Lesson 5: Continuity
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![Page 1: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/1.jpg)
Section 1.5Continuity
V63.0121, Calculus I
February 2–3, 2009
Announcements
I Quiz 1 is this week: Tuesday (Sections 11 and 12), Thursday(Section 4), Friday (Sections 13 and 14). 15 minutes, coversSections 1.1–1.2
I Fill your ALEKS pie by February 27, 11:59pm
I Congratulations to the Super Bowl XLIII ChampionPittsburgh Steelers!
![Page 2: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/2.jpg)
Outline
Warmup
Grader Notes
Continuity
The Intermediate Value Theorem
Back to the Questions
![Page 3: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/3.jpg)
Hatsumon
Here are some discussion questions to start.
I Were you ever exactly three feet tall?
I Was your height (in inches) ever equal to your weight (inpounds)?
I Is there a pair of points on opposite sides of the world at thesame temperature at the same time?
![Page 4: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/4.jpg)
Hatsumon
Here are some discussion questions to start.
I Were you ever exactly three feet tall?
I Was your height (in inches) ever equal to your weight (inpounds)?
I Is there a pair of points on opposite sides of the world at thesame temperature at the same time?
![Page 5: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/5.jpg)
Hatsumon
Here are some discussion questions to start.
I Were you ever exactly three feet tall?
I Was your height (in inches) ever equal to your weight (inpounds)?
I Is there a pair of points on opposite sides of the world at thesame temperature at the same time?
![Page 6: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/6.jpg)
Outline
Warmup
Grader Notes
Continuity
The Intermediate Value Theorem
Back to the Questions
![Page 7: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/7.jpg)
But first, a word from our friendly graders
I Please turn in neat problem sets: loose-leaf paper, stapled
I Label homework with Name, Section (10 or 4), Date, ProblemSet number
I Do not turn in scratch work
![Page 8: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/8.jpg)
Please label your graphs
Example: Graph F (x) = |2x + 1|.
incomplete
x
y
(−1/2, 0)
(0, 1)
better
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Example 6= explanation
ProblemIf f is even and g is odd, what can you say about fg?
Let f (x) = x2 (even) and g(x) = x3 (odd).Then (fg)(x) = x2 · x3 = x5, and that’s odd.So the product of an even function and an oddfunction is an odd function.
Dangerous!
![Page 10: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/10.jpg)
The trouble with proof by example
ProblemWhich odd numbers are prime?
The numbers 3, 5, and7 are odd, and all prime.So all odd numbers areprime.
Fallacious!
![Page 11: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/11.jpg)
Use the definitions
Let f be even and g be odd. Then
(fg)(−x) = f (−x)g(−x) = f (x)(−g(x))
= −f (x)g(x) = −(fg)(x)
So fg is odd.
Better
![Page 12: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/12.jpg)
Outline
Warmup
Grader Notes
Continuity
The Intermediate Value Theorem
Back to the Questions
![Page 13: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/13.jpg)
Recall: Direct Substitution Property
Theorem (The Direct Substitution Property)
If f is a polynomial or a rational function and a is in the domain off , then
limx→a
f (x) = f (a)
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Definition of Continuity
DefinitionLet f be a function defined near a. We say that f is continuous ata if
limx→a
f (x) = f (a).
![Page 15: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/15.jpg)
Free Theorems
Theorem
(a) Any polynomial is continuous everywhere; that is, it iscontinuous on R = (−∞,∞).
(b) Any rational function is continuous wherever it is defined; thatis, it is continuous on its domain.
![Page 16: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/16.jpg)
Showing a function is continuous
Example
Let f (x) =√
4x + 1. Show that f is continuous at 2.
SolutionWe have
limx→a
f (x) = limx→2
√4x + 1
=√
limx→2
(4x + 1)
=√
9 = 3.
Each step comes from the limit laws.
In fact, f is continuous on (−1/4,∞). The function f is rightcontinuous at the point −1/4.
![Page 17: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/17.jpg)
Showing a function is continuous
Example
Let f (x) =√
4x + 1. Show that f is continuous at 2.
SolutionWe have
limx→a
f (x) = limx→2
√4x + 1
=√
limx→2
(4x + 1)
=√
9 = 3.
Each step comes from the limit laws.
In fact, f is continuous on (−1/4,∞). The function f is rightcontinuous at the point −1/4.
![Page 18: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/18.jpg)
Showing a function is continuous
Example
Let f (x) =√
4x + 1. Show that f is continuous at 2.
SolutionWe have
limx→a
f (x) = limx→2
√4x + 1
=√
limx→2
(4x + 1)
=√
9 = 3.
Each step comes from the limit laws.
In fact, f is continuous on (−1/4,∞).
The function f is rightcontinuous at the point −1/4.
![Page 19: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/19.jpg)
Showing a function is continuous
Example
Let f (x) =√
4x + 1. Show that f is continuous at 2.
SolutionWe have
limx→a
f (x) = limx→2
√4x + 1
=√
limx→2
(4x + 1)
=√
9 = 3.
Each step comes from the limit laws.
In fact, f is continuous on (−1/4,∞). The function f is rightcontinuous at the point −1/4.
![Page 20: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/20.jpg)
The Limit Laws give Continuity Laws
TheoremIf f and g are continuous at a and c is a constant, then thefollowing functions are also continuous at a:
I f + g
I f − g
I cf
I fg
If
g(if g(a) 6= 0)
![Page 21: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/21.jpg)
Transcendental functions are continuous, too
TheoremThe following functions are continuous wherever they are defined:
1. sin, cos, tan, cot sec, csc
2. x 7→ ax , loga, ln
3. sin−1, tan−1, sec−1
![Page 22: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/22.jpg)
What could go wrong?
In what ways could a function f fail to be continuous at a point a?Look again at the definition:
limx→a
f (x) = f (a)
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Pitfall #1
: The limit does not exist
Example
Let
f (x) =
{x2 if 0 ≤ x ≤ 1
2x if 1 < x ≤ 2
At which points is f continuous?
SolutionAt any point a in [0, 2] besides 1, lim
x→af (x) = f (a) because f is
represented by a polynomial near a, and polynomials have thedirect substitution property. However,
limx→1−
f (x) = limx→1−
x2 = 12 = 1
limx→1+
f (x) = limx→1+
2x = 2(1) = 2
So f has no limit at 1. Therefore f is not continuous at 1.
![Page 24: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/24.jpg)
Pitfall #1: The limit does not exist
Example
Let
f (x) =
{x2 if 0 ≤ x ≤ 1
2x if 1 < x ≤ 2
At which points is f continuous?
SolutionAt any point a in [0, 2] besides 1, lim
x→af (x) = f (a) because f is
represented by a polynomial near a, and polynomials have thedirect substitution property. However,
limx→1−
f (x) = limx→1−
x2 = 12 = 1
limx→1+
f (x) = limx→1+
2x = 2(1) = 2
So f has no limit at 1. Therefore f is not continuous at 1.
![Page 25: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/25.jpg)
Graphical Illustration of Pitfall #1
x
y
−1 1 2
−1
1
2
3
4
![Page 26: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/26.jpg)
Pitfall #2
: The function has no value
Example
Let
f (x) =x2 + 2x + 1
x + 1
At which points is f continuous?
SolutionBecause f is rational, it is continuous on its whole domain. Notethat −1 is not in the domain of f , so f is not continuous there.
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Pitfall #2: The function has no value
Example
Let
f (x) =x2 + 2x + 1
x + 1
At which points is f continuous?
SolutionBecause f is rational, it is continuous on its whole domain. Notethat −1 is not in the domain of f , so f is not continuous there.
![Page 28: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/28.jpg)
Graphical Illustration of Pitfall #2
x
y
−1
1
f cannot be continuous where it has no value.
![Page 29: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/29.jpg)
Pitfall #3
: function value 6= limit
Example
Let
f (x) =
{7 if x 6= 1
π if x = 1
At which points is f continuous?
Solutionf is not continuous at 1 because f (1) = π but lim
x→1f (x) = 7.
![Page 30: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/30.jpg)
Pitfall #3: function value 6= limit
Example
Let
f (x) =
{7 if x 6= 1
π if x = 1
At which points is f continuous?
Solutionf is not continuous at 1 because f (1) = π but lim
x→1f (x) = 7.
![Page 31: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/31.jpg)
Graphical Illustration of Pitfall #3
x
y
π
7
1
![Page 32: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/32.jpg)
Special types of discontinuites
removable discontinuity The limit limx→a
f (x) exists, but f is not
defined at a or its value at a is not equal to the limitat a.
jump discontinuity The limits limx→a−
f (x) and limx→a+
f (x) exist, but
are different. f (a) is one of these limits.
![Page 33: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/33.jpg)
Graphical representations of discontinuities
x
y
π
7
1
removable
x
y
−1 1 2
−1
1
2
3
4
jump
![Page 34: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/34.jpg)
The greatest integer function
x
y
−2
−2
−1
−1
1
1
2
2
3
3
The greatest integer function f (x) = [[x ]] has jump discontinuities.
![Page 35: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/35.jpg)
The greatest integer function
x
y
−2
−2
−1
−1
1
1
2
2
3
3
The greatest integer function f (x) = [[x ]] has jump discontinuities.
![Page 36: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/36.jpg)
Outline
Warmup
Grader Notes
Continuity
The Intermediate Value Theorem
Back to the Questions
![Page 37: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/37.jpg)
A Big Time Theorem
Theorem (The Intermediate Value Theorem)
Suppose that f is continuous on the closed interval [a, b] and let Nbe any number between f (a) and f (b), where f (a) 6= f (b). Thenthere exists a number c in (a, b) such that f (c) = N.
![Page 38: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/38.jpg)
Illustrating the IVT
Suppose that f is continuous on the closed interval [a, b] and let Nbe any number between f (a) and f (b), where f (a) 6= f (b). Thenthere exists a number c in (a, b) such that f (c) = N.
x
f (x)
a b
f (a)
f (b)
N
cc1 c2 c3
![Page 39: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/39.jpg)
Illustrating the IVTSuppose that f is continuous on the closed interval [a, b]
and let Nbe any number between f (a) and f (b), where f (a) 6= f (b). Thenthere exists a number c in (a, b) such that f (c) = N.
x
f (x)
a b
f (a)
f (b)
N
cc1 c2 c3
![Page 40: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/40.jpg)
Illustrating the IVTSuppose that f is continuous on the closed interval [a, b]
and let Nbe any number between f (a) and f (b), where f (a) 6= f (b). Thenthere exists a number c in (a, b) such that f (c) = N.
x
f (x)
a b
f (a)
f (b)
N
cc1 c2 c3
![Page 41: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/41.jpg)
Illustrating the IVTSuppose that f is continuous on the closed interval [a, b] and let Nbe any number between f (a) and f (b), where f (a) 6= f (b).
Thenthere exists a number c in (a, b) such that f (c) = N.
x
f (x)
a b
f (a)
f (b)
N
cc1 c2 c3
![Page 42: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/42.jpg)
Illustrating the IVTSuppose that f is continuous on the closed interval [a, b] and let Nbe any number between f (a) and f (b), where f (a) 6= f (b). Thenthere exists a number c in (a, b) such that f (c) = N.
x
f (x)
a b
f (a)
f (b)
N
c
c1 c2 c3
![Page 43: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/43.jpg)
Illustrating the IVTSuppose that f is continuous on the closed interval [a, b] and let Nbe any number between f (a) and f (b), where f (a) 6= f (b). Thenthere exists a number c in (a, b) such that f (c) = N.
x
f (x)
a b
f (a)
f (b)
N
cc1 c2 c3
![Page 44: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/44.jpg)
Illustrating the IVTSuppose that f is continuous on the closed interval [a, b] and let Nbe any number between f (a) and f (b), where f (a) 6= f (b). Thenthere exists a number c in (a, b) such that f (c) = N.
x
f (x)
a b
f (a)
f (b)
N
c
c1 c2 c3
![Page 45: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/45.jpg)
Using the IVT
Example
Prove that the square root of two exists.
Proof.Let f (x) = x2, a continuous function on [1, 2]. Note f (1) = 1 andf (2) = 4. Since 2 is between 1 and 4, there exists a point c in(1, 2) such that
f (c) = c2 = 2.
In fact, we can “narrow in” on the square root of 2 by the methodof bisections.
![Page 46: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/46.jpg)
Using the IVT
Example
Prove that the square root of two exists.
Proof.Let f (x) = x2, a continuous function on [1, 2].
Note f (1) = 1 andf (2) = 4. Since 2 is between 1 and 4, there exists a point c in(1, 2) such that
f (c) = c2 = 2.
In fact, we can “narrow in” on the square root of 2 by the methodof bisections.
![Page 47: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/47.jpg)
Using the IVT
Example
Prove that the square root of two exists.
Proof.Let f (x) = x2, a continuous function on [1, 2]. Note f (1) = 1 andf (2) = 4. Since 2 is between 1 and 4, there exists a point c in(1, 2) such that
f (c) = c2 = 2.
In fact, we can “narrow in” on the square root of 2 by the methodof bisections.
![Page 48: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/48.jpg)
Using the IVT
Example
Prove that the square root of two exists.
Proof.Let f (x) = x2, a continuous function on [1, 2]. Note f (1) = 1 andf (2) = 4. Since 2 is between 1 and 4, there exists a point c in(1, 2) such that
f (c) = c2 = 2.
In fact, we can “narrow in” on the square root of 2 by the methodof bisections.
![Page 49: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/49.jpg)
Finding√
2 by bisections
x f (x) = x2
1 1
2 4
1.5 2.25
1.25 1.5625
1.375 1.8906251.4375 2.06640625
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Finding√
2 by bisections
x f (x) = x2
1 1
2 4
1.5 2.25
1.25 1.5625
1.375 1.8906251.4375 2.06640625
![Page 51: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/51.jpg)
Finding√
2 by bisections
x f (x) = x2
1 1
2 4
1.5 2.25
1.25 1.5625
1.375 1.8906251.4375 2.06640625
![Page 52: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/52.jpg)
Finding√
2 by bisections
x f (x) = x2
1 1
2 4
1.5 2.25
1.25 1.5625
1.375 1.8906251.4375 2.06640625
![Page 53: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/53.jpg)
Finding√
2 by bisections
x f (x) = x2
1 1
2 4
1.5 2.25
1.25 1.5625
1.375 1.890625
1.4375 2.06640625
![Page 54: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/54.jpg)
Finding√
2 by bisections
x f (x) = x2
1 1
2 4
1.5 2.25
1.25 1.5625
1.375 1.8906251.4375 2.06640625
![Page 55: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/55.jpg)
Using the IVT
Example
Let f (x) = x3 − x − 1. Show that there is a zero for f .
Solutionf (1) = −1 and f (2) = 5. So there is a zero between 1 and 2.
(More careful analysis yields 1.32472.)
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Using the IVT
Example
Let f (x) = x3 − x − 1. Show that there is a zero for f .
Solutionf (1) = −1 and f (2) = 5. So there is a zero between 1 and 2.(More careful analysis yields 1.32472.)
![Page 57: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/57.jpg)
Outline
Warmup
Grader Notes
Continuity
The Intermediate Value Theorem
Back to the Questions
![Page 58: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/58.jpg)
Back to the Questions
True or FalseAt one point in your life you were exactly three feet tall.
True or FalseAt one point in your life your height in inches equaled your weightin pounds.
True or FalseRight now there are two points on opposite sides of the Earth withexactly the same temperature.
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Question 1: True!
Let h(t) be height, which varies continuously over time. Thenh(birth) < 3 ft and h(now) > 3 ft. So there is a point c in(birth, now) where h(c) = 3.
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Back to the Questions
True or FalseAt one point in your life you were exactly three feet tall.
True or FalseAt one point in your life your height in inches equaled your weightin pounds.
True or FalseRight now there are two points on opposite sides of the Earth withexactly the same temperature.
![Page 61: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/61.jpg)
Question 2: True!
Let h(t) be height in inches and w(t) be weight in pounds, bothvarying continuously over time. Let f (t) = h(t)− w(t). For mostof us (call your mom), f (birth) > 0 and f (now) < 0. So there is apoint c in (birth, now) where f (c) = 0. In other words,
h(c)− w(c) = 0 ⇐⇒ h(c) = w(c).
![Page 62: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/62.jpg)
Back to the Questions
True or FalseAt one point in your life you were exactly three feet tall.
True or FalseAt one point in your life your height in inches equaled your weightin pounds.
True or FalseRight now there are two points on opposite sides of the Earth withexactly the same temperature.
![Page 63: Lesson 5: Continuity](https://reader033.fdocuments.net/reader033/viewer/2022060119/558e305e1a28ab48618b46d5/html5/thumbnails/63.jpg)
Question 3
I Let T (θ) be the temperature at the point on the equator atlongitude θ.
I How can you express the statement that the temperature onopposite sides is the same?
I How can you ensure this is true?