Lesson 5: Continuity (Section 41 slides)
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Transcript of Lesson 5: Continuity (Section 41 slides)
Section 1.5Continuity
V63.0121.041, Calculus I
New York University
September 20, 2010
Announcements
I Office Hours: Tuesday, Wednesday, 3pm–4pm
I TAs have office hours on website
Announcements
I Office Hours: Tuesday,Wednesday, 3pm–4pm
I TAs have office hours onwebsite
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 2 / 47
Grader’s corner
I HW Grades will be onblackboard this week, andthe papers will be returnedin recitation
I Remember units whencomputing slopes
I Remember to staple yourpapers—you have beenwarned.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 3 / 47
Objectives
I Understand and apply thedefinition of continuity for afunction at a point or on aninterval.
I Given a piecewise definedfunction, decide where it iscontinuous or discontinuous.
I State and understand theIntermediate ValueTheorem.
I Use the IVT to show that afunction takes a certainvalue, or that an equationhas a solution
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 4 / 47
Last time
Definition
We writelimx→a
f (x) = L
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 welike) by taking x to be sufficiently close to a (on either side of a) but notequal to a.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 5 / 47
Limit Laws for arithmetic
Theorem (Basic Limits)
I limx→a
x = a
I limx→a
c = c
Theorem (Limit Laws)
Let f and g be functions with limits at a point a. Then
I limx→a
(f (x) + g(x)) = limx→a
f (x) + limx→a
g(x)
I limx→a
(f (x)− g(x)) = limx→a
f (x)− limx→a
g(x)
I limx→a
(f (x) · g(x)) = limx→a
f (x) · limx→a
g(x)
I limx→a
f (x)
g(x)=
limx→a f (x)
limx→a g(x)if lim
x→ag(x) 6= 0
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 6 / 47
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 worldmeasuring the exact same temperature.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 7 / 47
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 worldmeasuring the exact same temperature.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 7 / 47
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 worldmeasuring the exact same temperature.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 7 / 47
Outline
Continuity
The Intermediate Value Theorem
Back to the Questions
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 8 / 47
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
limx→a
f (x) = f (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
Definition of Continuity
Definition
I Let f be a function definednear a. We say that f iscontinuous at a if
limx→a
f (x) = f (a).
I A function f is continuousif it is continuous at everypoint in its domain. x
y
a
f (a)
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 10 / 47
Definition of Continuity
Definition
I Let f be a function definednear a. We say that f iscontinuous at a if
limx→a
f (x) = f (a).
I A function f is continuousif it is continuous at everypoint in its domain. x
y
a
f (a)
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 10 / 47
Scholium
Definition
Let f be a function defined near a. We say that f is continuous at a if
limx→a
f (x) = f (a).
There are three important parts to this definition.
I The function has to have a limit at a,
I the function has to have a value at a,
I and these values have to agree.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 11 / 47
Free Theorems
Theorem
(a) Any polynomial is continuous everywhere; that is, it is continuous onR = (−∞,∞).
(b) Any rational function is continuous wherever it is defined; that is, it iscontinuous on its domain.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 12 / 47
Showing a function is continuous
Example
Let f (x) =√
4x + 1. Show that f is continuous at 2.
Solution
We want to show that limx→2
f (x) = f (2). We have
limx→a
f (x) = limx→2
√4x + 1 =
√limx→2
(4x + 1) =√
9 = 3 = f (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 sincelim
x→−1/4+f (x) = f (−1/4).
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 13 / 47
Showing a function is continuous
Example
Let f (x) =√
4x + 1. Show that f is continuous at 2.
Solution
We want to show that limx→2
f (x) = f (2). We have
limx→a
f (x) = limx→2
√4x + 1 =
√limx→2
(4x + 1) =√
9 = 3 = f (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 sincelim
x→−1/4+f (x) = f (−1/4).
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 13 / 47
Showing a function is continuous
Example
Let f (x) =√
4x + 1. Show that f is continuous at 2.
Solution
We want to show that limx→2
f (x) = f (2). We have
limx→a
f (x) = limx→2
√4x + 1 =
√limx→2
(4x + 1) =√
9 = 3 = f (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 sincelim
x→−1/4+f (x) = f (−1/4).
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 13 / 47
At which other points?
For reference: f (x) =√
4x + 1
I If a > −1/4, then limx→a
(4x + 1) = 4a + 1 > 0, so
limx→a
f (x) = limx→a
√4x + 1 =
√limx→a
(4x + 1) =√
4a + 1 = f (a)
and f is continuous at a.
I If a = −1/4, then 4x + 1 < 0 to the left of a, which means√
4x + 1 isundefined. Still,
limx→a+
f (x) = limx→a+
√4x + 1 =
√lim
x→a+(4x + 1) =
√0 = 0 = f (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
At which other points?
For reference: f (x) =√
4x + 1
I If a > −1/4, then limx→a
(4x + 1) = 4a + 1 > 0, so
limx→a
f (x) = limx→a
√4x + 1 =
√limx→a
(4x + 1) =√
4a + 1 = f (a)
and f is continuous at a.
I If a = −1/4, then 4x + 1 < 0 to the left of a, which means√
4x + 1 isundefined. Still,
limx→a+
f (x) = limx→a+
√4x + 1 =
√lim
x→a+(4x + 1) =
√0 = 0 = f (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
Showing a function is continuous
Example
Let f (x) =√
4x + 1. Show that f is continuous at 2.
Solution
We want to show that limx→2
f (x) = f (2). We have
limx→a
f (x) = limx→2
√4x + 1 =
√limx→2
(4x + 1) =√
9 = 3 = f (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 −1/4since lim
x→−1/4+f (x) = f (−1/4).
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 15 / 47
Showing a function is continuous
Example
Let f (x) =√
4x + 1. Show that f is continuous at 2.
Solution
We want to show that limx→2
f (x) = f (2). We have
limx→a
f (x) = limx→2
√4x + 1 =
√limx→2
(4x + 1) =√
9 = 3 = f (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 −1/4since lim
x→−1/4+f (x) = f (−1/4).
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 15 / 47
The Limit Laws give Continuity Laws
Theorem
If f (x) and g(x) are continuous at a and c is a constant, then thefollowing functions are also continuous at a:
I (f + g)(x)
I (f − g)(x)
I (cf )(x)
I (fg)(x)
If
g(x) (if g(a) 6= 0)
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 16 / 47
Why a sum of continuous functions is continuous
We want to show that
limx→a
(f + g)(x) = (f + g)(a).
We just follow our nose:
limx→a
(f + g)(x) = limx→a
[f (x) + g(x)] (def of f + g)
= limx→a
f (x) + limx→a
g(x) (if these limits exist)
= 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
Trigonometric functions are continuous
I sin and cos are continuous on R.
I tan =sin
cosand sec =
1
cosare
continuous on their domain,which isR \
{ π2
+ kπ∣∣∣ k ∈ Z
}.
I cot =cos
sinand csc =
1
sinare
continuous on their domain,which is R \ { kπ | k ∈ Z }.
sin
cos
tan sec
cot csc
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 18 / 47
Trigonometric functions are continuous
I sin and cos are continuous on R.
I tan =sin
cosand sec =
1
cosare
continuous on their domain,which isR \
{ π2
+ kπ∣∣∣ k ∈ Z
}.
I cot =cos
sinand csc =
1
sinare
continuous on their domain,which is R \ { kπ | k ∈ Z }.
sin
cos
tan sec
cot csc
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 18 / 47
Trigonometric functions are continuous
I sin and cos are continuous on R.
I tan =sin
cosand sec =
1
cosare
continuous on their domain,which isR \
{ π2
+ kπ∣∣∣ k ∈ Z
}.
I cot =cos
sinand csc =
1
sinare
continuous on their domain,which is R \ { kπ | k ∈ Z }.
sin
cos
tan
sec
cot csc
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 18 / 47
Trigonometric functions are continuous
I sin and cos are continuous on R.
I tan =sin
cosand sec =
1
cosare
continuous on their domain,which isR \
{ π2
+ kπ∣∣∣ k ∈ Z
}.
I cot =cos
sinand csc =
1
sinare
continuous on their domain,which is R \ { kπ | k ∈ Z }.
sin
cos
tan sec
cot csc
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 18 / 47
Trigonometric functions are continuous
I sin and cos are continuous on R.
I tan =sin
cosand sec =
1
cosare
continuous on their domain,which isR \
{ π2
+ kπ∣∣∣ k ∈ Z
}.
I cot =cos
sinand csc =
1
sinare
continuous on their domain,which is R \ { kπ | k ∈ Z }.
sin
cos
tan sec
cot
csc
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 18 / 47
Trigonometric functions are continuous
I sin and cos are continuous on R.
I tan =sin
cosand sec =
1
cosare
continuous on their domain,which isR \
{ π2
+ kπ∣∣∣ k ∈ Z
}.
I cot =cos
sinand csc =
1
sinare
continuous on their domain,which is R \ { kπ | k ∈ Z }.
sin
cos
tan sec
cot csc
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 18 / 47
Exponential and Logarithmic functions arecontinuous
For any base a > 1,
I the function x 7→ ax iscontinuous on R
I the function loga iscontinuous on its domain:(0,∞)
I In particular ex andln = loge are continuous ontheir domains
ax
loga x
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 19 / 47
Exponential and Logarithmic functions arecontinuous
For any base a > 1,
I the function x 7→ ax iscontinuous on R
I the function loga iscontinuous on its domain:(0,∞)
I In particular ex andln = loge are continuous ontheir domains
ax
loga x
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 19 / 47
Exponential and Logarithmic functions arecontinuous
For any base a > 1,
I the function x 7→ ax iscontinuous on R
I the function loga iscontinuous on its domain:(0,∞)
I In particular ex andln = loge are continuous ontheir domains
ax
loga x
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 19 / 47
Inverse trigonometric functions are mostlycontinuous
I sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, andright continuous at −1.
I sec−1 and csc−1 are continuous on (−∞,−1) ∪ (1,∞), leftcontinuous at −1, and right continuous at 1.
I tan−1 and cot−1 are continuous on R.
−π
−π/2
π/2
π
sin−1
cos−1 sec−1
csc−1tan−1
cot−1
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 20 / 47
Inverse trigonometric functions are mostlycontinuous
I sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, andright continuous at −1.
I sec−1 and csc−1 are continuous on (−∞,−1) ∪ (1,∞), leftcontinuous at −1, and right continuous at 1.
I tan−1 and cot−1 are continuous on R.
−π
−π/2
π/2
π
sin−1
cos−1
sec−1
csc−1tan−1
cot−1
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 20 / 47
Inverse trigonometric functions are mostlycontinuous
I sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, andright continuous at −1.
I sec−1 and csc−1 are continuous on (−∞,−1) ∪ (1,∞), leftcontinuous at −1, and right continuous at 1.
I tan−1 and cot−1 are continuous on R.
−π
−π/2
π/2
π
sin−1
cos−1 sec−1
csc−1tan−1
cot−1
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 20 / 47
Inverse trigonometric functions are mostlycontinuous
I sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, andright continuous at −1.
I sec−1 and csc−1 are continuous on (−∞,−1) ∪ (1,∞), leftcontinuous at −1, and right continuous at 1.
I tan−1 and cot−1 are continuous on R.
−π
−π/2
π/2
π
sin−1
cos−1 sec−1
csc−1
tan−1
cot−1
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 20 / 47
Inverse trigonometric functions are mostlycontinuous
I sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, andright continuous at −1.
I sec−1 and csc−1 are continuous on (−∞,−1) ∪ (1,∞), leftcontinuous at −1, and right continuous at 1.
I tan−1 and cot−1 are continuous on R.
−π
−π/2
π/2
π
sin−1
cos−1 sec−1
csc−1tan−1
cot−1
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 20 / 47
Inverse trigonometric functions are mostlycontinuous
I sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, andright continuous at −1.
I sec−1 and csc−1 are continuous on (−∞,−1) ∪ (1,∞), leftcontinuous at −1, and right continuous at 1.
I tan−1 and cot−1 are continuous on R.
−π
−π/2
π/2
π
sin−1
cos−1 sec−1
csc−1tan−1
cot−1
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 20 / 47
What could go wrong?
In what ways could a function f fail to be continuous at a point a? Lookagain at the definition:
limx→a
f (x) = f (a)
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 21 / 47
Continuity FAIL
: 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?
Solution
At any point a in [0, 2] besides 1, limx→a
f (x) = f (a) because f is represented by a
polynomial near a, and polynomials have the direct 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.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 22 / 47
Continuity FAIL: 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?
Solution
At any point a in [0, 2] besides 1, limx→a
f (x) = f (a) because f is represented by a
polynomial near a, and polynomials have the direct 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.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 22 / 47
Graphical Illustration of Pitfall #1
x
y
−1 1 2
−1
1
2
3
4
The function cannot becontinuous at a point if thefunction has no limit at thatpoint.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 23 / 47
Continuity FAIL
: The function has no value
Example
Let
f (x) =x2 + 2x + 1
x + 1
At which points is f continuous?
Solution
Because f is rational, it is continuous on its whole domain. Note that −1is 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
Continuity FAIL: The function has no value
Example
Let
f (x) =x2 + 2x + 1
x + 1
At which points is f continuous?
Solution
Because f is rational, it is continuous on its whole domain. Note that −1is 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
Graphical Illustration of Pitfall #2
x
y
−1
1 The function cannot becontinuous at a point outside itsdomain (that is, a point where ithas no value).
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 25 / 47
Continuity FAIL
: function value 6= limit
Example
Let
f (x) =
{7 if x 6= 1
π if x = 1
At which points is f continuous?
Solution
f is not continuous at 1 because f (1) = π but limx→1
f (x) = 7.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 26 / 47
Continuity FAIL: function value 6= limit
Example
Let
f (x) =
{7 if x 6= 1
π if x = 1
At which points is f continuous?
Solution
f is not continuous at 1 because f (1) = π but limx→1
f (x) = 7.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 26 / 47
Graphical Illustration of Pitfall #3
x
y
π
7
1
If the function has a limit and avalue at a point the two muststill agree.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 27 / 47
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 limit at a.
Byre-defining f (a) = lim
x→af (x), f can be made continuous at a
jump discontinuity The limits limx→a−
f (x) and limx→a+
f (x) exist, but are
different.
The function cannot be made continuous bychanging a single value.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 28 / 47
Graphical representations of discontinuities
x
y
π
7
1
Presto! continuous!
removable
x
y
1
1
2
continuous?
continuous?
continuous?
jump
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 29 / 47
Graphical representations of discontinuities
x
y
π
7
1
Presto! continuous!
removable
x
y
1
1
2
continuous?
continuous?
continuous?
jump
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 29 / 47
Graphical representations of discontinuities
x
y
π
7
1
Presto! continuous!
removable
x
y
1
1
2
continuous?
continuous?
continuous?
jump
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 29 / 47
Graphical representations of discontinuities
x
y
π
7
1
Presto! continuous!
removable
x
y
1
1
2
continuous?
continuous?
continuous?
jump
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 29 / 47
Graphical representations of discontinuities
x
y
π
7
1
Presto! continuous!
removable
x
y
1
1
2
continuous?
continuous?
continuous?
jump
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 29 / 47
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 limit at a. Byre-defining f (a) = lim
x→af (x), f can be made continuous at a
jump discontinuity The limits limx→a−
f (x) and limx→a+
f (x) exist, but are
different.
The function cannot be made continuous bychanging a single value.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 30 / 47
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 limit at a. Byre-defining f (a) = lim
x→af (x), f can be made continuous at a
jump discontinuity The limits limx→a−
f (x) and limx→a+
f (x) exist, but are
different. The function cannot be made continuous bychanging a single value.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 30 / 47
The greatest integer function
[[x ]] is the greatest integer ≤ x .
x [[x ]]
0 01 1
1.5 11.9 12.1 2−0.5 −1−0.9 −1−1.1 −2
x
y
−2
−2
−1
−1
1
1
2
2
3
3y = [[x ]]
This function has a jump discontinuity at each integer.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 31 / 47
The greatest integer function
[[x ]] is the greatest integer ≤ x .
x [[x ]]
0 01 1
1.5 11.9 12.1 2−0.5 −1−0.9 −1−1.1 −2
x
y
−2
−2
−1
−1
1
1
2
2
3
3y = [[x ]]
This function has a jump discontinuity at each integer.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 31 / 47
Outline
Continuity
The Intermediate Value Theorem
Back to the Questions
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 32 / 47
A Big Time Theorem
Theorem (The Intermediate Value Theorem)
Suppose that f is continuous on the closed interval [a, b] and let N be anynumber between f (a) and f (b), where f (a) 6= f (b). Then there exists anumber c in (a, b) such that f (c) = N.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 33 / 47
Illustrating the IVT
Suppose that f is continuous on the closed interval [a, b] and let N be anynumber between f (a) and f (b), where f (a) 6= f (b). Then there exists anumber c in (a, b) such that f (c) = N.
x
f (x)
a b
f (a)
f (b)
N
cc1 c2 c3
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 34 / 47
Illustrating the IVT
Suppose that f is continuous on the closed interval [a, b]
and let N be anynumber between f (a) and f (b), where f (a) 6= f (b). Then there exists anumber c in (a, b) such that f (c) = N.
x
f (x)
a b
f (a)
f (b)
N
cc1 c2 c3
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 34 / 47
Illustrating the IVT
Suppose that f is continuous on the closed interval [a, b]
and let N be anynumber between f (a) and f (b), where f (a) 6= f (b). Then there exists anumber c in (a, b) such that f (c) = N.
x
f (x)
a b
f (a)
f (b)
N
cc1 c2 c3
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 34 / 47
Illustrating the IVT
Suppose that f is continuous on the closed interval [a, b] and let N be anynumber between f (a) and f (b), where f (a) 6= f (b).
Then there exists anumber c in (a, b) such that f (c) = N.
x
f (x)
a b
f (a)
f (b)
N
cc1 c2 c3
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 34 / 47
Illustrating the IVT
Suppose that f is continuous on the closed interval [a, b] and let N be anynumber between f (a) and f (b), where f (a) 6= f (b). Then there exists anumber c in (a, b) such that f (c) = N.
x
f (x)
a b
f (a)
f (b)
N
c
c1 c2 c3
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 34 / 47
Illustrating the IVT
Suppose that f is continuous on the closed interval [a, b] and let N be anynumber between f (a) and f (b), where f (a) 6= f (b). Then there exists anumber c in (a, b) such that f (c) = N.
x
f (x)
a b
f (a)
f (b)
N
cc1 c2 c3
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 34 / 47
Illustrating the IVT
Suppose that f is continuous on the closed interval [a, b] and let N be anynumber between f (a) and f (b), where f (a) 6= f (b). Then there exists anumber c in (a, b) such that f (c) = N.
x
f (x)
a b
f (a)
f (b)
N
c
c1 c2 c3
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 34 / 47
What the IVT does not say
The Intermediate Value Theorem is an “existence” theorem.
I It does not say how many such c exist.
I It also does not say how to find c.
Still, it can be used in iteration or in conjunction with other theorems toanswer these questions.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 35 / 47
Using the IVT to find zeroes
Example
Let f (x) = x3− 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
Using the IVT to find zeroes
Example
Let f (x) = x3− 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
Finding a zero by bisection
x f (x)
1 − 11.25 − 0.296875
1.3125 − 0.05151371.375 0.224609
1.5 0.8752 5
(More careful analysis yields1.32472.)
x
y
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 37 / 47
Finding a zero by bisection
x f (x)
1 − 1
1.25 − 0.2968751.3125 − 0.0515137
1.375 0.2246091.5 0.875
2 5
(More careful analysis yields1.32472.)
x
y
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 37 / 47
Finding a zero by bisection
x f (x)
1 − 1
1.25 − 0.2968751.3125 − 0.0515137
1.375 0.2246091.5 0.875
2 5
(More careful analysis yields1.32472.)
x
y
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 37 / 47
Finding a zero by bisection
x f (x)
1 − 1
1.25 − 0.2968751.3125 − 0.0515137
1.375 0.224609
1.5 0.8752 5
(More careful analysis yields1.32472.)
x
y
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 37 / 47
Finding a zero by bisection
x f (x)
1 − 11.25 − 0.296875
1.3125 − 0.05151371.375 0.224609
1.5 0.8752 5
(More careful analysis yields1.32472.)
x
y
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 37 / 47
Finding a zero by bisection
x f (x)
1 − 11.25 − 0.296875
1.3125 − 0.0515137
1.375 0.2246091.5 0.875
2 5
(More careful analysis yields1.32472.)
x
y
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 37 / 47
Finding a zero by bisection
x f (x)
1 − 11.25 − 0.296875
1.3125 − 0.05151371.375 0.224609
1.5 0.8752 5
(More careful analysis yields1.32472.)
x
y
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 37 / 47
Finding a zero by bisection
x f (x)
1 − 11.25 − 0.296875
1.3125 − 0.05151371.375 0.224609
1.5 0.8752 5
(More careful analysis yields1.32472.)
x
y
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 37 / 47
Using the IVT to assert existence of numbers
Example
Suppose we are unaware of the square root function and that it’scontinuous. 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) suchthat
f (c) = c2 = 2.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 38 / 47
Using the IVT to assert existence of numbers
Example
Suppose we are unaware of the square root function and that it’scontinuous. 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) suchthat
f (c) = c2 = 2.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 38 / 47
Using the IVT to assert existence of numbers
Example
Suppose we are unaware of the square root function and that it’scontinuous. 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) suchthat
f (c) = c2 = 2.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 38 / 47
Outline
Continuity
The Intermediate Value Theorem
Back to the Questions
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 39 / 47
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 inpounds.
True or False
Right now there are two points on opposite sides of the Earth with exactlythe same temperature.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 40 / 47
Question 1
Answer
The answer is TRUE.
I Let h(t) be height, which varies continuously over time.
I Then h(birth) < 3 ft and h(now) > 3 ft.
I 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
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 inpounds.
True or False
Right now there are two points on opposite sides of the Earth with exactlythe same temperature.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 42 / 47
Question 2
Answer
The answer is TRUE.
I Let h(t) be height in inches and w(t) be weight in pounds, bothvarying continuously over time.
I Let f (t) = h(t)− w(t).
I For most of us (call your mom), f (birth) > 0 and f (now) < 0.
I So by the IVT there is a point c in (birth, now) where f (c) = 0.
I 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
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 inpounds.
True or False
Right now there are two points on opposite sides of the Earth with exactlythe same temperature.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 44 / 47
Question 3
Answer
The answer is TRUE.
I Let T (θ) be the temperature at the point on the equator at longitudeθ.
I How can you express the statement that the temperature on oppositesides is the same?
I How can you ensure this is true?
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 45 / 47
Question 3
I Let f (θ) = T (θ)− T (θ + 180◦)
I Thenf (0) = T (0)− T (180)
whilef (180) = T (180)− T (360) = −f (0)
I 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
What have we learned today?
I Definition: a function is continuous at a point if the limit of thefunction at that point agrees with the value of the function at thatpoint.
I We often make a fundamental assumption that functions we meet innature are continuous.
I The Intermediate Value Theorem is a basic property of real numbersthat we need and use a lot.
V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 47 / 47