This document contains lecture notes on continuity from a Calculus I class at New York University. It begins with announcements about office hours and homework grades. It then reviews the definition of a limit and introduces the definition of continuity as a function having a limit equal to its value at a point. Examples are provided to demonstrate showing a function is continuous. The document states that polynomials, rational functions, and trigonometric functions are continuous based on their definitions and limit properties. It concludes by explaining the continuity of inverse trigonometric functions.
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Lesson 5: Continuity (Section 41 slides)
1. Section 1.5
Continuity
V63.0121.041, Calculus I
New York University
September 20, 2010
Announcements
Office Hours: Tuesday, Wednesday, 3pm–4pm
TAs have office hours on website
2. Announcements
Office Hours: Tuesday,
Wednesday, 3pm–4pm
TAs have office hours on
website
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 2 / 47
3. Grader’s corner
HW Grades will be on
blackboard this week, and
the papers will be returned
in recitation
Remember units when
computing slopes
Remember to staple your
papers—you have been
warned.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 3 / 47
4. Objectives
Understand and apply the
definition of continuity for a
function at a point or on an
interval.
Given a piecewise defined
function, decide where it is
continuous or discontinuous.
State and understand the
Intermediate Value
Theorem.
Use the IVT to show that a
function takes a certain
value, or that an equation
has a solution
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 4 / 47
5. Last time
Definition
We write
lim f (x) = L
x→a
and say
“the limit of f (x), as x approaches a, equals L”
if we can make the values of f (x) arbitrarily close to L (as close to L as we
like) by taking x to be sufficiently close to a (on either side of a) but not
equal to a.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 5 / 47
6. Limit Laws for arithmetic
Theorem (Basic Limits)
lim x = a
x→a
lim c = c
x→a
Theorem (Limit Laws)
Let f and g be functions with limits at a point a. Then
lim (f (x) + g (x)) = lim f (x) + lim g (x)
x→a x→a x→a
lim (f (x) − g (x)) = lim f (x) − lim g (x)
x→a x→a x→a
lim (f (x) · g (x)) = lim f (x) · lim g (x)
x→a x→a x→a
f (x) limx→a f (x)
lim = if lim g (x) = 0
x→a g (x) limx→a g (x) x→a
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 6 / 47
7. Hatsumon
Here are some discussion questions to start.
True or False
At some point in your life you were exactly three feet tall.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 7 / 47
8. Hatsumon
Here are some discussion questions to start.
True or False
At some point in your life you were exactly three feet tall.
True or False
At some point in your life your height (in inches) was equal to your weight
(in pounds).
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 7 / 47
9. Hatsumon
Here are some discussion questions to start.
True or False
At some point in your life you were exactly three feet tall.
True or False
At some point in your life your height (in inches) was equal to your weight
(in pounds).
True or False
Right now there are a pair of points on opposite sides of the world
measuring the exact same temperature.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 7 / 47
11. 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
This property is so useful it’s worth naming.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 9 / 47
12. 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
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 10 / 47
13. Definition of Continuity
y
Definition
Let f be a function defined
near a. We say that f is f (a)
continuous at a if
lim f (x) = f (a).
x→a
A function f is continuous
if it is continuous at every
point in its domain. x
a
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 10 / 47
14. Scholium
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
There are three important parts to this definition.
The function has to have a limit at a,
the function has to have a value at a,
and these values have to agree.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 11 / 47
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.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 12 / 47
16. Showing a function is continuous
Example
√
Let f (x) = 4x + 1. Show that f is continuous at 2.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 13 / 47
17. Showing a function is continuous
Example
√
Let f (x) = 4x + 1. Show that f is continuous at 2.
Solution
We want to show that lim f (x) = f (2). We have
x→2
√ √
lim f (x) = lim 4x + 1 = lim (4x + 1) = 9 = 3 = f (2).
x→a x→2 x→2
Each step comes from the limit laws.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 13 / 47
18. Showing a function is continuous
Example
√
Let f (x) = 4x + 1. Show that f is continuous at 2.
Solution
We want to show that lim f (x) = f (2). We have
x→2
√ √
lim f (x) = lim 4x + 1 = lim (4x + 1) = 9 = 3 = f (2).
x→a x→2 x→2
Each step comes from the limit laws.
Question
At which other points is f continuous?
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 13 / 47
19. At which other points?
√
For reference: f (x) = 4x + 1
If a > −1/4, then lim (4x + 1) = 4a + 1 > 0, so
x→a
√ √
lim f (x) = lim 4x + 1 = lim (4x + 1) = 4a + 1 = f (a)
x→a x→a x→a
and f is continuous at a.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 14 / 47
20. At which other points?
√
For reference: f (x) = 4x + 1
If a > −1/4, then lim (4x + 1) = 4a + 1 > 0, so
x→a
√ √
lim f (x) = lim 4x + 1 = lim (4x + 1) = 4a + 1 = f (a)
x→a x→a x→a
and f is continuous at a.
√
If a = −1/4, then 4x + 1 < 0 to the left of a, which means 4x + 1 is
undefined. Still,
√ √
lim+ f (x) = lim+ 4x + 1 = lim+ (4x + 1) = 0 = 0 = f (a)
x→a x→a x→a
so f is continuous on the right at a = −1/4.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 14 / 47
21. Showing a function is continuous
Example
√
Let f (x) = 4x + 1. Show that f is continuous at 2.
Solution
We want to show that lim f (x) = f (2). We have
x→2
√ √
lim f (x) = lim 4x + 1 = lim (4x + 1) = 9 = 3 = f (2).
x→a x→2 x→2
Each step comes from the limit laws.
Question
At which other points is f continuous?
Answer
The function f is continuous on (−1/4, ∞).
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 15 / 47
22. Showing a function is continuous
Example
√
Let f (x) = 4x + 1. Show that f is continuous at 2.
Solution
We want to show that lim f (x) = f (2). We have
x→2
√ √
lim f (x) = lim 4x + 1 = lim (4x + 1) = 9 = 3 = f (2).
x→a x→2 x→2
Each step comes from the limit laws.
Question
At which other points is f continuous?
Answer
The function f is continuous on (−1/4, ∞). It is right continuous at −1/4
since lim f (x) = f (−1/4).
V63.0121.041, 1/4+
x→−Calculus I (NYU) Section 1.5 Continuity September 20, 2010 15 / 47
23. The Limit Laws give Continuity Laws
Theorem
If f (x) and g (x) are continuous at a and c is a constant, then the
following functions are also continuous at a:
(f + g )(x) (fg )(x)
(f − g )(x) f
(x) (if g (a) = 0)
(cf )(x) g
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 16 / 47
24. Why a sum of continuous functions is continuous
We want to show that
lim (f + g )(x) = (f + g )(a).
x→a
We just follow our nose:
lim (f + g )(x) = lim [f (x) + g (x)] (def of f + g )
x→a x→a
= lim f (x) + lim g (x) (if these limits exist)
x→a x→a
= f (a) + g (a) (they do; f and g are cts.)
= (f + g )(a) (def of f + g again)
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 17 / 47
25. Trigonometric functions are continuous
sin and cos are continuous on R.
sin
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 18 / 47
26. Trigonometric functions are continuous
sin and cos are continuous on R.
cos
sin
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 18 / 47
27. Trigonometric functions are continuous
tan
sin and cos are continuous on R.
sin 1
tan = and sec = are
cos cos cos
continuous on their domain,
which is
π sin
R + kπ k ∈ Z .
2
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 18 / 47
28. Trigonometric functions are continuous
tan sec
sin and cos are continuous on R.
sin 1
tan = and sec = are
cos cos cos
continuous on their domain,
which is
π sin
R + kπ k ∈ Z .
2
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 18 / 47
29. Trigonometric functions are continuous
tan sec
sin and cos are continuous on R.
sin 1
tan = and sec = are
cos cos cos
continuous on their domain,
which is
π sin
R + kπ k ∈ Z .
2
cos 1
cot = and csc = are
sin sin
continuous on their domain,
which is R { kπ | k ∈ Z }.
cot
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 18 / 47
30. Trigonometric functions are continuous
tan sec
sin and cos are continuous on R.
sin 1
tan = and sec = are
cos cos cos
continuous on their domain,
which is
π sin
R + kπ k ∈ Z .
2
cos 1
cot = and csc = are
sin sin
continuous on their domain,
which is R { kπ | k ∈ Z }.
cot csc
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 18 / 47
31. Exponential and Logarithmic functions are
continuous
For any base a > 1, ax
the function x → ax is
continuous on R
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 19 / 47
32. Exponential and Logarithmic functions are
continuous
For any base a > 1, ax
the function x → ax is loga x
continuous on R
the function loga is
continuous on its domain:
(0, ∞)
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 19 / 47
33. Exponential and Logarithmic functions are
continuous
For any base a > 1, ax
the function x → ax is loga x
continuous on R
the function loga is
continuous on its domain:
(0, ∞)
In particular e x and
ln = loge are continuous on
their domains
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 19 / 47
34. Inverse trigonometric functions are mostly
continuous
sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, and
right continuous at −1.
π
π/2
sin−1
V63.0121.041, Calculus I (NYU) −π/2
Section 1.5 Continuity September 20, 2010 20 / 47
35. Inverse trigonometric functions are mostly
continuous
sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, and
right continuous at −1.
π
cos−1
π/2
sin−1
V63.0121.041, Calculus I (NYU) −π/2
Section 1.5 Continuity September 20, 2010 20 / 47
36. Inverse trigonometric functions are mostly
continuous
sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, and
right continuous at −1.
sec−1 and csc−1 are continuous on (−∞, −1) ∪ (1, ∞), left
continuous at −1, and right continuous at 1.
π
cos−1 sec−1
π/2
sin−1
V63.0121.041, Calculus I (NYU) −π/2
Section 1.5 Continuity September 20, 2010 20 / 47
37. Inverse trigonometric functions are mostly
continuous
sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, and
right continuous at −1.
sec−1 and csc−1 are continuous on (−∞, −1) ∪ (1, ∞), left
continuous at −1, and right continuous at 1.
π
cos−1 sec−1
π/2
csc−1
sin−1
V63.0121.041, Calculus I (NYU) −π/2
Section 1.5 Continuity September 20, 2010 20 / 47
38. Inverse trigonometric functions are mostly
continuous
sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, and
right continuous at −1.
sec−1 and csc−1 are continuous on (−∞, −1) ∪ (1, ∞), left
continuous at −1, and right continuous at 1.
tan−1 and cot−1 are continuous on R.
π
cos−1 sec−1
π/2
tan−1
csc−1
sin−1
V63.0121.041, Calculus I (NYU) −π/2
Section 1.5 Continuity September 20, 2010 20 / 47
39. Inverse trigonometric functions are mostly
continuous
sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, and
right continuous at −1.
sec−1 and csc−1 are continuous on (−∞, −1) ∪ (1, ∞), left
continuous at −1, and right continuous at 1.
tan−1 and cot−1 are continuous on R.
π
cot−1
cos−1 sec−1
π/2
tan−1
csc−1
sin−1
V63.0121.041, Calculus I (NYU) −π/2
Section 1.5 Continuity September 20, 2010 20 / 47
40. 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
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 21 / 47
41. Continuity FAIL
Example
Let
x2 if 0 ≤ x ≤ 1
f (x) =
2x if 1 < x ≤ 2
At which points is f continuous?
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 22 / 47
42. Continuity FAIL: The limit does not exist
Example
Let
x2 if 0 ≤ x ≤ 1
f (x) =
2x if 1 < x ≤ 2
At which points is f continuous?
Solution
At any point a in [0, 2] besides 1, lim f (x) = f (a) because f is represented by a
x→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.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 22 / 47
43. Graphical Illustration of Pitfall #1
y
4
3
The function cannot be
2 continuous at a point if the
function has no limit at that
point.
1
x
−1 1 2
−1
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 23 / 47
44. Continuity FAIL
Example
Let
x 2 + 2x + 1
f (x) =
x +1
At which points is f continuous?
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 24 / 47
45. Continuity FAIL: 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.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 24 / 47
46. Graphical Illustration of Pitfall #2
y
1 The function cannot be
continuous at a point outside its
domain (that is, a point where it
x has no value).
−1
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 25 / 47
47. Continuity FAIL
Example
Let
7 if x = 1
f (x) =
π if x = 1
At which points is f continuous?
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 26 / 47
48. Continuity FAIL: 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
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 26 / 47
49. Graphical Illustration of Pitfall #3
y
7 If the function has a limit and a
value at a point the two must
π still agree.
x
1
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 27 / 47
50. Special types of discontinuites
removable discontinuity The limit lim f (x) exists, but f is not defined
x→a
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 are
x→a− x→a
different.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 28 / 47
51. Graphical representations of discontinuities
y
y
7
2
π
1
x
x
1
1
removable
jump
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 29 / 47
52. Graphical representations of discontinuities
y
y
Presto! continuous!
7
2
π
1
x
x
1
1
removable
jump
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 29 / 47
53. Graphical representations of discontinuities
y
y
Presto! continuous!
7
2
π
1 continuous?
x
x
1
1
removable
jump
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 29 / 47
54. Graphical representations of discontinuities
y
y
Presto! continuous!
7
2 continuous?
π
1
x
x
1
1
removable
jump
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 29 / 47
55. Graphical representations of discontinuities
y
y
Presto! continuous!
7
2
continuous?
π
1
x
x
1
1
removable
jump
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 29 / 47
56. Special types of discontinuites
removable discontinuity The limit lim f (x) exists, but f is not defined
x→a
at a or its value at a is not equal to the limit at a. By
re-defining f (a) = lim f (x), f can be made continuous at a
x→a
jump discontinuity The limits lim f (x) and lim+ f (x) exist, but are
x→a− x→a
different.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 30 / 47
57. Special types of discontinuites
removable discontinuity The limit lim f (x) exists, but f is not defined
x→a
at a or its value at a is not equal to the limit at a. By
re-defining f (a) = lim f (x), f can be made continuous at a
x→a
jump discontinuity The limits lim f (x) and lim+ f (x) exist, but are
x→a− x→a
different. The function cannot be made continuous by
changing a single value.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 30 / 47
58. The greatest integer function
[[x]] is the greatest integer ≤ x.
y
3
x [[x]] y = [[x]]
0 0 2
1 1
1.5 1 1
1.9 1
2.1 2 x
−0.5 −1 −2 −1 1 2 3
−0.9 −1 −1
−1.1 −2
−2
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 31 / 47
59. The greatest integer function
[[x]] is the greatest integer ≤ x.
y
3
x [[x]] y = [[x]]
0 0 2
1 1
1.5 1 1
1.9 1
2.1 2 x
−0.5 −1 −2 −1 1 2 3
−0.9 −1 −1
−1.1 −2
−2
This function has a jump discontinuity at each integer.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 31 / 47
61. 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.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 33 / 47
62. Illustrating the IVT
f (x)
x
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 34 / 47
63. Illustrating the IVT
Suppose that f is continuous on the closed interval [a, b]
f (x)
x
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 34 / 47
64. Illustrating the IVT
Suppose that f is continuous on the closed interval [a, b]
f (x)
f (b)
f (a)
a b x
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 34 / 47
65. 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)
a b x
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 34 / 47
66. 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)
a c b x
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 34 / 47
67. 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)
a b x
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 34 / 47
68. 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)
a c1 c2 c3 b x
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 34 / 47
69. What the IVT does not say
The Intermediate Value Theorem is an “existence” theorem.
It does not say how many such c exist.
It also does not say how to find c.
Still, it can be used in iteration or in conjunction with other theorems to
answer these questions.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 35 / 47
70. Using the IVT to find zeroes
Example
Let f (x) = x 3 − x − 1. Show that there is a zero for f on the interval [1, 2].
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 36 / 47
71. Using the IVT to find zeroes
Example
Let f (x) = x 3 − x − 1. Show that there is a zero for f on the interval [1, 2].
Solution
f (1) = −1 and f (2) = 5. So there is a zero between 1 and 2.
In fact, we can “narrow in” on the zero by the method of bisections.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 36 / 47
72. Finding a zero by bisection
y
x f (x)
x
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 37 / 47
73. Finding a zero by bisection
y
x f (x)
1 −1
x
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 37 / 47
74. Finding a zero by bisection
y
x f (x)
1 −1
2 5
x
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 37 / 47
75. Finding a zero by bisection
y
x f (x)
1 −1
1.5 0.875
2 5
x
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 37 / 47
76. Finding a zero by bisection
y
x f (x)
1 −1
1.25 − 0.296875
1.5 0.875
2 5
x
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 37 / 47
77. Finding a zero by bisection
y
x f (x)
1 −1
1.25 − 0.296875
1.375 0.224609
1.5 0.875
2 5
x
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 37 / 47
78. Finding a zero by bisection
y
x f (x)
1 −1
1.25 − 0.296875
1.3125 − 0.0515137
1.375 0.224609
1.5 0.875
2 5
x
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 37 / 47
79. Finding a zero by bisection
y
x f (x)
1 −1
1.25 − 0.296875
1.3125 − 0.0515137
1.375 0.224609
1.5 0.875
2 5
(More careful analysis yields x
1.32472.)
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 37 / 47
80. Using the IVT to assert existence of numbers
Example
Suppose we are unaware of the square root function and that it’s
continuous. Prove that the square root of two exists.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 38 / 47
81. Using the IVT to assert existence of numbers
Example
Suppose we are unaware of the square root function and that it’s
continuous. Prove that the square root of two exists.
Proof.
Let f (x) = x 2 , a continuous function on [1, 2].
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 38 / 47
82. Using the IVT to assert existence of numbers
Example
Suppose we are unaware of the square root function and that it’s
continuous. 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.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 38 / 47
84. Back to the Questions
True or False
At one point in your life you were exactly three feet tall.
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85. Question 1
Answer
The answer is TRUE.
Let h(t) be height, which varies continuously over time.
Then h(birth) < 3 ft and h(now) > 3 ft.
So by the IVT there is a point c in (birth, now) where h(c) = 3.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 41 / 47
86. 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.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 42 / 47
87. Question 2
Answer
The answer is 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 by the IVT there is a point c in (birth, now) where f (c) = 0.
In other words,
h(c) − w (c) = 0 ⇐⇒ h(c) = w (c).
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 43 / 47
88. 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.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 44 / 47
89. Question 3
Answer
The answer is TRUE.
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?
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 45 / 47
90. Question 3
Let f (θ) = T (θ) − T (θ + 180◦ )
Then
f (0) = T (0) − T (180)
while
f (180) = T (180) − T (360) = −f (0)
So somewhere between 0 and 180 there is a point θ where f (θ) = 0!
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 46 / 47
91. What have we learned today?
Definition: a function is continuous at a point if the limit of the
function at that point agrees with the value of the function at that
point.
We often make a fundamental assumption that functions we meet in
nature are continuous.
The Intermediate Value Theorem is a basic property of real numbers
that we need and use a lot.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 47 / 47