continuity says that nearby points go to nearby values, and the Intermediate Value Theorem is an important property of continuous functions

1. Section 1.5
Continuity
V63.0121, Calculus I
February 2–3, 2009
Announcements
Quiz 1 is this week: Tuesday (Sections 11 and 12), Thursday
(Section 4), Friday (Sections 13 and 14). 15 minutes, covers
Sections 1.1–1.2
Fill your ALEKS pie by February 27, 11:59pm
Congratulations to the Super Bowl XLIII Champion
Pittsburgh Steelers!

2. Outline
Warmup
Grader Notes
Continuity
The Intermediate Value Theorem
Back to the Questions

3. Hatsumon
Here are some discussion questions to start.
Were you ever exactly three feet tall?

4. Hatsumon
Here are some discussion questions to start.
Were you ever exactly three feet tall?
Was your height (in inches) ever equal to your weight (in
pounds)?

5. Hatsumon
Here are some discussion questions to start.
Were you ever exactly three feet tall?
Was your height (in inches) ever equal to your weight (in
pounds)?
Is there a pair of points on opposite sides of the world at the
same temperature at the same time?

6. Outline
Warmup
Grader Notes
Continuity
The Intermediate Value Theorem
Back to the Questions

7. But first, a word from our friendly graders
Please turn in neat problem sets: loose-leaf paper, stapled
Label homework with Name, Section (10 or 4), Date, Problem
Set number
Do not turn in scratch work

8. Please label your graphs
Example: Graph F (x) = |2x + 1|.
y
(0, 1)
x
(−1/2, 0)
incomplete better

9. Example = explanation
Problem
If f is even and g is odd, what can you say about fg ?
Let f (x) = x 2 (even) and g (x) = x 3 (odd).
Then (fg )(x) = x 2 · x 3 = x 5 , and that’s odd.
So the product of an even function and an odd
function is an odd function.
Dangerous!

10. The trouble with proof by example
Problem
Which odd numbers are prime?
The numbers 3, 5, and
7 are odd, and all prime.
So all odd numbers are
prime.
Fallacious!

11. 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

12. Outline
Warmup
Grader Notes
Continuity
The Intermediate Value Theorem
Back to the Questions

13. Recall: Direct Substitution Property
Theorem (The Direct Substitution Property)
If f is a polynomial or a rational function and a is in the domain of
f , then
lim f (x) = f (a)
x→a

14. Definition of Continuity
Definition
Let f be a function defined near a. We say that f is continuous at
a if
lim f (x) = f (a).
x→a

15. Free Theorems
Theorem
(a) Any polynomial is continuous everywhere; that is, it is
continuous on R = (−∞, ∞).
(b) Any rational function is continuous wherever it is defined; that
is, it is continuous on its domain.

16. Showing a function is continuous
Example
√
Let f (x) = 4x + 1. Show that f is continuous at 2.

17. Showing a function is continuous
Example
√
Let f (x) = 4x + 1. Show that f is continuous at 2.
Solution
We have
√
lim f (x) = lim 4x + 1
x→a x→2
= lim (4x + 1)
x→2
√
= 9 = 3.
Each step comes from the limit laws.

18. Showing a function is continuous
Example
√
Let f (x) = 4x + 1. Show that f is continuous at 2.
Solution
We have
√
lim f (x) = lim 4x + 1
x→a x→2
= lim (4x + 1)
x→2
√
= 9 = 3.
Each step comes from the limit laws.
In fact, f is continuous on (−1/4, ∞).

19. Showing a function is continuous
Example
√
Let f (x) = 4x + 1. Show that f is continuous at 2.
Solution
We have
√
lim f (x) = lim 4x + 1
x→a x→2
= lim (4x + 1)
x→2
√
= 9 = 3.
Each step comes from the limit laws.
In fact, f is continuous on (−1/4, ∞). The function f is right
continuous at the point −1/4.

20. The Limit Laws give Continuity Laws
Theorem
If f and g are continuous at a and c is a constant, then the
following functions are also continuous at a:
f +g
f −g
cf
fg
f
(if g (a) = 0)
g

21. Transcendental functions are continuous, too
Theorem
The following functions are continuous wherever they are defined:
1. sin, cos, tan, cot sec, csc
2. x → ax , loga , ln
3. sin−1 , tan−1 , sec−1

22. What could go wrong?
In what ways could a function f fail to be continuous at a point a?
Look again at the definition:
lim f (x) = f (a)
x→a

23. Pitfall #1
Example
Let
x2 if 0 ≤ x ≤ 1
f (x) =
if 1 < x ≤ 2
2x
At which points is f continuous?

24. Pitfall #1: The limit does not exist
Example
Let
x2 if 0 ≤ x ≤ 1
f (x) =
if 1 < x ≤ 2
2x
At which points is f continuous?
Solution
At any point a in [0, 2] besides 1, lim f (x) = f (a) because f is
x→a
represented by a polynomial near a, and polynomials have the
direct substitution property. However,
lim f (x) = lim x 2 = 12 = 1
x→1− x→1−
lim f (x) = lim+ 2x = 2(1) = 2
x→1+ x→1
So f has no limit at 1. Therefore f is not continuous at 1.

25. Graphical Illustration of Pitfall #1
y
4
3
2
1
x
−1 1 2
−1

26. Pitfall #2
Example
Let
x 2 + 2x + 1
f (x) =
x +1
At which points is f continuous?

27. Pitfall #2: The function has no value
Example
Let
x 2 + 2x + 1
f (x) =
x +1
At which points is f continuous?
Solution
Because f is rational, it is continuous on its whole domain. Note
that −1 is not in the domain of f , so f is not continuous there.

28. Graphical Illustration of Pitfall #2
y
1
x
−1
f cannot be continuous where it has no value.

29. Pitfall #3
Example
Let
7 if x = 1
f (x) =
π if x = 1
At which points is f continuous?

30. Pitfall #3: function value = limit
Example
Let
7 if x = 1
f (x) =
π if x = 1
At which points is f continuous?
Solution
f is not continuous at 1 because f (1) = π but lim f (x) = 7.
x→1

31. Graphical Illustration of Pitfall #3
y
7
π
x
1

32. Special types of discontinuites
removable discontinuity The limit lim f (x) exists, but f is not
x→a
defined at a or its value at a is not equal to the limit
at a.
jump discontinuity The limits lim f (x) and lim+ f (x) exist, but
x→a− x→a
are different. f (a) is one of these limits.

33. Graphical representations of discontinuities
y
4
y
3
7
2
π
1
x
x
1
−1 1 2
removable −1
jump

34. The greatest integer function
y
3
2
1
x
−2 −1 1 2 3
−1
−2

35. The greatest integer function
y
3
2
1
x
−2 −1 1 2 3
−1
−2
The greatest integer function f (x) = [[x]] has jump discontinuities.

36. Outline
Warmup
Grader Notes
Continuity
The Intermediate Value Theorem
Back to the Questions

37. A Big Time Theorem
Theorem (The Intermediate Value Theorem)
Suppose that f is continuous on the closed interval [a, b] and let N
be any number between f (a) and f (b), where f (a) = f (b). Then
there exists a number c in (a, b) such that f (c) = N.

38. Illustrating the IVT
f (x)
x

39. Illustrating the IVT
Suppose that f is continuous on the closed interval [a, b]
f (x)
x

40. Illustrating the IVT
Suppose that f is continuous on the closed interval [a, b]
f (x)
f (b)
f (a)
x
a b

41. Illustrating the IVT
Suppose that f is continuous on the closed interval [a, b] and let N
be any number between f (a) and f (b), where f (a) = f (b).
f (x)
f (b)
N
f (a)
x
a b

42. Illustrating the IVT
Suppose that f is continuous on the closed interval [a, b] and let N
be any number between f (a) and f (b), where f (a) = f (b). Then
there exists a number c in (a, b) such that f (c) = N.
f (x)
f (b)
N
f (a)
x
a c b

43. Illustrating the IVT
Suppose that f is continuous on the closed interval [a, b] and let N
be any number between f (a) and f (b), where f (a) = f (b). Then
there exists a number c in (a, b) such that f (c) = N.
f (x)
f (b)
N
f (a)
x
a b

44. Illustrating the IVT
Suppose that f is continuous on the closed interval [a, b] and let N
be any number between f (a) and f (b), where f (a) = f (b). Then
there exists a number c in (a, b) such that f (c) = N.
f (x)
f (b)
N
f (a)
x
a c1 c2 c3 b

45. Using the IVT
Example
Prove that the square root of two exists.

46. Using the IVT
Example
Prove that the square root of two exists.
Proof.
Let f (x) = x 2 , a continuous function on [1, 2].

47. Using the IVT
Example
Prove that the square root of two exists.
Proof.
Let f (x) = x 2 , a continuous function on [1, 2]. Note f (1) = 1 and
f (2) = 4. Since 2 is between 1 and 4, there exists a point c in
(1, 2) such that
f (c) = c 2 = 2.

48. Using the IVT
Example
Prove that the square root of two exists.
Proof.
Let f (x) = x 2 , a continuous function on [1, 2]. Note f (1) = 1 and
f (2) = 4. Since 2 is between 1 and 4, there exists a point c in
(1, 2) such that
f (c) = c 2 = 2.
In fact, we can “narrow in” on the square root of 2 by the method
of bisections.

49. √
Finding 2 by bisections
x f (x) = x 2
24
11

50. √
Finding 2 by bisections
x f (x) = x 2
24
11

51. √
Finding 2 by bisections
x f (x) = x 2
24
1.5 2.25
11

52. √
Finding 2 by bisections
x f (x) = x 2
24
1.5 2.25
1.25 1.5625
11

53. √
Finding 2 by bisections
x f (x) = x 2
24
1.5 2.25
1.375 1.890625
1.25 1.5625
11

54. √
Finding 2 by bisections
x f (x) = x 2
24
1.5 2.25
1.4375 2.06640625
1.375 1.890625
1.25 1.5625
11

55. Using the IVT
Example
Let f (x) = x 3 − x − 1. Show that there is a zero for f .
Solution
f (1) = −1 and f (2) = 5. So there is a zero between 1 and 2.

56. Using the IVT
Example
Let f (x) = x 3 − x − 1. Show that there is a zero for f .
Solution
f (1) = −1 and f (2) = 5. So there is a zero between 1 and 2.
(More careful analysis yields 1.32472.)

57. Outline
Warmup
Grader Notes
Continuity
The Intermediate Value Theorem
Back to the Questions

58. Back to the Questions
True or False
At one point in your life you were exactly three feet tall.

59. Question 1: True!
Let h(t) be height, which varies continuously over time. Then
h(birth) < 3 ft and h(now) > 3 ft. So there is a point c in
(birth, now) where h(c) = 3.

60. Back to the Questions
True or False
At one point in your life you were exactly three feet tall.
True or False
At one point in your life your height in inches equaled your weight
in pounds.

61. Question 2: True!
Let h(t) be height in inches and w (t) be weight in pounds, both
varying continuously over time. Let f (t) = h(t) − w (t). For most
of us (call your mom), f (birth) > 0 and f (now) < 0. So there is a
point c in (birth, now) where f (c) = 0. In other words,
h(c) − w (c) = 0 ⇐⇒ h(c) = w (c).

62. Back to the Questions
True or False
At one point in your life you were exactly three feet tall.
True or False
At one point in your life your height in inches equaled your weight
in pounds.
True or False
Right now there are two points on opposite sides of the Earth with
exactly the same temperature.

63. Question 3
Let T (θ) be the temperature at the point on the equator at
longitude θ.
How can you express the statement that the temperature on
opposite sides is the same?
How can you ensure this is true?