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Section3.1ExponentialFunctions
V63.0121, CalculusI
March2/3/9, 2009
Announcements
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.
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“Isthereacurve?”
I MidtermI Meanwas70%and
standarddeviationwas12%
I Soscoresabove70%(52.5/75)aregood
I Scoresabove82%(61.5/75)aregreat
I Forfinallettergrades,refertosyllabus
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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
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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
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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
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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
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. . . . . .
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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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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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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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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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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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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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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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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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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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Outline
Definitionofexponentialfunctions
PropertiesofexponentialFunctions
Thenumber e andthenaturalexponentialfunctionCompoundInterestThenumber eA limit
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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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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
aynegativeexponentsmeanreciprocals.
I (ax)y = axy
fractionalexponentsmeanroots
I (ab)x = axbx
Proof.
I Thisistrueforpositiveintegerexponentsbynaturaldefinition
I Ourconventionaldefinitionsmakethesetrueforrationalexponents
I Ourlimitdefinitionmaketheseforirrationalexponents, too
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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
aynegativeexponentsmeanreciprocals.
I (ax)y = axy fractionalexponentsmeanrootsI (ab)x = axbx
Proof.
I Thisistrueforpositiveintegerexponentsbynaturaldefinition
I Ourconventionaldefinitionsmakethesetrueforrationalexponents
I Ourlimitdefinitionmaketheseforirrationalexponents, too
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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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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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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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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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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
.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
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Outline
Definitionofexponentialfunctions
PropertiesofexponentialFunctions
Thenumber e andthenaturalexponentialfunctionCompoundInterestThenumber eA limit
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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
QuestionSupposeyousave$100at10%annualinterest, withinterestcompoundedonceayear. Howmuchdoyouhave
I Afteroneyear?I Aftertwoyears?I after t years?
Answer
I $100 + 10% = $110
I $110 + 10% = $110 + $11 = $121I $100(1.1)t.
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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 = $121
I $100(1.1)t.
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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: quarterly
QuestionSupposeyousave$100at10%annualinterest, withinterestcompounded fourtimes ayear. Howmuchdoyouhave
I Afteroneyear?I Aftertwoyears?I after t years?
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: quarterly
QuestionSupposeyousave$100at10%annualinterest, withinterestcompounded fourtimes ayear. Howmuchdoyouhave
I Afteroneyear?I Aftertwoyears?I after t years?
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: quarterly
QuestionSupposeyousave$100at10%annualinterest, withinterestcompounded fourtimes ayear. Howmuchdoyouhave
I Afteroneyear?I Aftertwoyears?I after t years?
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: quarterly
QuestionSupposeyousave$100at10%annualinterest, withinterestcompounded fourtimes ayear. Howmuchdoyouhave
I Afteroneyear?I Aftertwoyears?I after t years?
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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CompoundedInterest: quarterly
QuestionSupposeyousave$100at10%annualinterest, withinterestcompounded fourtimes ayear. Howmuchdoyouhave
I Afteroneyear?I Aftertwoyears?I after t years?
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: 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: 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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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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Themagicnumber
Definition
e = limn→∞
(1 +
1n
)n
Sonowcontinuously-compoundedinterestcanbeexpressedas
B(t) = Pert.
Page 59
. . . . . .
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
Page 60
. . . . . .
Existenceof eSeeAppendixB
I Wecanexperimentallyverifythatthisnumberexistsandis
e ≈ 2.718281828459045 . . .
I e isirrationalI e is transcendental
n(1 +
1n
)n
1 22 2.253 2.37037
10 2.59374100 2.704811000 2.71692106 2.71828
Page 61
. . . . . .
Existenceof eSeeAppendixB
I Wecanexperimentallyverifythatthisnumberexistsandis
e ≈ 2.718281828459045 . . .
I e isirrationalI e is transcendental
n(1 +
1n
)n
1 22 2.253 2.3703710 2.59374
100 2.704811000 2.71692106 2.71828
Page 62
. . . . . .
Existenceof eSeeAppendixB
I Wecanexperimentallyverifythatthisnumberexistsandis
e ≈ 2.718281828459045 . . .
I e isirrationalI e is transcendental
n(1 +
1n
)n
1 22 2.253 2.3703710 2.59374100 2.70481
1000 2.71692106 2.71828
Page 63
. . . . . .
Existenceof eSeeAppendixB
I Wecanexperimentallyverifythatthisnumberexistsandis
e ≈ 2.718281828459045 . . .
I e isirrationalI e is transcendental
n(1 +
1n
)n
1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692
106 2.71828
Page 64
. . . . . .
Existenceof eSeeAppendixB
I Wecanexperimentallyverifythatthisnumberexistsandis
e ≈ 2.718281828459045 . . .
I e isirrationalI e is transcendental
n(1 +
1n
)n
1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828
Page 65
. . . . . .
Existenceof eSeeAppendixB
I Wecanexperimentallyverifythatthisnumberexistsandis
e ≈ 2.718281828459045 . . .
I e isirrationalI e is transcendental
n(1 +
1n
)n
1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828
Page 66
. . . . . .
Existenceof eSeeAppendixB
I Wecanexperimentallyverifythatthisnumberexistsandis
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
Page 67
. . . . . .
Existenceof eSeeAppendixB
I Wecanexperimentallyverifythatthisnumberexistsandis
e ≈ 2.718281828459045 . . .
I e isirrationalI e is transcendental
n(1 +
1n
)n
1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828
Page 68
. . . . . .
MeettheMathematician: LeonhardEuler
I BorninSwitzerland,livedinPrussia(Germany)andRussia
I Eyesighttroubleallhislife, blindfrom1766onward
I Hundredsofcontributionstocalculus, numbertheory,graphtheory, fluidmechanics, optics, andastronomy
LeonhardPaulEulerSwiss, 1707–1783
Page 69
. . . . . .
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
Page 70
. . . . . .
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
Page 71
. . . . . .
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