. . . . . .

Section1.5Continuity

V63.0121.027, CalculusI

September17, 2009

Announcements

I PleaseputyourhomeworkintheenvelopebylastnameI QuiznextweekinRecitationon§§1.1–1.3I WebAssignments1and2dueTuesday

. . . . . .

Hatsumon

Herearesomediscussionquestionstostart.I Wereyoueverexactlythreefeettall?

I Wasyourheight(ininches)everequaltoyourweight(inpounds)?

I Isthereapairofpointsonoppositesidesoftheworldatthesametemperatureatthesametime?

. . . . . .

Hatsumon

Herearesomediscussionquestionstostart.I Wereyoueverexactlythreefeettall?I Wasyourheight(ininches)everequaltoyourweight(inpounds)?

I Isthereapairofpointsonoppositesidesoftheworldatthesametemperatureatthesametime?

. . . . . .

Hatsumon

Herearesomediscussionquestionstostart.I Wereyoueverexactlythreefeettall?I Wasyourheight(ininches)everequaltoyourweight(inpounds)?

I Isthereapairofpointsonoppositesidesoftheworldatthesametemperatureatthesametime?

. . . . . .

Outline

Continuity

TheIntermediateValueTheorem

BacktotheQuestions

. . . . . .

Recall: DirectSubstitutionProperty

Theorem(TheDirectSubstitutionProperty)If f isapolynomialorarationalfunctionand a isinthedomainoff, then

limx→a

f(x) = 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)

Therearethreeimportantpartstothisdefinition.I ThelimithastoexistI thefunctionhastobedefinedI andthesevalueshavetoagree.

. . . . . .

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)

Therearethreeimportantpartstothisdefinition.I ThelimithastoexistI thefunctionhastobedefinedI 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 rightcontinuousatthepoint −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 rightcontinuousatthepoint −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 rightcontinuousatthepoint −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 rightcontinuousatthepoint −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 rightcontinuousatthepoint −1/4 since lim

x→−1/4+f(x) = f(−1/4).

. . . . . .

TheLimitLawsgiveContinuityLaws

TheoremIf f and g arecontinuousat a and c isaconstant, thenthefollowingfunctionsarealsocontinuousat a:

I f + gI f− gI cfI fg

I fg(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)]

= limx→a

f(x) + limx→a

g(x) (iftheselimitsexist)

= f(a) + g(a) (theydo; f and g arects)

= (f + g)(a)

. . . . . .

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)

. . . . . .

Pitfall#1

: 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.

. . . . . .

Pitfall#1: 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

.

.

.

.

. . . . . .

Pitfall#2

: Thefunctionhasnovalue

ExampleLet

f(x) =x2 + 2x + 1

x + 1

Atwhichpointsis f continuous?

SolutionBecause f isrational, itiscontinuousonitswholedomain. Notethat −1 isnotinthedomainof f, so f isnotcontinuousthere.

. . . . . .

Pitfall#2: 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.

. . . . . .

Pitfall#3

: 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.

. . . . . .

Pitfall#3: 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.

jumpdiscontinuity Thelimits limx→a−

f(x) and limx→a+

f(x) exist, but

aredifferent. f(a) isoneoftheselimits.

. . . . . .

Graphicalrepresentationsofdiscontinuities

. .x

.y

..π

..7

..1

.

.

removable

. .x

.y

..−1

..1

..2

..−1

..1

..2

..3

..4

.

.

.

.

jump

. . . . . .

Thegreatestintegerfunction[[x]] isthegreatestinteger ≤ x.

. .x

.y

..−2

..−2

..−1

..−1

..1

..1

..2

..2

..3

..3

. .

. .

. .

. .

. ..y = [[x]]

Thisfunctionhasajumpdiscontinuityateachinteger.

. . . . . .

Thegreatestintegerfunction[[x]] isthegreatestinteger ≤ x.

. .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!

Let h(t) beheight, whichvariescontinuouslyovertime. Thenh(birth) < 3 ft and h(now) > 3 ft. Sothereisapoint c in(birth, now) where h(c) = 3.

. . . . . .

BacktotheQuestions

TrueorFalseAtonepointinyourlifeyouwereexactlythreefeettall.

TrueorFalseAtonepointinyourlifeyourheightininchesequaledyourweightinpounds.

TrueorFalseRightnowtherearetwopointsonoppositesidesoftheEarthwithexactlythesametemperature.

. . . . . .

Question2: True!

Let h(t) beheightininchesand w(t) beweightinpounds, bothvaryingcontinuouslyovertime. Let f(t) = h(t) −w(t). Formostofus(callyourmom), f(birth) > 0 and f(now) < 0. Sothereisapoint c in (birth, now) where f(c) = 0. 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!