lesson 5: continuity
DESCRIPTION
Continuous function have an important property that small changes in input do not produce large changes in output. The Intermediate Value Theorem shows that a continuous function takes all values between any two values. From this we know that your height and weight were once the same, and right now there are two points on opposite sides of the world with the same temperature!TRANSCRIPT
. . . . . .
Section1.5Continuity
V63.0121.006/016, CalculusI
February2, 2010
Announcements
I OfficeHours: M,W 1:30–2:30, R 9–10(CIWW 726)I WrittenAssignment#2dueThursdayI FirstQuiz: FridayFebruary12inrecitation(§§1.1–1.4)
. . . . . .
Hatsumon
Herearesomediscussionquestionstostart.
TrueorFalseAtsomepointinyourlifeyouwereexactlythreefeettall.
TrueorFalseAtsomepointinyourlifeyourheight(ininches)wasequaltoyourweight(inpounds).
TrueorFalseRightnowthereareapairofpointsonoppositesidesoftheworldmeasuringtheexactsametemperature.
. . . . . .
Hatsumon
Herearesomediscussionquestionstostart.
TrueorFalseAtsomepointinyourlifeyouwereexactlythreefeettall.
TrueorFalseAtsomepointinyourlifeyourheight(ininches)wasequaltoyourweight(inpounds).
TrueorFalseRightnowthereareapairofpointsonoppositesidesoftheworldmeasuringtheexactsametemperature.
. . . . . .
Hatsumon
Herearesomediscussionquestionstostart.
TrueorFalseAtsomepointinyourlifeyouwereexactlythreefeettall.
TrueorFalseAtsomepointinyourlifeyourheight(ininches)wasequaltoyourweight(inpounds).
TrueorFalseRightnowthereareapairofpointsonoppositesidesoftheworldmeasuringtheexactsametemperature.
. . . . . .
Outline
Continuity
TheIntermediateValueTheorem
BacktotheQuestions
. . . . . .
Recall: DirectSubstitutionProperty
Theorem(TheDirectSubstitutionProperty)If f isapolynomialorarationalfunctionand a isinthedomainoff, then
limx→a
f(x) = f(a)
Thispropertyissousefulit’sworthnaming.
. . . . . .
DefinitionofContinuity
Definition
I Let f beafunctiondefinednear a. Wesaythat f is continuousat aif
limx→a
f(x) = f(a).
I A function f iscontinuous ifitiscontinuousateverypointinitsdomain.
. .x
.y
.
.a
.f(a)
. . . . . .
DefinitionofContinuity
Definition
I Let f beafunctiondefinednear a. Wesaythat f is continuousat aif
limx→a
f(x) = f(a).
I A function f iscontinuous ifitiscontinuousateverypointinitsdomain.
. .x
.y
.
.a
.f(a)
. . . . . .
Scholium
DefinitionLet f beafunctiondefinednear a. Wesaythat f is continuousata if
limx→a
f(x) = f(a).
Therearethreeimportantpartstothisdefinition.I Thefunctionhastohavealimitat a,I thefunctionhastohaveavalueat a,I andthesevalueshavetoagree.
. . . . . .
FreeTheorems
Theorem
(a) Anypolynomialiscontinuouseverywhere; thatis, itiscontinuouson R = (−∞,∞).
(b) Anyrationalfunctioniscontinuouswhereveritisdefined;thatis, itiscontinuousonitsdomain.
. . . . . .
Showingafunctioniscontinuous
ExampleLet f(x) =
√4x+ 1. Showthat f iscontinuousat 2.
SolutionWewanttoshowthat lim
x→2f(x) = f(2). Wehave
limx→a
f(x) = limx→2
√4x+ 1 =
√limx→2
(4x+ 1) =√9 = 3 = f(2).
Eachstepcomesfromthelimitlaws.
QuestionAtwhichotherpointsis f continuous?
AnswerThefunction f iscontinuouson (−1/4,∞). Itis rightcontinuousat −1/4 since lim
x→−1/4+f(x) = f(−1/4).
. . . . . .
Showingafunctioniscontinuous
ExampleLet f(x) =
√4x+ 1. Showthat f iscontinuousat 2.
SolutionWewanttoshowthat lim
x→2f(x) = f(2). Wehave
limx→a
f(x) = limx→2
√4x+ 1 =
√limx→2
(4x+ 1) =√9 = 3 = f(2).
Eachstepcomesfromthelimitlaws.
QuestionAtwhichotherpointsis f continuous?
AnswerThefunction f iscontinuouson (−1/4,∞). Itis rightcontinuousat −1/4 since lim
x→−1/4+f(x) = f(−1/4).
. . . . . .
Showingafunctioniscontinuous
ExampleLet f(x) =
√4x+ 1. Showthat f iscontinuousat 2.
SolutionWewanttoshowthat lim
x→2f(x) = f(2). Wehave
limx→a
f(x) = limx→2
√4x+ 1 =
√limx→2
(4x+ 1) =√9 = 3 = f(2).
Eachstepcomesfromthelimitlaws.
QuestionAtwhichotherpointsis f continuous?
AnswerThefunction f iscontinuouson (−1/4,∞). Itis rightcontinuousat −1/4 since lim
x→−1/4+f(x) = f(−1/4).
. . . . . .
Atwhichotherpoints?
Forreference: f(x) =√4x+ 1
I If a > −1/4, then limx→a
(4x+ 1) = 4a+ 1 > 0, so
limx→a
f(x) = limx→a
√4x+ 1 =
√limx→a
(4x+ 1) =√4a+ 1 = f(a)
and f iscontinuousat a.
I If a = −1/4, then 4x+ 1 < 0 totheleftof a, whichmeans√4x+ 1 isundefined. Still,
limx→a+
f(x) = limx→a+
√4x+ 1 =
√limx→a+
(4x+ 1) =√0 = 0 = f(a)
so f iscontinuousontherightat a = −1/4.
. . . . . .
Atwhichotherpoints?
Forreference: f(x) =√4x+ 1
I If a > −1/4, then limx→a
(4x+ 1) = 4a+ 1 > 0, so
limx→a
f(x) = limx→a
√4x+ 1 =
√limx→a
(4x+ 1) =√4a+ 1 = f(a)
and f iscontinuousat a.I If a = −1/4, then 4x+ 1 < 0 totheleftof a, whichmeans√
4x+ 1 isundefined. Still,
limx→a+
f(x) = limx→a+
√4x+ 1 =
√limx→a+
(4x+ 1) =√0 = 0 = f(a)
so f iscontinuousontherightat a = −1/4.
. . . . . .
Showingafunctioniscontinuous
ExampleLet f(x) =
√4x+ 1. Showthat f iscontinuousat 2.
SolutionWewanttoshowthat lim
x→2f(x) = f(2). Wehave
limx→a
f(x) = limx→2
√4x+ 1 =
√limx→2
(4x+ 1) =√9 = 3 = f(2).
Eachstepcomesfromthelimitlaws.
QuestionAtwhichotherpointsis f continuous?
AnswerThefunction f iscontinuouson (−1/4,∞).
Itis rightcontinuousat −1/4 since lim
x→−1/4+f(x) = f(−1/4).
. . . . . .
Showingafunctioniscontinuous
ExampleLet f(x) =
√4x+ 1. Showthat f iscontinuousat 2.
SolutionWewanttoshowthat lim
x→2f(x) = f(2). Wehave
limx→a
f(x) = limx→2
√4x+ 1 =
√limx→2
(4x+ 1) =√9 = 3 = f(2).
Eachstepcomesfromthelimitlaws.
QuestionAtwhichotherpointsis f continuous?
AnswerThefunction f iscontinuouson (−1/4,∞). Itis rightcontinuousat −1/4 since lim
x→−1/4+f(x) = f(−1/4).
. . . . . .
TheLimitLawsgiveContinuityLaws
TheoremIf f(x) and g(x) arecontinuousat a and c isaconstant, thenthefollowingfunctionsarealsocontinuousat a:
I (f+ g)(x)I (f− g)(x)I (cf)(x)I (fg)(x)
Ifg(x) (if g(a) ̸= 0)
. . . . . .
Whyasumofcontinuousfunctionsiscontinuous
Wewanttoshowthat
limx→a
(f+ g)(x) = (f+ g)(a).
Wejustfollowournose:
limx→a
(f+ g)(x) = limx→a
[f(x) + g(x)] (defof f+ g)
= limx→a
f(x) + limx→a
g(x) (iftheselimitsexist)
= f(a) + g(a) (theydo; f and g arects.)
= (f+ g)(a) (defof f+ g again)
. . . . . .
Trigonometricfunctionsarecontinuous
I sin and cos arecontinuouson R.
I tan =sincos
and sec =1cos
arecontinuousontheirdomain, whichisR \
{ π
2+ kπ
∣∣∣ k ∈ Z}.
I cot =cossin
and csc =1sin
arecontinuousontheirdomain, whichisR \ { kπ | k ∈ Z }.
..sin
.cos
.tan .sec
.cot .csc
. . . . . .
Trigonometricfunctionsarecontinuous
I sin and cos arecontinuouson R.
I tan =sincos
and sec =1cos
arecontinuousontheirdomain, whichisR \
{ π
2+ kπ
∣∣∣ k ∈ Z}.
I cot =cossin
and csc =1sin
arecontinuousontheirdomain, whichisR \ { kπ | k ∈ Z }.
..sin
.cos
.tan .sec
.cot .csc
. . . . . .
Trigonometricfunctionsarecontinuous
I sin and cos arecontinuouson R.
I tan =sincos
and sec =1cos
arecontinuousontheirdomain, whichisR \
{ π
2+ kπ
∣∣∣ k ∈ Z}.
I cot =cossin
and csc =1sin
arecontinuousontheirdomain, whichisR \ { kπ | k ∈ Z }.
..sin
.cos
.tan
.sec
.cot .csc
. . . . . .
Trigonometricfunctionsarecontinuous
I sin and cos arecontinuouson R.
I tan =sincos
and sec =1cos
arecontinuousontheirdomain, whichisR \
{ π
2+ kπ
∣∣∣ k ∈ Z}.
I cot =cossin
and csc =1sin
arecontinuousontheirdomain, whichisR \ { kπ | k ∈ Z }.
..sin
.cos
.tan .sec
.cot .csc
. . . . . .
Trigonometricfunctionsarecontinuous
I sin and cos arecontinuouson R.
I tan =sincos
and sec =1cos
arecontinuousontheirdomain, whichisR \
{ π
2+ kπ
∣∣∣ k ∈ Z}.
I cot =cossin
and csc =1sin
arecontinuousontheirdomain, whichisR \ { kπ | k ∈ Z }.
..sin
.cos
.tan .sec
.cot
.csc
. . . . . .
Trigonometricfunctionsarecontinuous
I sin and cos arecontinuouson R.
I tan =sincos
and sec =1cos
arecontinuousontheirdomain, whichisR \
{ π
2+ kπ
∣∣∣ k ∈ Z}.
I cot =cossin
and csc =1sin
arecontinuousontheirdomain, whichisR \ { kπ | k ∈ Z }.
..sin
.cos
.tan .sec
.cot .csc
. . . . . .
ExponentialandLogarithmicfunctionsarecontinuous
Foranybase a > 1,
I thefunction x 7→ ax iscontinuouson R
I thefunction loga iscontinuousonitsdomain: (0,∞)
I Inparticular ex andln = loge arecontinuousontheirdomains
.
.ax
.loga x
. . . . . .
ExponentialandLogarithmicfunctionsarecontinuous
Foranybase a > 1,
I thefunction x 7→ ax iscontinuouson R
I thefunction loga iscontinuousonitsdomain: (0,∞)
I Inparticular ex andln = loge arecontinuousontheirdomains
.
.ax
.loga x
. . . . . .
ExponentialandLogarithmicfunctionsarecontinuous
Foranybase a > 1,
I thefunction x 7→ ax iscontinuouson R
I thefunction loga iscontinuousonitsdomain: (0,∞)
I Inparticular ex andln = loge arecontinuousontheirdomains
.
.ax
.loga x
. . . . . .
InversetrigonometricfunctionsaremostlycontinuousI sin−1 and cos−1 arecontinuouson (−1,1), leftcontinuousat 1, andrightcontinuousat −1.
I sec−1 and csc−1 arecontinuouson (−∞,−1) ∪ (1,∞), leftcontinuousat −1, andrightcontinuousat 1.
I tan−1 and cot−1 arecontinuouson R.
.
.−π
.−π/2
.π/2
.π
.
.
.sin−1
.
.
.cos−1.sec−1
.
..csc−1
.
.
.tan−1
.cot−1
. . . . . .
InversetrigonometricfunctionsaremostlycontinuousI sin−1 and cos−1 arecontinuouson (−1,1), leftcontinuousat 1, andrightcontinuousat −1.
I sec−1 and csc−1 arecontinuouson (−∞,−1) ∪ (1,∞), leftcontinuousat −1, andrightcontinuousat 1.
I tan−1 and cot−1 arecontinuouson R.
.
.−π
.−π/2
.π/2
.π
.
.
.sin−1.
.
.cos−1
.sec−1
.
..csc−1
.
.
.tan−1
.cot−1
. . . . . .
InversetrigonometricfunctionsaremostlycontinuousI sin−1 and cos−1 arecontinuouson (−1,1), leftcontinuousat 1, andrightcontinuousat −1.
I sec−1 and csc−1 arecontinuouson (−∞,−1) ∪ (1,∞), leftcontinuousat −1, andrightcontinuousat 1.
I tan−1 and cot−1 arecontinuouson R.
.
.−π
.−π/2
.π/2
.π
.
.
.sin−1.
.
.cos−1.sec−1
.
.
.csc−1
.
.
.tan−1
.cot−1
. . . . . .
InversetrigonometricfunctionsaremostlycontinuousI sin−1 and cos−1 arecontinuouson (−1,1), leftcontinuousat 1, andrightcontinuousat −1.
I sec−1 and csc−1 arecontinuouson (−∞,−1) ∪ (1,∞), leftcontinuousat −1, andrightcontinuousat 1.
I tan−1 and cot−1 arecontinuouson R.
.
.−π
.−π/2
.π/2
.π
.
.
.sin−1.
.
.cos−1.sec−1
.
..csc−1
.
.
.tan−1
.cot−1
. . . . . .
InversetrigonometricfunctionsaremostlycontinuousI sin−1 and cos−1 arecontinuouson (−1,1), leftcontinuousat 1, andrightcontinuousat −1.
I sec−1 and csc−1 arecontinuouson (−∞,−1) ∪ (1,∞), leftcontinuousat −1, andrightcontinuousat 1.
I tan−1 and cot−1 arecontinuouson R.
.
.−π
.−π/2
.π/2
.π
.
.
.sin−1.
.
.cos−1.sec−1
.
..csc−1
.
.
.tan−1
.cot−1
. . . . . .
InversetrigonometricfunctionsaremostlycontinuousI sin−1 and cos−1 arecontinuouson (−1,1), leftcontinuousat 1, andrightcontinuousat −1.
I sec−1 and csc−1 arecontinuouson (−∞,−1) ∪ (1,∞), leftcontinuousat −1, andrightcontinuousat 1.
I tan−1 and cot−1 arecontinuouson R.
.
.−π
.−π/2
.π/2
.π
.
.
.sin−1.
.
.cos−1.sec−1
.
..csc−1
.
.
.tan−1
.cot−1
. . . . . .
Whatcouldgowrong?
Inwhatwayscouldafunction f failtobecontinuousatapoint a?Lookagainatthedefinition:
limx→a
f(x) = f(a)
. . . . . .
ContinuityFAIL
:Thelimitdoesnotexist
ExampleLet
f(x) =
{x2 if 0 ≤ x ≤ 1
2x if 1 < x ≤ 2
Atwhichpointsis f continuous?
SolutionAtanypoint a in [0, 2] besides 1, lim
x→af(x) = f(a) because f is
representedbyapolynomialnear a, andpolynomialshavethedirectsubstitutionproperty. However,
limx→1−
f(x) = limx→1−
x2 = 12 = 1
limx→1+
f(x) = limx→1+
2x = 2(1) = 2
So f hasnolimitat 1. Therefore f isnotcontinuousat 1.
. . . . . .
ContinuityFAIL:Thelimitdoesnotexist
ExampleLet
f(x) =
{x2 if 0 ≤ x ≤ 1
2x if 1 < x ≤ 2
Atwhichpointsis f continuous?
SolutionAtanypoint a in [0, 2] besides 1, lim
x→af(x) = f(a) because f is
representedbyapolynomialnear a, andpolynomialshavethedirectsubstitutionproperty. However,
limx→1−
f(x) = limx→1−
x2 = 12 = 1
limx→1+
f(x) = limx→1+
2x = 2(1) = 2
So f hasnolimitat 1. Therefore f isnotcontinuousat 1.
. . . . . .
GraphicalIllustrationofPitfall#1
. .x
.y
..−1
..1
..2
..−1
..1
..2
..3
..4
.
.
.
.
. . . . . .
ContinuityFAIL
:Thefunctionhasnovalue
ExampleLet
f(x) =x2 + 2x+ 1
x+ 1
Atwhichpointsis f continuous?
SolutionBecause f isrational, itiscontinuousonitswholedomain. Notethat −1 isnotinthedomainof f, so f isnotcontinuousthere.
. . . . . .
ContinuityFAIL:Thefunctionhasnovalue
ExampleLet
f(x) =x2 + 2x+ 1
x+ 1
Atwhichpointsis f continuous?
SolutionBecause f isrational, itiscontinuousonitswholedomain. Notethat −1 isnotinthedomainof f, so f isnotcontinuousthere.
. . . . . .
GraphicalIllustrationofPitfall#2
. .x
.y
...−1
. .1
f cannotbecontinuouswhereithasnovalue.
. . . . . .
ContinuityFAIL
:functionvalue ̸= limit
ExampleLet
f(x) =
{7 if x ̸= 1
π if x = 1
Atwhichpointsis f continuous?
Solutionf isnotcontinuousat 1 because f(1) = π but lim
x→1f(x) = 7.
. . . . . .
ContinuityFAIL:functionvalue ̸= limit
ExampleLet
f(x) =
{7 if x ̸= 1
π if x = 1
Atwhichpointsis f continuous?
Solutionf isnotcontinuousat 1 because f(1) = π but lim
x→1f(x) = 7.
. . . . . .
GraphicalIllustrationofPitfall#3
. .x
.y
..π
..7
..1
.
.
. . . . . .
Specialtypesofdiscontinuites
removablediscontinuity Thelimit limx→a
f(x) exists, but f isnot
definedat a oritsvalueat a isnotequaltothelimitat a.
Byre-defining f(a) = limx→a
f(x), f canbemade
continuousat a
jumpdiscontinuity Thelimits limx→a−
f(x) and limx→a+
f(x) exist, but
aredifferent.
Thefunctioncannotbemadecontinuousbychangingasinglevalue.
. . . . . .
Graphicalrepresentationsofdiscontinuities
. .x
.y
..π
..7
..1
.
.
..Presto! continuous!
removable
. .x
.y
..1
..1
..2
.
.
.
.
. .continuous?.
. .continuous?
. .continuous?
jump
. . . . . .
Graphicalrepresentationsofdiscontinuities
. .x
.y
..π
..7
..1
.
.
..Presto! continuous!
removable
. .x
.y
..1
..1
..2
.
.
.
.
. .continuous?.
. .continuous?
. .continuous?
jump
. . . . . .
Graphicalrepresentationsofdiscontinuities
. .x
.y
..π
..7
..1
.
.
..Presto! continuous!
removable
. .x
.y
..1
..1
..2
.
.
.
.
. .continuous?
.
. .continuous?
. .continuous?
jump
. . . . . .
Graphicalrepresentationsofdiscontinuities
. .x
.y
..π
..7
..1
.
.
..Presto! continuous!
removable
. .x
.y
..1
..1
..2
.
.
.
.
. .continuous?
.
. .continuous?
. .continuous?
jump
. . . . . .
Graphicalrepresentationsofdiscontinuities
. .x
.y
..π
..7
..1
.
.
..Presto! continuous!
removable
. .x
.y
..1
..1
..2
.
.
.
.
. .continuous?
.
. .continuous?
. .continuous?
jump
. . . . . .
Specialtypesofdiscontinuites
removablediscontinuity Thelimit limx→a
f(x) exists, but f isnot
definedat a oritsvalueat a isnotequaltothelimitat a. Byre-defining f(a) = lim
x→af(x), f canbemade
continuousat a
jumpdiscontinuity Thelimits limx→a−
f(x) and limx→a+
f(x) exist, but
aredifferent.
Thefunctioncannotbemadecontinuousbychangingasinglevalue.
. . . . . .
Specialtypesofdiscontinuites
removablediscontinuity Thelimit limx→a
f(x) exists, but f isnot
definedat a oritsvalueat a isnotequaltothelimitat a. Byre-defining f(a) = lim
x→af(x), f canbemade
continuousat a
jumpdiscontinuity Thelimits limx→a−
f(x) and limx→a+
f(x) exist, but
aredifferent. Thefunctioncannotbemadecontinuousbychangingasinglevalue.
. . . . . .
Thegreatestintegerfunction
[[x]] isthegreatestinteger ≤ x.
x [[x]]0 01 1
1.5 11.9 12.1 2
−0.5 −1−0.9 −1−1.1 −2
. .x
.y
..−2
..−2
..−1
..−1
..1
..1
..2
..2
..3
..3
. .
. .
. .
. .
. ..y = [[x]]
Thisfunctionhasajumpdiscontinuityateachinteger.
. . . . . .
Thegreatestintegerfunction
[[x]] isthegreatestinteger ≤ x.
x [[x]]0 01 1
1.5 11.9 12.1 2
−0.5 −1−0.9 −1−1.1 −2
. .x
.y
..−2
..−2
..−1
..−1
..1
..1
..2
..2
..3
..3
. .
. .
. .
. .
. ..y = [[x]]
Thisfunctionhasajumpdiscontinuityateachinteger.
. . . . . .
Outline
Continuity
TheIntermediateValueTheorem
BacktotheQuestions
. . . . . .
A BigTimeTheorem
Theorem(TheIntermediateValueTheorem)Supposethat f iscontinuousontheclosedinterval [a,b] andlet Nbeanynumberbetween f(a) and f(b), where f(a) ̸= f(b). Thenthereexistsanumber c in (a,b) suchthat f(c) = N.
. . . . . .
IllustratingtheIVT
Supposethat f iscontinuousontheclosedinterval [a,b] andlet Nbeanynumberbetween f(a) and f(b), where f(a) ̸= f(b). Thenthereexistsanumber c in (a,b) suchthat f(c) = N.
. .x
.f(x)
.
.
.a .b
.f(a)
.f(b)
.N
.c
..
.c1
.
.c2
.
.c3
. . . . . .
IllustratingtheIVTSupposethat f iscontinuousontheclosedinterval [a,b]
andlet Nbeanynumberbetween f(a) and f(b), where f(a) ̸= f(b). Thenthereexistsanumber c in (a,b) suchthat f(c) = N.
. .x
.f(x)
.
.
.a .b
.f(a)
.f(b)
.N
.c
..
.c1
.
.c2
.
.c3
. . . . . .
IllustratingtheIVTSupposethat f iscontinuousontheclosedinterval [a,b]
andlet Nbeanynumberbetween f(a) and f(b), where f(a) ̸= f(b). Thenthereexistsanumber c in (a,b) suchthat f(c) = N.
. .x
.f(x)
.
.
.a .b
.f(a)
.f(b)
.N
.c
..
.c1
.
.c2
.
.c3
. . . . . .
IllustratingtheIVTSupposethat f iscontinuousontheclosedinterval [a,b] andlet Nbeanynumberbetween f(a) and f(b), where f(a) ̸= f(b).
Thenthereexistsanumber c in (a,b) suchthat f(c) = N.
. .x
.f(x)
.
.
.a .b
.f(a)
.f(b)
.N
.c
..
.c1
.
.c2
.
.c3
. . . . . .
IllustratingtheIVTSupposethat f iscontinuousontheclosedinterval [a,b] andlet Nbeanynumberbetween f(a) and f(b), where f(a) ̸= f(b). Thenthereexistsanumber c in (a,b) suchthat f(c) = N.
. .x
.f(x)
.
.
.a .b
.f(a)
.f(b)
.N
.c
.
.
.c1
.
.c2
.
.c3
. . . . . .
IllustratingtheIVTSupposethat f iscontinuousontheclosedinterval [a,b] andlet Nbeanynumberbetween f(a) and f(b), where f(a) ̸= f(b). Thenthereexistsanumber c in (a,b) suchthat f(c) = N.
. .x
.f(x)
.
.
.a .b
.f(a)
.f(b)
.N
.c
..
.c1
.
.c2
.
.c3
. . . . . .
IllustratingtheIVTSupposethat f iscontinuousontheclosedinterval [a,b] andlet Nbeanynumberbetween f(a) and f(b), where f(a) ̸= f(b). Thenthereexistsanumber c in (a,b) suchthat f(c) = N.
. .x
.f(x)
.
.
.a .b
.f(a)
.f(b)
.N
.c
.
.
.c1
.
.c2
.
.c3
. . . . . .
WhattheIVT doesnotsay
TheIntermediateValueTheoremisan“existence”theorem.I Itdoesnotsayhowmanysuch c exist.I Italsodoesnotsayhowtofind c.
Still, itcanbeusediniterationorinconjunctionwithothertheoremstoanswerthesequestions.
. . . . . .
UsingtheIVT
ExampleSupposeweareunawareofthesquarerootfunctionandthatit’scontinuous. Provethatthesquarerootoftwoexists.
Proof.Let f(x) = x2, acontinuousfunctionon [1, 2]. Note f(1) = 1 andf(2) = 4. Since 2 isbetween 1 and 4, thereexistsapoint c in(1, 2) suchthat
f(c) = c2 = 2.
Infact, wecan“narrowin”onthesquarerootof 2 by themethodofbisections.
. . . . . .
UsingtheIVT
ExampleSupposeweareunawareofthesquarerootfunctionandthatit’scontinuous. Provethatthesquarerootoftwoexists.
Proof.Let f(x) = x2, acontinuousfunctionon [1, 2].
Note f(1) = 1 andf(2) = 4. Since 2 isbetween 1 and 4, thereexistsapoint c in(1, 2) suchthat
f(c) = c2 = 2.
Infact, wecan“narrowin”onthesquarerootof 2 by themethodofbisections.
. . . . . .
UsingtheIVT
ExampleSupposeweareunawareofthesquarerootfunctionandthatit’scontinuous. Provethatthesquarerootoftwoexists.
Proof.Let f(x) = x2, acontinuousfunctionon [1, 2]. Note f(1) = 1 andf(2) = 4. Since 2 isbetween 1 and 4, thereexistsapoint c in(1, 2) suchthat
f(c) = c2 = 2.
Infact, wecan“narrowin”onthesquarerootof 2 by themethodofbisections.
. . . . . .
UsingtheIVT
ExampleSupposeweareunawareofthesquarerootfunctionandthatit’scontinuous. Provethatthesquarerootoftwoexists.
Proof.Let f(x) = x2, acontinuousfunctionon [1, 2]. Note f(1) = 1 andf(2) = 4. Since 2 isbetween 1 and 4, thereexistsapoint c in(1, 2) suchthat
f(c) = c2 = 2.
Infact, wecan“narrowin”onthesquarerootof 2 by themethodofbisections.
. . . . . .
Finding√2 bybisections
.
.x .f(x) = x2
..1 .1
..2 .4
..1.5 .2.25
..1.25 .1.5625
..1.375 .1.890625
..1.4375 .2.06640625
. . . . . .
Finding√2 bybisections
.
.x .f(x) = x2
..1 .1
..2 .4
..1.5 .2.25
..1.25 .1.5625
..1.375 .1.890625
..1.4375 .2.06640625
. . . . . .
Finding√2 bybisections
.
.x .f(x) = x2
..1 .1
..2 .4
..1.5 .2.25
..1.25 .1.5625
..1.375 .1.890625
..1.4375 .2.06640625
. . . . . .
Finding√2 bybisections
.
.x .f(x) = x2
..1 .1
..2 .4
..1.5 .2.25
..1.25 .1.5625
..1.375 .1.890625
..1.4375 .2.06640625
. . . . . .
Finding√2 bybisections
.
.x .f(x) = x2
..1 .1
..2 .4
..1.5 .2.25
..1.25 .1.5625
..1.375 .1.890625
..1.4375 .2.06640625
. . . . . .
Finding√2 bybisections
.
.x .f(x) = x2
..1 .1
..2 .4
..1.5 .2.25
..1.25 .1.5625
..1.375 .1.890625
..1.4375 .2.06640625
. . . . . .
UsingtheIVT
ExampleLet f(x) = x3 − x− 1. Showthatthereisazerofor f.
Solutionf(1) = −1 and f(2) = 5. Sothereisazerobetween 1 and 2.
(Morecarefulanalysisyields 1.32472.)
. . . . . .
UsingtheIVT
ExampleLet f(x) = x3 − x− 1. Showthatthereisazerofor f.
Solutionf(1) = −1 and f(2) = 5. Sothereisazerobetween 1 and 2.(Morecarefulanalysisyields 1.32472.)
. . . . . .
Outline
Continuity
TheIntermediateValueTheorem
BacktotheQuestions
. . . . . .
BacktotheQuestions
TrueorFalseAtonepointinyourlifeyouwereexactlythreefeettall.
TrueorFalseAtonepointinyourlifeyourheightininchesequaledyourweightinpounds.
TrueorFalseRightnowtherearetwopointsonoppositesidesoftheEarthwithexactlythesametemperature.
. . . . . .
Question1: True!
I Let h(t) beheight, whichvariescontinuouslyovertime.I Then h(birth) < 3 ft and h(now) > 3 ft.I SobytheIVT thereisapoint c in (birth, now) where
h(c) = 3.
. . . . . .
BacktotheQuestions
TrueorFalseAtonepointinyourlifeyouwereexactlythreefeettall.
TrueorFalseAtonepointinyourlifeyourheightininchesequaledyourweightinpounds.
TrueorFalseRightnowtherearetwopointsonoppositesidesoftheEarthwithexactlythesametemperature.
. . . . . .
Question2: True!
I Let h(t) beheightininchesand w(t) beweightinpounds,bothvaryingcontinuouslyovertime.
I Let f(t) = h(t)−w(t).I Formostofus(callyourmom), f(birth) > 0 and f(now) < 0.I SobytheIVT thereisapoint c in (birth, now) where
f(c) = 0.I Inotherwords,
h(c)−w(c) = 0 ⇐⇒ h(c) = w(c).
. . . . . .
BacktotheQuestions
TrueorFalseAtonepointinyourlifeyouwereexactlythreefeettall.
TrueorFalseAtonepointinyourlifeyourheightininchesequaledyourweightinpounds.
TrueorFalseRightnowtherearetwopointsonoppositesidesoftheEarthwithexactlythesametemperature.
. . . . . .
Question3
I Let T(θ) bethetemperatureatthepointontheequatoratlongitude θ.
I Howcanyouexpressthestatementthatthetemperatureonoppositesidesisthesame?
I Howcanyouensurethisistrue?
. . . . . .
Question3: True!
I Let f(θ) = T(θ)− T(θ + 180◦)I Then
f(0) = T(0)− T(180)
whilef(180) = T(180)− T(360) = −f(0)
I Sosomewherebetween 0 and 180 thereisapoint θ wheref(θ) = 0!
. . . . . .
Whathavewelearnedtoday?
I Definition: afunctioniscontinuousatapointifthelimitofthefunctionatthatpointagreeswiththevalueofthefunctionatthatpoint.
I Weoftenmakeafundamentalassumptionthatfunctionswemeetinnaturearecontinuous.
I TheIntermediateValueTheoremisabasicpropertyofrealnumbersthatweneedandusealot.