Chapter 11: Introduction to Continuitymigomendoza.weebly.com/uploads/5/4/7/4/54745209/... · Chapter 11: Introduction to Continuity ... front of the class. Pointing System: 5 points

Introduction to

Continuity

Chapter 11:

SSMTH1: Precalculus

Science and Technology, Engineering and Mathematics (STEM) Strands

Mr. Migo M. Mendoza

Chapter 11: Introduction to Continuity

Lecture 11.1: Introduction to Continuity

Lecture 11.2: Removable Discontinuity

Lecture 11.3: Jump Discontinuity

Chapter 11: Introduction to Continuity

Lecture 11.4: Properties of Continuity

Lecture 11.5: Intermediate Value Theorem

Lecture 11.6: Extreme Value Theorem

Introduction to ContinuityLecture 11.1:

SSMTH1: Precalculus


Mr. Migo M. Mendoza

The Motion of ContinuityFamily Activity 3

SSMTH1: Precalculus


Mr. Migo M. Mendoza

Instructions: Together with your family members, do this

activity by communicating with your family members through sign language. This is a

time-pressure task. The moment the music stops you are expected that you have

finished doing what is instructed your family to do.

Instructions: Redraw the following graph (you have seen from your family activity sheet 3) on one whole cartolina

and post it on the board. Please make sure to occupy the whole cartolina with your graph.

Afterwards, have a reflection by answering the questions below. Be ready to present your work in

front of the class.

Pointing System:

5 points – Time Management

5 points – Creativity

5 points – Presentation and Communication

Skills

5 points – Behavior and Teamwork

Total: 20 points

Something to think about…

Did you lift the pentelpen when you redraw

the graph?

Something to think about…Can you say that the

function is continuous for the set of real numbers? Why or

why not?


Which of the function(s) is/are

continuous?


For what value(s) of x is/are the function

discontinuous?

Our Observation:Intuitively speaking, a function is

continuous at a point if we can draw the graph of the function near that point without lifting our pencil

from the paper.

Our Observation:Conversely, a function is

discontinuous at a point if the pencil must be lifted from the paper in order to draw the graph on both

sides of the point.

Figure 11.1.1:

Reason:

In Figure 11.1, is not defined (because there is a hole in the

graph).

)2(f

Something to think about…From the previous graph,

what should be the first condition before we can say a

function is continuous?

Continuity at a PointA function f is continuous at x = c,

when:

FIRST CONDITION:

1. is defined;)(cf

Figure 11.1.2:

Reason:In Figure 11.2, does not exist (because the left-hand limit does not equal

the right-hand limit).

)(lim2

xfx

Something to think about…From the previous graph, what should be the second

condition before we can say a function is continuous?


when:

SECOND CONDITION:

2. exists)(lim xfcx

Figure 11.1.3:

Reason:In Figure 11.3, is defined

and exists, but

(because

whereas .

)(lim2

xfx

)2(f

)2()(lim2

fxfx

4)(lim2

xfx

6)2( f

Something to think about…From the previous graph,

what should be the third condition before we can say a

function is continuous?


when:

THIRD CONDITION:

3. H).()(lim cfxfcx

Continuity at a Point A function f is continuous at x = c, when:

1. is defined;

2. exists; and

3. k

Otherwise, f is said to be

discontinuous at x = c.

)(cf

)(lim xfcx

).()(lim cfxfcx

Example 11.1.1:Using the definition of

continuity, determine whether the function is

continuous or not at:12

1)(

2

xx

xxf

0)( ca 1)( cb

Condition Number 1:Since which is a

real number, is defined.

1)0( f

)0(f

Condition Number 2:Using properties of

limits, we see that

exists.

)(lim0

xfx

Condition Number 3:Thus, as x gets closer to 0, the corresponding values of f(x) get closer to the function value at 0:

)0()(lim0

fxfx

Final Answer:The three conditions were

satisfied, so we conclude that f is continuous at 0.

Condition Number 1:Since division by zero is

undefined, the domain of f is

.Thus, f is not defined at 1.

1,2

1: xxx

Final Answer:One of the three conditions is

not satisfied, so we conclude that f is not continuous at 1.

Equivalently, we can say that fis discontinuous at 1.

Figure 11.1.4:

Asymptotic DiscontinuityAlso called as “infinite discontinuity”.

They are located by finding the values of x that yield a divide by zero error. In

other words, a function will yield a

result of where .0

a0a

Example 11.1.2:Determine the asymptotic discontinuity of the function:

.4

3)(

2

x

xxf

Final Answer:The values that will make the

denominator 0 are -2 and 2. Hence, the asymptotic discontinuities are at

.22 andxx

Example 11.1.3:Test the continuity of each of the following functions at . If a function is not continuous at

give a reason why it is not continuous.

2x

,2x

Example 11.1.3(a):

1)( 3 xxxf

Final Answer:The function f is a polynomial function. Hence, .

Thus, f is continuous at

.2x

)(lim7)2(2

xffx

Example 11.1.3(b):

2

6)(

2

x

xxxg

Final Answer:The function g is not

continuous at because g is not defined at

this point.

2x

Example 11.1.3(c):

2;1

2;3)(

2

xx

xxxh

Final Answer:By definition of continuity,

the three conditions are satisfied, so h is continuous

at .2x

Did you know? Recall that the limit of a polynomial

function as x approaches c is the

polynomial function evaluated at c. That is,

for any number c. This

means that a polynomial function is continuous at every number.

)()(lim cfxfcx

Did you know? Rational functions are continuous at every

value, except any value at which they are not defined. At values that are not in the

domain of a rational function, a “hole”

or “open dot” in the graph or an

asymptote appears.

Did you know? Exponential, logarithmic, sine, and cosine

functions are continuous at every value in their domain. Like rational functions, the tangent, cotangent, secant, and cosecant functions are

continuous at every number, except any number at which they are not defined. At values that are

not in the domain of these trigonometric functions, an asymptote occurs.

Removable DiscontinuityLecture 11.2:

SSMTH1: Precalculus


Mr. Migo M. Mendoza

Example 11.1.4:Determine where the given function below is continuous:

3

32)(

2

x

xxxf

Figure 11.1.5:

Final Answer:So, f is discontinuous

at but continuous elsewhere.

3x

Removable Discontinuity The type of discontinuity shown in the previous example is called a “removable

discontinuity” because we could remove the discontinuity by redefining f at the point of discontinuity. This could be done by finding

).(lim3

xfx

Classroom Task:Redefine the function

to removed the discontinuity at

3

32)(

2

x

xxxf

.3x

Final Answer:Hence, if we redefine the function f as

we have removed the discontinuity at since3x

3;4

3;3

32)(

2

x

xx

xxxf

).3(4)(lim3

fxfx

Take Note: In general, removable discontinuities

often occur at a “hole” in the graph. They are located by finding the values

where both the numerator and the denominator equal zero.

Example 11.1.5:Determine where the removable

discontinuity exists in

and redefine the function to remove the discontinuity.

9

12)(

2

2

x

xxxf

Take Note:Since we redefine the function

this way:6

7)(lim

3

xf

x

3;6

7

3;9

12

)(2

2

x

xx

xx

xf

Jump DiscontinuityLecture 11.3:

SSMTH1: Precalculus


Mr. Migo M. Mendoza

Example 11.1.4:Use the definition of

continuity to discuss the continuity of the function

whose graph is shown on the next slide.

Figure 11.1.6:

Our First Observation: Discontinuity at

f(-4) does not exist. The hole at

indicates that 5 is not the value of f

at -4. Since there is no solid dot elsewhere

on the vertical line x = -4, f(-4) is not

defined.

:4x

5,4

Final Answer:So f is not continuous at

because condition 1

is not satisfied.

4x

Our Second Observation: Discontinuity at

does not exist.

:1x2)1( f

2)(lim1

xfx

4)(lim1

xfx

)(lim1

xfx

Final Answer:This time, even is

defined, f is not continuous at

because condition 2 is not satisfied.

)1(f

1x

Our Third Observation: Discontinuity at

; the solid dot at (2, 2) indicates that

:2x2)2( f

1)(lim2

xfx

1)(lim2

xfx

1)(lim2

xfx

2)2( f

Final Answer:Conditions 1 and 2 are

satisfied, but f is not continuous at because

condition 3 is not satisfied.2x

Jump Discontinuity From the previous example, .

Hence, does not exist and f is discontinuous at . This type of

discontinuity is called a jump discontinuity. Jump discontinuities occur where the

function approaches two different values from either side of the discontinuity.

)(lim)(lim11

xfxfxx

)(lim1

xfx

1x

Example 11.1.7:Determine where the jump

discontinuity exists in:

3;2

3;9)(

2

xx

xxxf

Final Answer:Hence, the jump

discontinuity exists at

.3x

Classroom Task 11.1:

Please answer "Let's Practice (LP)"

Number 41 and 42.

Properties of ContinuityLecture 11.4:

SSMTH1: Precalculus


Mr. Migo M. Mendoza

Properties of Continuous Functions

If the function f and

g are continuous at cthen:

Property Number 1: If the function f and g are continuous at

c then:

is continuous at c;

gf


c then:

is continuous at c;

gf


c then:

is continuous at c;

gf


c then:

is continuous at c if

and is discontinuous at c if

g

f

0)( cg

.0)( cg

To sum it up…Simply put, the sum, difference, or

product of continuous function is continuous, while the quotient of two continuous functions is continuous at any point at which the denominator is

nonzero.

Example 11.4.1:Reason out why the polynomial

function below is continuous:

2)( 2 xxxf

Reason 1:To illustrate, the polynomial function

may be considered as sum and difference of three

functions: and f

which are all continuous.

2)( 2 xxxf

,)( 2xxg ,)( xxh 2)( xp

Reason 2:The same function may be

considered as a product of two continuous functions since:

.122)( 2 xxxxxf

Example 11.4.2:Reason out why the polynomial


12

1)(

2

xx

xxf

Reason:The function is

continuous for all x ∈ ℝ except at and which are the

values that make the denominator zero.

12

1)(

2

xx

xxf

1x2

1x

Did you know? The properties of continuous functions,

along with the continuity properties of specific functions, enable us to determine

intervals of continuity for some important classes of functions without having to look at their graphs or use the three conditions

in the definition of continuity.

of Some Specific FunctionsContinuity Properties

SSMTH1: Precalculus


Mr. Migo M. Mendoza

Continuity Property 1 of Some Specific Functions

A constant function

where k is a constant, is continuous

for all x.

kxf )(


For positive integer n, is

continuous for all x.

nxxf )(


A polynomial function is



A rational function is continuous for all x in its

domain, that is, except those values that will make the

denominator 0.


For an odd positive integer n

greater than 1, is

continuous wherever f(x) is

continuous.

n xf )(


For an even positive integer

n, is continuous

wherever f(x) is continuous

and nonnegative.

n xf )(

Example 11.4.3:Using properties of continuity, determine where the function is

continuous:

12)( 2 xxxf

Final Answer:Since f is a polynomial function, f is continuous

for all x.


continuous:

32

1)(

xx

xxf

Final Answer:Since f is a rational

function, f is continuous

for all x except 2 and -3.


continuous:

3 2 9)( xxf

Final Answer:The polynomial function

is continuous for all x. Since n is odd, f is


92 x


continuous:

xxf 3)(

Final Answer:The polynomial function is

continuous for all x and nonnegative for

Since is even, f is continuous for or on the interval

x3

.3x

.3,

3x2n

Did you know?Properties 5 and 6 considered the

continuity of which is a composition of function where

In general, composition of functions are continuous at every point

at which they are defined.

)(xfg

.)( n xxg

n xf )(

Composition of Continuous Functions

If f is continuous at c, and g

is continuous at f(c), then

the composition is

continuous at c.

gf

Example 11.4.7:Determine where the


2cos)( 2 xxxh

Final Answer:We note that where

and Since both f and g are continuous for all x, h

is continuous for all x, by the continuity property of the composition

of functions.

)()( xfgxh

2)( 2 xxxf .cos)( xxg



Number 43.

The Intermediate Value Theorem

Lecture 11.5:

SSMTH1: Precalculus


Mr. Migo M. Mendoza

Classroom Task 11.5.1:Determine the closed interval

intervals for which each function is continuous. Write the intervals in

the box below each function:

Example 11.5.1 (a):

Example 11.5.1 (b):

Example 11.5.1 (c):

Classroom Task 11.5.2:Given that a and b are the

endpoints of each closed interval, determine some f(x) between f(a)

and f(b). For every f(x), find its corresponding x-value.

Example 11.5.1 (d):

Did you know?Functions that are continuous on intervals have properties that make

them particularly useful in mathematics and its applications. One of these is the intermediate

value property.

Intermediate Value PropertyA function f is said to have the

intermediate value property if,

whenever f is defined on then

f(x) takes every value between f(a)

and f(b).

ba,

The Intermediate Value TheoremA function that is continuous

on a closed interval takes on

every value between f(a) and f(b),

In other words, if y0 is

between f(a) and f(b), then k

for some c in

ba,

)(xfy

).()( bfaf

)(0 cfy

.,ba

The Intermediate Value TheoremSuppose f is continuous on a closed

interval and q is any number between f(a) and f(b). If .

or then there exists a number m in such that

ba,

)()( bfqaf

.)( qmf

),()( afqbf

Figure 11.5.1: The Intermediate Value Theorem

Take Note:

The continuity of f on the interval is

essential to the Intermediate Value

Theorem. If f is discontinuous at even

one point of the interval, the theorem’s conclusion may fail.

Example 11.5.1:The graph of the function

which has a jump discontinuity at

x = 2.

62,2

20,3)(

x

xxxf

Figure 11.5.2:

Thus,

The function does not take on

all values between -2 and 3;

it misses all values between -

2 and 1.

Take Note:

If a function f(x) is continuous on a

closed interval what the Intermediate Value Theorem really

says is that a continuous function will

take on all values between f(a) and

f(b).

,,ba

Figure 11.5.3:

Take Note: It is also important to note that the

Intermediate Value Theorem only states

that the function will take on the value

of y0 somewhere between a and b. It

does not states what that value will be. It only states that it exists.

Take Note:So, the Intermediate Value Theorem tells

us that a function will take the value y0

of somewhere between a and b but it

does not tell us where it will take the value nor does it tell us how many

times it will take the value.

Example 11.5.1:Show that the intermediate value theorem holds for .

on the interval for

12)( xxf

8,3

.3)( xf

Take Note: In order to apply Intermediate Value Theorem, we need to

verify that the two conditions in the theorem are satisfied, namely the:

(1)f(x) is continuous in the specified closed interval [3, 8]; and

(2)the value 0 is between f(3) and f(8).

Final Answer:Since 6 is between 3 and 8, then intermediate value theorem holds

for on the interval

for

12)( xxf

8,3 .3)( xf

Did you know?An application of the Intermediate

Value Theorem is to provide the existence of roots of equations which happens if the values at the endpoints

of the interval differ in sign.

Existence of RootsFor a function that is

continuous on a closed interval g

suppose f(a) and f(b) have different

signs, then there exists c between a and

b such that

)(xfy ,,ba

.0)( cf

Intermediate Zero TheoremLet f be continuous on a closed interval

If f(a) and f(b) have opposite signs, then there is at least one number

c between a and b such that . and c is a zero of f.

.,ba

0)( cf

Example 11.5.2:Use the Intermediate Value Theorem to

show that the polynomial function

below has a root between 1 and 2:

2)( 3 xxxf



(1)f(x) is continuous in the specified closed interval [1. 2]; and

(2)the value 0 is between f(1) and f(2).

Figure 11.5.2:

Final Answer:Since and the

sign change shows that the polynomial

function has a root between 1 and 2.

This root is actually irrational and is approximated using a graphing utility

as f

02)1( f ,04)2( f

.5214.1x

Example 11.5.3:Use the Intermediate Value Theorem to

show that the two functions below

intersect at a point between 0 and 1:

andxxf cos)( xxg )(

Our Process: To find the intersection of these two functions, we

set f(x) = g(x). Hence, we have cos x = x. We

create a new function by subtracting the right hand side of the equation from the left hand side, i.e.

h(x) = cos x – x. Now, the problem is equivalent

to finding a value c for x between 0 and 1 such

that h(c) = 0.



(1)h(x) is continuous in the specified closed interval [0, 1]; and

(2)the value 0 is between h(0) and h(1).

Final Answer:Since h(x) is continuous on the closed

interval and 0 is between h(0) and

h(1), by Intermediate Value Theorem,

we can find a number c between 0 and

1, such that that is, ,0)( ch

1,0

.0)cos( cc

Final Answer:So, the number c satisfies g

which is what we want. Using a graphing utility, the intersection of the

graphs of f(x) and g(x) is at

.7391.0,7391.0

,)cos( cc

Figure 11.5.3:


Please answer "Let's Practice (LP)"Number 44.

The Extreme Value TheoremLecture 11.6:

SSMTH1: Precalculus


Mr. Migo M. Mendoza

The Extreme Value PropertyAnother property of continuous

functions is the extreme value property which tells us that a continuous

function has a maximum and minimum value on a closed interval.

The Maximum ValueThe maximum value of a

function on an interval is the largest value the function takes

on within the interval.

The Minimum ValueThe minimum value of a

function on an interval is the smallest value the function takes

on within the interval.

Figure 11.6.1: The Extreme Value Theorem

The Extreme Value Theorem

If f(x) is continuous on a closed

interval then f has both a

maximum and a minimum value on the interval. In other words, it must have at

least two extreme values.

,,ba

Take Note:

If f(x) is discontinuous at

even point of the interval, the conclusion of the theorem

may not be true.

Example 11.6.1:

The function on the interval has

infinite/ asymptotic discontinuity at x = 0.

xxf

1)(

5,5

Figure 11.6.2:

Thus,

The function has neither a maximum nor a minimum

value on the interval.

Take Note:

If f is continuous on an

interval but the interval is not

closed, the f might not have a

maximum or a minimum value.

Example 11.6.2:

Let us consider the function

on the open

interval .

xxf tan)(

2,

2

Figure 11.6.3:

Did you know?

Maxima and Minima can occur at interior points or

at the endpoints.

Figure 11.6.4: Maximum and Minimum Point at Interior Points

Maximum and Minimum Point at Interior Points

Figure 11.6.5: Maximum and Minimum at Endpoints

Maximum and Minimum Point at Endpoints

Figure 11.6.6: Maximum at Interior Point and Minimum at Endpoint

Maximum at Interior Point and Minimum at Endpoint

Example 11.6.3.1:

Given the graph of the function f on

the next slide, approximate the

maximum and minimum values of f on the given interval:

0,1)( a

Figure 11.6.7:

Example 11.6.3.2:




1,5.0)( b

Figure 11.6.7:

Example 11.6.3.3:




5.1,0)(c

Figure 11.6.7:

Example 11.6.4:

Find the maximum and minimum values of

0n1|2|)( xxf .4,1

Figure 11.6.7:

Example 11.6.5:

What are the extreme values of on

?

xxf sin)(

2,2

Figure 11.6.8:



Number 45.

Documents

Chapter 11: Introduction to Continuitymigomendoza.weebly.com/uploads/5/4/7/4/54745209/... · Chapter 11: Introduction to Continuity ... front of the class. Pointing System: 5 points