3.
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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 3 / 46
4.
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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 4 / 46
5.
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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 5 / 46
6.
Hatsumon
Here are some discussion questions to start.
True or False
At some point in your life you were exactly three feet tall.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 6 / 46
7.
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).
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 6 / 46
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).
True or False
Right now there are a pair of points on opposite sides of the world
measuring the exact same temperature.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 6 / 46
9.
Outline
Continuity
The Intermediate Value Theorem
Back to the Questions
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 7 / 46
10.
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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 8 / 46
11.
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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 9 / 46
12.
Definition of Continuity
y
.
Definition
Let f be a function defined
f
.(a) .
near a. We say that f is
continuous at a if
lim f(x) = f(a).
x→a
A function f is continuous
if it is continuous at every
. x
.
point in its domain. a
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 9 / 46
13.
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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 10 / 46
14.
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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 11 / 46
15.
Showing a function is continuous
.
Example
√
Let f(x) = 4x + 1. Show that f is continuous at 2.
. . . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 12 / 46
16.
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.
. . . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 12 / 46
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.
Question
At which other points is f continuous?
. . . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 12 / 46
18.
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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 13 / 46
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.
√
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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 13 / 46
20.
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, ∞). . . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 14 / 46
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, ∞). It is right continuous at
. . . . . .
−1/4 since lim f(x)
V63.0121.002.2010Su, Calculus I (NYU) = f(−1 Section 1.5 Continuity
/4). May 20, 2010 14 / 46
22.
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)
(f − g)(x)
(cf)(x)
(fg)(x)
f
(x) (if g(a) ̸= 0)
g
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 15 / 46
23.
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)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 16 / 46
24.
Trigonometric functions are continuous
sin and cos are continuous on
R.
.
s
. in
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 17 / 46
25.
Trigonometric functions are continuous
sin and cos are continuous on
R.
c
. os
.
s
. in
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 17 / 46
26.
Trigonometric functions are continuous
t
.an
sin and cos are continuous on
R.
sin 1
tan = and sec = are c
. os
cos cos
continuous on their domain,
{π
which is } .
s
. in
R + kπ k ∈ Z .
2
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 17 / 46
27.
Trigonometric functions are continuous
t
.an s
. ec
sin and cos are continuous on
R.
sin 1
tan = and sec = are c
. os
cos cos
continuous on their domain,
{π
which is } .
s
. in
R + kπ k ∈ Z .
2
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 17 / 46
28.
Trigonometric functions are continuous
t
.an s
. ec
sin and cos are continuous on
R.
sin 1
tan = and sec = are c
. os
cos cos
continuous on their domain,
{π
which is } .
s
. in
R + kπ k ∈ Z .
2
cos 1
cot = and csc = are
sin sin
continuous on their domain,
which is R { kπ | k ∈ Z }.
c
. ot
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 17 / 46
29.
Trigonometric functions are continuous
t
.an s
. ec
sin and cos are continuous on
R.
sin 1
tan = and sec = are c
. os
cos cos
continuous on their domain,
{π
which is } .
s
. in
R + kπ k ∈ Z .
2
cos 1
cot = and csc = are
sin sin
continuous on their domain,
which is R { kπ | k ∈ Z }.
c
. ot . sc
c
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 17 / 46
30.
Exponential and Logarithmic functions are continuous
For any base a > 1, .x
a
the function x → ax is
continuous on R
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 18 / 46
31.
Exponential and Logarithmic functions are continuous
For any base a > 1, .x
a
the function x → ax is l
.oga x
continuous on R
the function loga is
continuous on its domain: .
(0, ∞)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 18 / 46
32.
Exponential and Logarithmic functions are continuous
For any base a > 1, .x
a
the function x → ax is l
.oga x
continuous on R
the function loga is
continuous on its domain: .
(0, ∞)
In particular ex and
ln = loge are continuous
on their domains
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 18 / 46
33.
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
.
. /2
π
.
. in−1
s
.
−
. π/2
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 19 / 46
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
. os−1
c .
. /2
π
. .
. in−1
s
.
−
. π/2
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 19 / 46
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.
sec−1 and csc−1 are continuous on (−∞, −1) ∪ (1, ∞), left
continuous at −1, and right continuous at 1.
.
.
π
. os−1
c . . ec−1
s
. /2
π
. .
. in−1
s
.
−
. π/2
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 19 / 46
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
. os−1
c . . ec−1
s
. /2
π
. sc−1
c
. .
. in−1
s
.
−
. π/2
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 19 / 46
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.
tan−1 and cot−1 are continuous on R.
.
.
π
. os−1
c . . ec−1
s
. /2
π
.an−1
t
. sc−1
c
. .
. in−1
s
.
−
. π/2
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 19 / 46
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
. ot−1
c
. os−1
c . . ec−1
s
. /2
π
.an−1
t
. sc−1
c
. .
. in−1
s
.
−
. π/2
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 19 / 46
39.
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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 20 / 46
40.
Continuity FAIL
.
Example
Let {
x2 if 0 ≤ x ≤ 1
f(x) =
2x if 1 < x ≤ 2
At which points is f continuous?
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 21 / 46
41.
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 x2 = 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.
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 21 / 46
43.
Continuity FAIL
Example
Let
x2 + 2x + 1
f(x) =
x+1
At which points is f continuous?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 23 / 46
44.
Continuity FAIL: The function has no value
Example
Let
x2 + 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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 23 / 46
45.
Graphical Illustration of Pitfall #2
y
.
. .
1
. . x
.
−
. 1
f cannot be continuous where it has no value.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 24 / 46
46.
Continuity FAIL
Example
Let {
7 if x ̸= 1
f(x) =
π if x = 1
At which points is f continuous?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 25 / 46
47.
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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 25 / 46
48.
Graphical Illustration of Pitfall #3
y
.
. .
7 .
. .
π .
. . x
.
1
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 26 / 46
49.
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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 27 / 46
50.
Graphical representations of discontinuities
.
y
.
y
.
. .
7 .
. .
2 .
. .
π .
. .
1 .
. . x
. . . x
.
1
.
1
.
removable
jump
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 28 / 46
51.
Graphical representations of discontinuities
.
y
.
y
.
P
. resto! continuous!
. .
7 .
. .
2 .
. .
π
. .
1 .
. . x
. . . x
.
1
.
1
.
removable
jump
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 28 / 46
52.
Graphical representations of discontinuities
.
y
.
y
.
P
. resto! continuous!
. .
7 .
. .
2 .
. .
π
. .
1 . . ontinuous?
c
. . x
. . . x
.
1
.
1
.
removable
jump
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 28 / 46
53.
Graphical representations of discontinuities
.
y
.
y
.
P
. resto! continuous!
. .
7 .
. .
2 . . ontinuous?
c
. .
π
. .
1 .
. . x
. . . x
.
1
.
1
.
removable
jump
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 28 / 46
54.
Graphical representations of discontinuities
.
y
.
y
.
P
. resto! continuous!
. .
7 .
. .
2 .
. . ontinuous?
c
. .
π
. .
1 .
. . x
. . . x
.
1
.
1
.
removable
jump
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 28 / 46
55.
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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 29 / 46
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. The function cannot be made continuous by
changing a single value.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 29 / 46
57.
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 .
−
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 30 / 46
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 .
−
This function has a jump discontinuity at each integer.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 30 / 46
59.
Outline
Continuity
The Intermediate Value Theorem
Back to the Questions
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 31 / 46
60.
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.
. . . . . .
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61.
Illustrating the IVT
f
.(x)
. x
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 33 / 46
62.
Illustrating the IVT
Suppose that f is continuous on the closed interval [a, b]
f
.(x)
.
.
. x
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 33 / 46
63.
Illustrating the IVT
Suppose that f is continuous on the closed interval [a, b]
f
.(x)
f
.(b) .
f
.(a) .
. x
.
a
. b
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 33 / 46
64.
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
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 33 / 46
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). 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
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 33 / 46
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) .
. x
.
a
. b
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 33 / 46
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) .
. x
.
ac
. .1 c
.2 c b
.3 .
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 33 / 46
68.
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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 34 / 46
69.
Using the IVT
Example
Suppose we are unaware of the square root function and that it’s
continuous. Prove that the square root of two exists.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 35 / 46
70.
Using the IVT
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) = x2 , a continuous function on [1, 2].
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 35 / 46
71.
Using the IVT
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) = x2 , 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) = c2 = 2.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 35 / 46
72.
Using the IVT
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) = x2 , 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) = c2 = 2.
In fact, we can “narrow in” on the square root of 2 by the method of
bisections.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 35 / 46
73.
√
Finding 2 by bisections
. .(x) = x2
x f
. ..
2 4
. ..
1 1
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 36 / 46
74.
√
Finding 2 by bisections
. .(x) = x2
x f
. ..
2 4
. ..
1 1
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 36 / 46
75.
√
Finding 2 by bisections
. .(x) = x2
x f
. ..
2 4
. .5 . . .25
1 2
. ..
1 1
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 36 / 46
79.
Using the IVT
Example
Let f(x) = x3 − 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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 37 / 46
80.
Using the IVT
Example
Let f(x) = x3 − 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.)
. . . . . .
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81.
Outline
Continuity
The Intermediate Value Theorem
Back to the Questions
. . . . . .
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82.
Back to the Questions
True or False
At one point in your life you were exactly three feet tall.
. . . . . .
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83.
Question 1: 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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 40 / 46
84.
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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 41 / 46
85.
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 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).
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 42 / 46
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.
True or False
Right now there are two points on opposite sides of the Earth with
exactly the same temperature.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 43 / 46
87.
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?
. . . . . .
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88.
Question 3: True!
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!
. . . . . .
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89.
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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 46 / 46
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