2)x.y = (1/3)x .y = (1/10)x.y = (2/3)x

. . . . . .


. .x

.y

.y = 1x

.y = 2x.y = 3x.y = 10x .y = 1.5x.y = (1/2)x

.y = (1/3)x .y = (1/10)x.y = (2/3)x

. . . . . .


. .x

.y

.y = 1x

.y = 2x.y = 3x.y = 10x .y = 1.5x.y = (1/2)x.y = (1/3)x

.y = (1/10)x.y = (2/3)x

. . . . . .


. .x

.y

.y = 1x

.y = 2x.y = 3x.y = 10x .y = 1.5x.y = (1/2)x.y = (1/3)x .y = (1/10)x

.y = (2/3)x

. . . . . .


. .x

.y

.y = 1x


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Outline




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PropertiesofexponentialFunctionsTheoremIf a > 0 and a ̸= 1, then f(x) = ax isacontinuousfunctionwithdomain R andrange (0,∞). Inparticular, ax > 0 forall x. Ifa,b > 0 and x, y ∈ R, then

I ax+y = axay

I ax−y =ax

ay

negativeexponentsmeanreciprocals.

I (ax)y = axy

fractionalexponentsmeanroots

I (ab)x = axbx

Proof.

I Thisistrueforpositiveintegerexponentsbynaturaldefinition

I Ourconventionaldefinitionsmakethesetrueforrationalexponents

I Ourlimitdefinitionmaketheseforirrationalexponents, too

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I ax+y = axay

I ax−y =ax

aynegativeexponentsmeanreciprocals.

I (ax)y = axy

fractionalexponentsmeanroots

I (ab)x = axbx

Proof.




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I ax+y = axay

I ax−y =ax

aynegativeexponentsmeanreciprocals.

I (ax)y = axy fractionalexponentsmeanrootsI (ab)x = axbx

Proof.




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ExampleSimplify: 82/3

Solution

I 82/3 =3√82 =

3√64 = 4

I Or,(

3√8)2

= 22 = 4.

Example

Simplify:

√8

21/2

Answer2

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Fact(Limitsofexponentialfunctions)

I If a > 1, thenlimx→∞

ax = ∞ and

limx→−∞

ax = 0

I If 0 < a < 1, thenlimx→∞

ax = 0 and

limx→−∞

ax = ∞ . .x

.y

.y = 1x


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Outline




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CompoundedInterest

QuestionSupposeyousave$100at10%annualinterest, withinterestcompoundedonceayear. Howmuchdoyouhave

I Afteroneyear?I Aftertwoyears?I after t years?

Answer

I $100 + 10% = $110I $110 + 10% = $110 + $11 = $121I $100(1.1)t.

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CompoundedInterest



Answer

I $100 + 10% = $110

I $110 + 10% = $110 + $11 = $121I $100(1.1)t.

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CompoundedInterest



Answer

I $100 + 10% = $110I $110 + 10% = $110 + $11 = $121

I $100(1.1)t.

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CompoundedInterest



Answer

I $100 + 10% = $110I $110 + 10% = $110 + $11 = $121I $100(1.1)t.

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CompoundedInterest: quarterly

QuestionSupposeyousave$100at10%annualinterest, withinterestcompounded fourtimes ayear. Howmuchdoyouhave


Answer

I $100(1.025)4 = $110.38, not $100(1.1)4!I $100(1.025)8 = $121.84I $100(1.025)4t.

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Answer

I $100(1.025)4 = $110.38,

not $100(1.1)4!I $100(1.025)8 = $121.84I $100(1.025)4t.

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Answer

I $100(1.025)4 = $110.38, not $100(1.1)4!

I $100(1.025)8 = $121.84I $100(1.025)4t.

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Answer

I $100(1.025)4 = $110.38, not $100(1.1)4!I $100(1.025)8 = $121.84

I $100(1.025)4t.

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Answer

I $100(1.025)4 = $110.38, not $100(1.1)4!I $100(1.025)8 = $121.84I $100(1.025)4t.

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CompoundedInterest: monthly

QuestionSupposeyousave$100at10%annualinterest, withinterestcompounded twelvetimes ayear. Howmuchdoyouhaveafter tyears?

Answer$100(1 + 10%/12)12t

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CompoundedInterest: general

QuestionSupposeyousave P atinterestrate r, withinterestcompounded ntimes ayear. Howmuchdoyouhaveafter t years?

Answer

B(t) = P(1 +

rn

)nt

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CompoundedInterest: continuous

QuestionSupposeyousave P atinterestrate r, withinterestcompoundedeveryinstant. Howmuchdoyouhaveafter t years?

Answer

B(t) = limn→∞

P(1 +

rn

)nt= lim

n→∞P

(1 +

1n

)rnt

= P[

limn→∞

(1 +

1n

)n

︸︷︷︸independentof P, r, or t

]rt

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Themagicnumber

Definition

e = limn→∞

(1 +

1n

)n

Sonowcontinuously-compoundedinterestcanbeexpressedas

B(t) = Pert.

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Existenceof eSeeAppendixB

I Wecanexperimentallyverifythatthisnumberexistsandis

e ≈ 2.718281828459045 . . .

I e isirrationalI e is transcendental

n(1 +

1n

)n

1 22 2.25

3 2.3703710 2.59374100 2.704811000 2.71692106 2.71828

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e ≈ 2.718281828459045 . . .


n(1 +

1n

)n

1 22 2.253 2.37037

10 2.59374100 2.704811000 2.71692106 2.71828

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e ≈ 2.718281828459045 . . .


n(1 +

1n

)n

1 22 2.253 2.3703710 2.59374

100 2.704811000 2.71692106 2.71828

. . . . . .



e ≈ 2.718281828459045 . . .


n(1 +

1n

)n

1 22 2.253 2.3703710 2.59374100 2.70481

1000 2.71692106 2.71828

. . . . . .



e ≈ 2.718281828459045 . . .


n(1 +

1n

)n

1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692

106 2.71828

. . . . . .



e ≈ 2.718281828459045 . . .


n(1 +

1n

)n

1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828

. . . . . .



e ≈ 2.718281828459045 . . .

I e isirrational

I e is transcendental

n(1 +

1n

)n

1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828

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e ≈ 2.718281828459045 . . .


n(1 +

1n

)n

1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828

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MeettheMathematician: LeonhardEuler

I BorninSwitzerland,livedinPrussia(Germany)andRussia

I Eyesighttroubleallhislife, blindfrom1766onward

I Hundredsofcontributionstocalculus, numbertheory,graphtheory, fluidmechanics, optics, andastronomy

LeonhardPaulEulerSwiss, 1707–1783

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A limit

Question

Whatis limh→0

eh − 1h

?

Answer

I If h issmallenough, e ≈ (1 + h)1/h. So

eh − 1h

≈ 1

I Infact, limh→0

eh − 1h

= 1.

I Thiscanbeusedtocharacterize e:

limh→0

2h − 1h

= 0.693 · · · < 1 and limh→0

3h − 1h

= 1.099 · · · < 1

Lesson 14: Exponential Functions

Education

Transcript of Lesson 14: Exponential Functions